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 1 answerLast reply by: Professor Selhorst-JonesSun Mar 8, 2015 9:10 PMPost by Jamal Tischler on March 8, 2015If we try to evaluate the integral of 1/x from a=1 to b=2 with the summation we get the limit of n*(1/(n+1)+1/(n+2)+...+1/(n+n)). How do we evaluate this ? 1 answerLast reply by: Professor Selhorst-JonesMon Aug 26, 2013 11:09 PMPost by Richard Gregory on August 22, 2013Thank you very much Vincent. I really enjoyed this course and will be moving on to further maths classes. Your teaching style is excellent and I admire your enthusiasm for the topic. I always struggled to understand the meaning of calculus in school and I believe the meaning is more important than the technical application. Your lectures have filled in all the holes school created.

### Area Under a Curve (Integrals)

• The integral is a way to find the area underneath some portion of a curve.
• We can approximate an integral by using rectangles. We can break an interval up into sub-intervals, and then put a rectangle the width of each sub-interval in place. By taking the area of each rectangle and summing them all up, we have an approximation of the area under the curve in that interval.
• The width of each rectangle is the same, because we cut the sub-intervals evenly. However, the height of each rectangle can vary depending on the height of the function in that sub-interval. Furthermore, the function has different heights throughout the sub-interval, so we have to come up with some method to choose an xi in each sub-interval to find the height of its associated rectangle: f(xi).
• Here are some of the most common methods for choosing the xi in each sub-interval:
• Left-Most Point: We choose xi such that it is the left-most point in each sub-interval. The height of the rectangle is based on its left side.
• Right-Most Point: We choose xi such that it is the right-most point in each sub-interval. The height of the rectangle is based on its right side.
• Mid-Point: We choose xi such that it is the mid-point in each sub-interval. The height of the rectangle is based on its middle.
• Maximum (Upper Sum): We choose xi such that the rectangle is the highest possible for the sub-interval. The height of the rectangle is the highest place the function achieves in that sub-interval.
• Minimum (Lower Sum): We choose xi such that the rectangle is the lowest possible for the sub-interval. The height of the rectangle is the lowest place the function achieves in that sub-interval.
• We can evaluate the area approximation by summing up all of the rectangles. This comes out to be
 ∑ ⎛⎝ b−a n ⎞⎠ ·f(xi).
[However, if you are told to approximate area, you normally won't need the above formula. Just figure out the area for each rectangle, then add them all together.]
• Since the above approximation becomes more accurate as n→ ∞, we can take the limit at infinity to find area.
 Area under f(x)from a tob: lim ∑ ⎛⎝ b−a n ⎞⎠ ·f(xi)
• If the above limit exists, it is called the integral from a to b. It is denoted by
 ⌠⌡ f(x)   dx
The process of finding integrals is called integration.
• The really amazing part is that the integral of f(x) is based on the antiderivative of f(x): that is, the derivative process done in reverse on f(x). We can symbolize the antiderivative of f(x) with F(x). With this notation, we have
 ⌠⌡ f(x)   dx    =   F(b) − F(a).
• We can see the above is true, because we can think of the area underneath the curve as a function A(x). Notice that the rate of change for the area is based on the height of the function: f(x). Thus, height is the derivative of area, so area is based on the antiderivative of height.

### Area Under a Curve (Integrals)

Find an approximation to the area under the curve of f(x) = 10−x2 from a=0 to b=3 by using three equal width rectangles (n=3), where the height of each rectangle is determined by the left-hand side of each interval.
• If you didn't watch the video lesson, make sure to watch it first. The idea of finding area under a curve is difficult to explain just using words, but the video has a lot of helpful diagrams to clarify the idea.
• Start off by figuring out the width of the intervals you'll be using. It's the total length being covered, divided by the number of rectangles to be used:
 Width of each rectangle: b−a n ⇒ 3−0 3 =     1
Thus the width of each interval/rectangle is 1 for this problem. Next, figure out where you'll be determining the height for each rectangle. The problem told us to determine the heights using the "left-hand side of each interval". Since we have a starting location of a=0 and an interval width of 1, our intervals are:
 [0,  1]               [1,  2]               [2,  3]
The left-hand side of each rectangle are the locations of x=0,   1,  2. We will use these locations to determine the heights.
• Compute the height of each rectangle by applying the function to each of the locations we're using:
 f(0)
 f(1)
 f(2)
 10
 9
 6
Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width (1), then sum them up to find the approximate area:
 Aapprox     =     10 ·1   +   9 ·1   +   6 ·1

 =     25
25
Find an approximation to the area under the curve of f(x) = 10−x2 from a=0 to b=3 by using three equal width rectangles (n=3), where the height of each rectangle is determined by the right-hand side of each interval.
• If you didn't watch the video lesson, make sure to watch it first. The idea of finding area under a curve is difficult to explain just using words, but the video has a lot of helpful diagrams to clarify the idea.
• Start off by figuring out the width of the intervals you'll be using. It's the total length being covered, divided by the number of rectangles to be used:
 Width of each rectangle: b−a n ⇒ 3−0 3 =     1
Thus the width of each interval/rectangle is 1 for this problem. Next, figure out where you'll be determining the height for each rectangle. The problem told us to determine the heights using the "right-hand side of each interval". Since we have a starting location of a=0 and an interval width of 1, our intervals are:
 [0,  1]               [1,  2]               [2,  3]
The right-hand side of each rectangle are the locations of x=1,   2,  3. We will use these locations to determine the heights.
• Compute the height of each rectangle by applying the function to each of the locations we're using:
 f(1)
 f(2)
 f(3)
 9
 6
 1
Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width (1), then sum them up to find the approximate area:
 Aapprox     =     9 ·1   +   6 ·1   +   1 ·1

 =     16
16
Find an approximation to the area under the curve of f(x) = 10−x2 from a=0 to b=3 by using six equal width rectangles (n=6), where the height of each rectangle is determined by the left-hand side of each interval.
• If you didn't watch the video lesson, make sure to watch it first. The idea of finding area under a curve is difficult to explain just using words, but the video has a lot of helpful diagrams to clarify the idea.
• Start off by figuring out the width of the intervals you'll be using. It's the total length being covered, divided by the number of rectangles to be used:
 Width of each rectangle: b−a n ⇒ 3−0 6 = 1 2
Thus the width of each interval/rectangle is [1/2] for this problem. Next, figure out where you'll be determining the height for each rectangle. The problem told us to determine the heights using the "left-hand side of each interval". Since we have a starting location of a=0 and an interval width of [1/2], our intervals are:
 [0, 1 2 ]            [ 1 2 ,  1]            [1, 3 2 ]            [ 3 2 ,  2]            [2, 5 2 ]            [ 5 2 ,  3]
The left-hand side of each rectangle are the locations of x=0,   [1/2],  1,   [3/2],   2,   [5/2]. We will use these locations to determine the heights.
• Compute the height of each rectangle by applying the function to each of the locations we're using:
 f(0)
 f( 1 2 )
 f(1)
 f( 3 2 )
 f(2)
 f( 5 2 )
 10
 9.75
 9
 7.75
 6
 3.75
Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width ([1/2]), then sum them up to find the approximate area:
 Aapprox     =     10 · 1 2 +   9.75 · 1 2 +   9 · 1 2 +   7.75 · 1 2 +   6 · 1 2 +   3.75 · 1 2

 = ⎛⎝ 10  +  9.75  +  9  +  7.75  +  6  +  3.75 ⎞⎠ · 1 2

 =     46.25 · 1 2

 =     23.125
23.125
Find an approximation to the area under the curve of g(x) = cos(x) + 1 from a=0 to b=2π by using four equal width rectangles (n=4), where the height of each rectangle is determined by the maximum height in each interval (also called the upper sum).
• To help us understand what we're doing, begin by graphing the function. Seeing it graphed will assist us later on by making it easier to find where we should determine the heights of the rectangles. Also, to help with that, let's mark the intervals that will be used by each rectangle. There should be four rectangles, so we mark out four intervals (the dashed blue lines, below).
• Notice that the problem told us to determine the rectangle heights by using the "maximum height in each interval." This means that the place we use to determine rectangle height can vary from one interval to the next. Look at the graph from the previous step to figure out where the locations should be. In the first two intervals, the maximum height occurs at the left of each interval, while the last two intervals have the maximum height at the right of each interval. This means we'll use the following x-values to determine the height of each rectangle:
 x=   0, π2 , 3π2 ,     2 π
• Compute the height of each rectangle by applying the function to each of the locations we're using:
 g(0)
 g( π2 )
 g( 3π2 )
 g(2π)
 2
 1
 1
 2
Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width ([(π)/2]), then sum them up to find the approximate area:
 Aapprox     =     2 · π2 +   1 · π2 +   1 · π2 +   2 · π2

 = ⎛⎝ 2  +  1  +  1  +  2 ⎞⎠ · π2

 =     6 · π2

 =     3π
Find an approximation to the area under the curve of g(x) = cos(x) + 1 from a=0 to b=2π by using four equal width rectangles (n=4), where the height of each rectangle is determined by the minimum height in each interval (also called the lower sum).
• To help us understand what we're doing, begin by graphing the function. Seeing it graphed will assist us later on by making it easier to find where we should determine the heights of the rectangles. Also, to help with that, let's mark the intervals that will be used by each rectangle. There should be four rectangles, so we mark out four intervals (the dashed blue lines, below).
• Notice that the problem told us to determine the rectangle heights by using the "minimum height in each interval." This means that the place we use to determine rectangle height can vary from one interval to the next. Look at the graph from the previous step to figure out where the locations should be. In the first two intervals, the minimum height occurs at the right of each interval, while the last two intervals have the minimum height at the left of each interval. This means we'll use the following x-values to determine the height of each rectangle:
 x= π2 ,    π,     π, 3π2
• Compute the height of each rectangle by applying the function to each of the locations we're using:
 g( π2 )
 g(π)
 g(π)
 g( 3π2 )
 1
 0
 0
 1
Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width ([(π)/2]), then sum them up to find the approximate area:
 Aapprox     =     1 · π2 +   0 · π2 +   0 · π2 +   1 · π2

 = ⎛⎝ 1  +  0  +  0  +  1 ⎞⎠ · π2

 =     2 · π2

 =     π
π
Consider approximating the area under the curve f(x) = 10−x2 from a=0 to b=3 by using equal-width rectangles. Find a formula that gives the area approximation for using n rectangles, where the height of each rectangle is determined by the right-hand side of each interval.

Hint: In doing this problem, you will come up with a summation of all the rectangles added together. To convert this into an easily usable formula, you will need to use the below identities:
 n∑i=1 kai   =  k n∑i=1 ai,    where k is a constant

 n∑i=1 ⎛⎝ ai ±bi ⎞⎠ = n∑i=1 ai  ± n∑i=1 bi

 n∑i=1 c  = cn,    where c is a constant

 n∑i=1 i2  = n(n+1)(2n+1) 6
• Notice this problem has not given us a specific number of rectangles to work with. Instead, we need to come up with a formula that gives the approximate area for an arbitrary number n of rectangles. We are going to make a formula that will tell us the approximate area for any number of rectangles we choose to plug in for n. With this in mind, we need to figure out a way to talk about how wide and tall the rectangles we will be adding up are. Since the number of rectangles is a variable (n), these things will be based off the variable as well.
• Begin by figuring out how wide each rectangle must be. The total length being covered is from a to b, and the number of rectangles is n. Since the rectangles are all the same width, we have
 Width of each rectangle: b−a n ⇒ 3−0 n = 3 n
• Next, start working out the heights of the rectangles. This is a little bit more difficult since the height of the rectangle varies depending on which one we consider. Still, we can at least figure out where we will determine their heights from. The problem said we should base the heights off of the "right-hand side of each interval," so we start from the right side of our first interval. The first interval begins at a=0 and has the width we figured out above, so it will be [0, [3/n]]. The right side of that gives [3/n] for our first location. The second interval will be based off the right side of the next interval, so one width forward: 2·[3/n]. Similarly, the third interval will be based off the right side again, so another width forward to 3·[3/n]. This pattern continues until we've got all n locations:
 x = 3 n ,        2· 3 n ,        3· 3 n ,        4· 3 n ,     …    n· 3 n
• Now that we know the locations to find the rectangle heights from, we can find the area of each rectangle by multiplying the height (plug location in to f(x)) by the width (what we figured out before, [3/n]):
 Aapprox =     f ⎛⎝ 3 n ⎞⎠ · 3 n +   f ⎛⎝ 2· 3 n ⎞⎠ · 3 n +   f ⎛⎝ 3· 3 n ⎞⎠ · 3 n +  …  +  f ⎛⎝ n · 3 n ⎞⎠ · 3 n
This is an awful lot to write out, so let's instead use sigma summation notation. [If you're unfamiliar with summation notation, watch the lesson Introduction to Series.]
 Aapprox = n∑i=1 f ⎛⎝ i · 3 n ⎞⎠ · 3 n
Make sure that the above sigma notation matches the original, expanded version. It does, so we can continue.
• We can now apply the function and simplify to get
 Aapprox = n∑i=1 f ⎛⎝ i · 3 n ⎞⎠ · 3 n = n∑i=1 ⎡⎣ 10− ⎛⎝ i · 3 n ⎞⎠ 2 ⎤⎦ · 3 n

 = n∑i=1 ⎡⎣ 10− ⎛⎝ 3i n ⎞⎠ 2

 · 3 n

 = n∑i=1 ⎡⎣ 10− 9i2n2 ⎤⎦ · 3 n
• Great, we have a formula in summation notation now:
 Aapprox = n∑i=1 ⎡⎣ 10− 9i2n2 ⎤⎦ · 3 n
However, it would be very difficult to actually compute a value using the summation notation. This is where all the identities given in the problem hint come into play. We apply these identities to break the summation into something that we could calculate more easily. First, notice that while n is a variable, it doesn't actually change. It's assumed to be fixed while we work with it, so we can treat it as a constant. This means that the [3/n] multiplied on the right can be "pulled out" since it's just multiplying by a constant. Identity: ∑i=1n kai   =  k ∑i=1n ai,    where k is a constant
 n∑i=1 ⎡⎣ 10− 9i2n2 ⎤⎦ · 3 n = 3 n · ⎡⎣ n∑i=1 ⎡⎣ 10− 9i2n2 ⎤⎦ ⎤⎦
Next, we have two pieces of the summation separated by subtraction, so we can split them into two different summations: Identity: ∑i=1n ( ai ±bi )  = ∑i=1n ai  ± ∑i=1n bi
 3 n · ⎡⎣ n∑i=1 ⎡⎣ 10− 9i2n2 ⎤⎦ ⎤⎦ = 3 n · ⎡⎣ n∑i=1 10    − n∑i=1 9i2n2 ⎤⎦
• Continue to break apart the expression by using the identities. Notice that [9/(n2)] is also a constant (remember that n is constant for the sum, even if it is a variable), so we can pull it out of its sum, too:
 3 n · ⎡⎣ n∑i=1 10    − n∑i=1 9i2n2 ⎤⎦ = 3 n · ⎡⎣ n∑i=1 10    − 9 n2 · n∑i=1 i2 ⎤⎦
Finally, we can use the last two identities from the problem to convert the summations entirely: Identity: ∑i=1n c  = cn,    where c is a constant
 3 n · ⎡⎣ n∑i=1 10    − 9 n2 · n∑i=1 i2 ⎤⎦ = 3 n · ⎡⎣ 10 ·n    − 9 n2 · n∑i=1 i2 ⎤⎦
Identity: ∑i=1n i2  = [(n(n+1)(2n+1))/6]
 3 n · ⎡⎣ 10 ·n    − 9 n2 · n∑i=1 i2 ⎤⎦ = 3 n · ⎡⎣ 10 ·n    − 9 n2 · n(n+1)(2n+1) 6 ⎤⎦
• Finally, now that our expression for area has no sum, we can simplify to finish finding our formula:
 Aapprox = 3 n · ⎡⎣ 10 ·n    − 9 n2 · n(n+1)(2n+1) 6 ⎤⎦

 = 30   − 27(n)(n+1)(2n+1) 6n3

 =30   − 9(n)(2n2+3n+1) 2n3

 =30   − 9(2n3+3n2+n) 2n3

[Remark: In a previous problem, we figured out the approximate area under f(x) = 10−x2 from a=0 to b=3 using three right-side rectangles. The approximation we got was 16. We can use this to check the formula we just made. We should be able to plug in n=3 and get the same value:
 30  − 9(2(3)3+3(3)2+(3)) 2(3)3 =     30  − 9(84) 54 =    30−14     =     16
Great! Using our newly created formula gives us the same value for n=3 compared to when we found it before, so our formula checks out against our previous work!]
30  −  [(9(2n3+3n2+n))/(2n3)]
Consider the area under the curve of f(x) = 10 −x2 from a=0 to b=3. In the previous problem, you found a formula for the right-side approximation using n equal-width rectangles. This formula is
 30  − 9(2n3+3n2+n) 2n3 .
Using this formula, find the exact value of the area under the curve by taking the limit as the number of rectangles goes to infinity.
• The area approximation of a curve becomes more and more accurate as the number of rectangles (n) being used to approximate the area becomes larger and larger. Thus, if we consider the limit as n→ ∞, we can find the exact area under the curve.
• Since we found a formula for the approximate area using n rectangles in the previous problem, we can take the limit of that as n→ ∞ to find the precise area:
 limn→ ∞ 30  − 9(2n3+3n2+n) 2n3
• If you're unfamiliar with taking the limit as a variable approaches infinity, check out the lesson Limits at Infinity and Limits of Sequences. Distribute the 9 to make it easier to see what happens:
 limn→ ∞ 30  − 9(2n3+3n2+n) 2n3 = limn→ ∞ 30  − 18n3+27n2+9n 2n3
Remember, as n→ ∞, the only "important" parts of the fraction are the n3 parts, since they are the fastest growing parts:
 limn→ ∞ 30  − 18n3+27n2+9n 2n3 = limn→ ∞ 30  − 18n32n3
We can now simplify, canceling out the n3 on top and bottom:
 limn→ ∞ 30  − 18n32n3 = limn→ ∞ 30  − 18 2
Finally, since these are now just constants, they are unaffected by the limit:
 limn→ ∞ 30  − 18 2 =     30  − 18 2 =     30  −  9     =     21
21
From the lesson, we learned that the area under the curve of f(x) = 10−x2 from a=0 to b=3 can be represented by the integral below:
 ⌠⌡ 30 10−x2   dx
Furthermore, we also learned about the fundamental theorem of calculus. Given some function f(x) where F(x) is its antiderivative, then
 ⌠⌡ ba f(x)  dx    =   F(b) − F(a).
Use the above theorem to find the area under f(x) = 10−x2 from a=0 to b=3 and confirm the value you found for the area in the previous problem.
• To use the fundamental theorem of calculus we first need to understand what an antiderivative is. The antiderivative to a function is the opposite of its derivative: in effect, it is using the derivative process in "reverse" on the function f(x). The function F(x) is set up in such a way that when we take its derivative, it produces f(x). Thus, we need to think about what expressions would produce each part of f(x) when we take their derivatives. Notice that both 10 and x2 could have the power rule [which we learned about at the very end of the previous lesson] applied to them. This realization allows us to see that we can create them with the power rule in "reverse", so to speak. Instead of multiplying by the exponent and then subtracting 1 from the exponent, we will add 1 to the exponent and divide by the new exponent:
 10
 x2
 ⇓
 Antiderivative
 ⇓
 10x
 1 3 x3
• Combining these two pieces, we have found our antiderivative:
 f(x) = 10−x2        ⇒        F(x) = 10x − x33
Do a quick check on F(x) to make sure that if we take its derivative, we do indeed get f(x). This is the case, so we are ready to move on.

[Remark: Don't worry if the above process was a little confusing-if you take a calculus class, lots of time will be spent on this idea to make it clear. The important part to realize is that finding the integral ( ∫ ) of a function is based around figuring out its antiderivative. Also, it should be pointed out that F(x) above is technically not the only antiderivative to f(x). Notice that we could add any constant (+C) to F(x) and, since constants "disappear" when we take the derivative, it would still produce the same f(x). Don't worry: this will not affect using the integral to find the area under f(x). Still, the idea of adding an unknown constant (+C) is an important point that will be addressed if/when you take a calculus class.]
• Once we know the antiderivative F(x), we're ready to apply the fundamental theorem of calculus. Just plug in based on how the theorem is set up:
 ⌠⌡ ba f(x)  dx    =   F(b) − F(a)

 ⌠⌡ 30 10−x2   dx    =   F(3) − F(0)

 = ⎡⎣ 10(3)− (3)33 ⎤⎦ − ⎡⎣ 10(0) − (0)33 ⎤⎦

 = ⎡⎣ 30−32 ⎤⎦ − ⎡⎣ 0 ⎤⎦

 =   30 − 9

 =   21
Therefore, by using the fundamental theorem of calculus, we have found that the area underneath the curve of f(x) = 10−x2 from a=0 to b=3 is 21. In the previous problem, we considered approximating the area under the curve by using n rectangles, then, since the approximation grows more accurate as n becomes larger, we looked at what happened as n→ ∞. Doing so we found that the area under the curve was 21. Thus the two different ways of approaching the problem give the same answer, so they check out against each other.
21
Evaluate the integral ∫−31 −2x+3   dx  by using a graph of f(x) = −2x+3.
• Remember what the integral (∫) means: it's a way of talking about the area underneath a curve. In general,
 ⌠⌡ ba f(x)   dx        ⇔        Area under f(x) from a to b
That means, for the problem we're working on, we want to find the area under f(x) = −2x+3 from the starting location of a=−3 to the ending location of b=1.
• The problem told us to find the value of the integral by using a graph of the function, so graph f(x)=−2x+3. We also mark where the integral starts (a=−3) and stops (b=1) on the graph, then shade in the area we're looking to find.
• Notice that we can make it easier to find the area by splitting it into two chunks: an upper triangle and a lower rectangle, as in the below picture:
• With the area now split into two pieces, we can easily find the area of each piece, then add them together to find the total area. For the upper triangle, we have a height of 8 (bottom at 1, top at 9) and a width of 4 (left at −3, right at 1). This gives us
 Atriangle     = h·w 2 = 8·4 2 =     16
For the lower rectangle, we have a height of 1 (bottom at 0, top at 1) and a width of 4 (left at −3, right at 1). This gives us
 Arectangle     =     h·w     =     1 ·4     =     4
Since the total area is just made up of these two pieces, we add them together to find the area:
 Atotal     =     16 + 4     =     20
Since   ∫−31 −2x+3   dx  just represents this area, we've found its value.
• If we want to, we can also check this answer by using the fundamental theorem of calculus (which appeared in the previous problem), which says
 ⌠⌡ ba f(x)  dx    =   F(b) − F(a),
where F(x) is the antiderivative of f(x). Using it, we start by finding the antiderivative of f(x):
 ⌠⌡ 1−3 −2x+3   dx     =     −x2 +3x ⎢⎢ 1−3
Then we plug in the endpoints of b=1 and a=−3 as done in the theorem (the bar with the numbers on the right of it indicates that this plugging in still needs to be done):
 ⌠⌡ 1−3 −2x+3   dx    =   −x2 +3x ⎢⎢ 1−3

 = ⎡⎣ −(1)2 +3(1) ⎤⎦ − ⎡⎣ −(−3)2 +3(−3) ⎤⎦

 = ⎡⎣ −1 +3 ⎤⎦ − ⎡⎣ −9 −9 ⎤⎦

 = ⎡⎣ 2 ⎤⎦ − ⎡⎣ −18 ⎤⎦

 =   20
Thus the fundamental theorem of calculus gives the same value for the area, which confirms our previous answer.
20
A large water tank currently stands empty. At time t=0 (where t is measured in hours), water begins to flow into the tank from a pipe. However, the rate of flow from the pipe varies moment by moment. The below table tells us the flow rate at the beginning of every hour:
 t (in hours)
 0
 1
 2
 3
 4
 5
 6
 7
 V′ (liters/hour)
 1250
 1134
 1018
 902
 786
 670
 554
 438
Using the table, approximate the amount of water in the tank eight hours after the water begins to flow from the pipe.
• While we can't figure out the precise amount of water that enters the tank from the pipe for every hour, we can come up with an approximate value for each hour by using the rate at the beginning of the hour. For example, for the first hour of t=0↔1, we know that the hour begins at t=0 with the water flowing at a rate of 1250 liters per hour. Since we're interested in knowing the total volume during that hour, we multiply the rate by the amount of time (1 hour):
 V0, approx     =     1250 ·1     =     1250
• We can carry this process out for each hour, finding the approximate volume to flow through for every hour. Since we're multiplying 1 hour by each of the flow rates (which are in liters/hour), the approximate volume is the same number as the flow rate at the beginning of the hour. Once we know the approximate volumes for each hour, we sum them all up to find the total approximate volume:
 Vtotal, approx   =  1250 + 1134 + 1018 + 902 + 786 + 670 + 554 + 438

 =  6752
Thus the approximate volume in the tank at the end of eight hours is 6752 liters.
• Remark: You might wonder at this point, "How is this problem connected to calculus and integration?" To answer this, notice that the problem was very similar to approximating the area under a curve by using equal-width rectangles on the left side. While we never mentioned finding the area under the curve, that is effectively what we did. We found an approximation to the area under the curve of the flow rate. (Although we weren't explicitly given the flow rate as a function, we were given all the points we needed to make an approximation with rectangles.) We found (an approximation of) the integral for the flow rate from a=0 to b=8. Furthermore, notice that this integral gave us the total volume. In other words, by taking the integral of a function that describes how some quantity changes, we find the total quantity at the end of that period. This is a crucial idea in calculus and one you should hold on to. You can visualize the integral as the area under a curve, but you can also think of it as a way of "totaling up" the change over time.
Vapprox =  6752 liters [Note: If you're not sure how this problem is connected to calculus or the idea of an integral, check out the last step to the problem. It's a remark that explains how this word problem is based on the same ideas.]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Area Under a Curve (Integrals)

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Introduction 0:06
• Integral
• Idea of Area Under a Curve 1:18
• Approximation by Rectangles 2:12
• The Easiest Way to Find Area is With a Rectangle
• Various Methods for Choosing Rectangles 4:30
• Rectangle Method - Left-Most Point 5:12
• The Left-Most Point
• Rectangle Method - Right-Most Point 5:58
• The Right-Most Point
• Rectangle Method - Mid-Point 6:42
• Horizontal Mid-Point
• Rectangle Method - Maximum (Upper Sum) 7:34
• Maximum Height
• Rectangle Method - Minimum 8:54
• Minimum Height
• Evaluating the Area Approximation 10:08
• Split the Interval Into n Sub-Intervals
• More Rectangles, Better Approximation 12:14
• The More We Us , the Better Our Approximation Becomes
• Our Approximation Becomes More Accurate as the Number of Rectangles n Goes Off to Infinity
• Finding Area with a Limit 13:08
• If This Limit Exists, It Is Called the Integral From a to b
• The Process of Finding Integrals is Called Integration
• The Big Reveal 14:40
• The Integral is Based on the Antiderivative
• The Big Reveal - Wait, Why? 16:28
• The Rate of Change for the Area is Based on the Height of the Function
• Height is the Derivative of Area, So Area is Based on the Antiderivative of Height
• Example 1 19:06
• Example 2 22:48
• Example 3 29:06
• Example 3, cont.
• Example 4 40:14

### Transcription: Area Under a Curve (Integrals)

Hi--welcome back to Educator.com.0000

Today, we are going to talk about area under a curve--also known as integrals.0002

The other major idea in calculus is the notion of the integral, a way to find the area underneath some portion of a curve.0006

Like the derivative, at first glance, this might not seem terribly useful.0013

Of course, it is somewhat interesting to be able to know the area under a curve.0017

But what can we do with it that has any real use?--as it turns out, a huge amount.0020

It allows us to consider what a function has done, in total, over the span of some interval.0025

For example, if we have a function that gives the velocity of an object, how fast an object is moving around, the integral will tell us the object's location.0031

The velocity of an object, as a function--we can take the integral of that, and we can get location out of it.0039

Why is that? Well, velocity tells us motion--how fast we are moving around.0045

Well, if we look at how fast something has moved around in total, that ends up telling us where the thing lands at the end.0049

Motion, looked at over the long term--if we put all of the motion together--if we look at all it has done in total,0055

all of the motion in total, we end up seeing where we end up; we end up seeing location out of it.0062

So, that is just one example of how useful integrals can be.0067

Trust me on this: this is really useful stuff--let's check it out.0070

Consider if we had some function f(x) that has the graph right here.0075

We could ask, "What is the area between the curve and the x-axis on the interval going from a to b?"0079

So, we can fill that in; and the shaded-in portion that we have right here represents that area.0086

But how could we go about finding how much area that actually is?0090

Let's look for some way to approximate it; and notice that that is what we did with derivatives, as well.0095

What we did is thought, "Is there some way to approximate close to what the slope is?" and then,0101

"Is there a way to improve on that approximation?" and then,0105

"Is there a way to basically see what that improvement will eventually go to in the infinite version?"0107

As we got closer and closer to being right up against it with the derivative, as h went to 0, we were able to see0115

what it became perfectly--what the slope was in that instant.0120

And we will see sort of a similar idea as we work with integrals.0123

All right, we are getting a little ahead of ourselves; so let's start looking at our approximation.0126

The easiest way to find area is with a rectangle; rectangles are a great way to find area, because it is so easy to figure out what the area of a rectangle is.0130

It is simply width times height: area is just width times height.0138

Let's approximate the area under our curve with rectangles.0143

We begin by breaking our interval into n equal sub-intervals.0146

In this case, we have broken it into n = 4; here are 1, 2, 3, 4 equal-width sub-intervals.0150

So, we go from a to b, and we break it into n equal-width chunks.0158

In each one of these sub-intervals, we will base the height of each rectangle off of some xi in each sub-interval.0164

That is to say, we will choose some horizontal location in the sub-interval; and then we will see what height that horizontal location goes to.0170

That horizontal location, as a point--what height is that point at?0178

And then, that is going to set the height of that rectangle for that sub-interval.0183

So, for the ith sub-interval, we will use xi to determine that height.0189

The ith rectangle has a height of f evaluated at xi;0195

so xi is the horizontal reference location that we use to determine the height for the ith rectangle.0200

For example, in our third sub-interval, we might look at x3, the horizontal reference location for our third sub-interval.0207

We check, and we see what height that ends up going to; and so, that would be f evaluated at x3.0215

And then, that determines the height for our third rectangle.0222

We use xi to determine the horizontal reference location for our ith sub-interval,0226

which then determines the height for our ith rectangle.0232

We use this xi to determine height by getting f(xi) to tell us what that rectangle's height is.0238

Now, there is a variety of ways to choose our xi.0244

In any given sub-interval, there is a continuum: there are infinitely many different locations we could pick to be our x that we are going to choose.0247

We could choose this location, but we could also choose this location.0255

Or we could choose the location between them, or the location in between them, or over here,0258

or at the very far edge, or the other very far edge, or somewhere else.0261

There are all sorts of different ways that we can choose this.0265

And notice: the way we decide to choose our xi, for each one of these sub-intervals:0268

because the xi, that horizontal reference location, determines the height for that rectangle,0272

it is going to affect the height of the ith rectangle,0278

because height is just f evaluated at whatever horizontal reference location we chose, our xi.0281

By using a different method to choose our xi, we will get a different approximation for the area,0287

because we are going to change the height of that rectangle; and if we change the heights of all of our rectangles,0294

we will end up having a totally new version for our area approximation.0298

So, we are going to now look at some of the most common methods to choose xi for creating the heights of our rectangles.0302

Our first common method that we see is the leftmost point method,0309

where we end up evaluating the height of each rectangle by just evaluating where the leftmost point gets mapped to.0313

How high is the left side of each sub-interval?0320

We can see an example of this right here: what we do for our first sub-interval--0323

we look: where would the leftmost point of that sub-interval get mapped to?0327

It gets mapped to a height of here; and so, that determines the height of our entire rectangle.0330

For our second sub-interval, we look at the leftmost location here; it gets mapped to this height, and that determines the height of that rectangle.0335

For our third sub-interval, we look at this leftmost location; that determines that height.0342

Our fourth and final: we look at the leftmost location, and that determines that height.0347

We are determining the height of each rectangle based on the leftmost location in each of the rectangles.0351

Basically the same idea, but flipped: we can look at the rightmost point.0357

We can say, "For each sub-interval, what is the height that the rightmost horizontal location gets mapped to?"0361

For our first sub-interval, we look at the rightmost location for that one;0367

that ends up getting mapped to here, so that determines the height of our first rectangle.0370

For our second sub-interval, we look at the rightmost location in that one; that is here, so that determines the height of our second rectangle.0374

For our third sub-interval, we look at the rightmost location of that sub-interval; that determines the height of our third rectangle.0382

And for our fourth and final one, we look at the rightmost location, and that determines the height of that fourth and final rectangle.0389

We are just looking at the rightmost horizontal location to determine the height for each one of our rectangles.0396

We could also do this based on the horizontal midpoint.0402

So far, we have looked at the two extremes: the far left side of the sub-interval and the far right side of the sub-interval.0404

But we could also ask what happens to the one in the middle.0409

For our first sub-interval, we look at the one that is in the middle of that; we go up, and the midpoint gets mapped to this height.0412

So, that determines the height of that first rectangle.0419

For our second sub-interval, we look at the midpoint of that sub-interval.0422

We go up, and that tells us the height for our rectangle here.0426

For our third sub-interval, we look at the midpoint; we see what height that horizontal midpoint gets mapped to.0430

And that determines the height of that entire third rectangle.0437

And the same thing for our fourth one: we look at the midpoint.0440

What value does that midpoint get mapped to? That determines the height of that entire rectangle.0442

So, so far, we have looked at the far left side, the far right side, and the middle.0448

But there is also another way of looking at this; we can also look at the maximum height in each one of these.0452

We look at the sub-interval, and we see which one of these points gets mapped to the highest possible location.0457

We look over the entire sub-interval, and we see which one is highest in this portion.0463

So, in this case, the highest location that we get for this sub-interval is here; so that determines the height of the entire rectangle.0468

For our second sub-interval, the highest possible location we reach in this sub-interval is this location right here.0474

So, that determines the height of that entire second rectangle.0480

For our third sub-interval, we see that the highest possible height we reach in that sub-interval is here.0483

So, that determines the height of the entire third rectangle.0488

And for our fourth and final one, the highest possible point we get is here; so that determines the height of the entire rectangle.0491

The maximum height method is also sometimes known as the upper sum, because, notice:0497

by choosing the maximum height in each of our sub-intervals, we will always end up getting an approximation0502

that is going to be at least the value of the area under the curve, and more than that area, usually.0508

We can see all of the places that we ended up going above it.0514

And so, we end up having something that is above what we are actually going to end up having as the area underneath the curve.0518

And so, we call it an upper sum, because we are getting something that is above the value of the actual area underneath the curve.0525

We can also talk about the minimum height for each one of our sub-intervals.0533

We can look and see, for our first sub-interval, what is the lowest possible height that we have in here.0537

We see that it is here, and so we end up getting this as the height for it.0542

We can also do this for each one of our sub-intervals.0547

We can have our second sub-interval; the lowest possible height in that second sub-interval is right here,0551

so we end up evaluating that as the height of our second rectangle.0558

For our third sub-interval, the lowest possible height is right here; so that gives us the height of our third rectangle.0561

And for our fourth and final sub-interval, the lowest possible height over that sub-interval is right there.0566

And so, that gives us the height of our fourth rectangle.0571

This one is sometimes called the lower sum, because it is always going to end up giving us a value0574

that is below the actual area underneath the curve, because, since we are choosing based on minimum heights,0580

each of our rectangles is actually under-cutting the area.0586

We can see area that we are missing each time.0589

For each of our rectangles, we are not fully filling out that area, because we are always below it.0592

So, we end up having less than the area that is actually underneath the curve.0596

And so, it is sometimes also called the lower sum.0601

How can we find what this approximation actually ends up giving us?0605

How can we find this area? Well, first we want to figure out what the width of each one of these rectangles is.0610

That is the easier thing to figure out: we split the interval into n sub-intervals.0615

That is how we did this, right from the beginning; we split it into n equal-width sub-intervals.0620

So, what is the total length of the interval we have?0624

Well, that is going to be b - a, where we end minus where we started.0627

If we have split it into n equal-width sub-intervals, the width of each one of them must be the total length, divided by how many intervals we have.0632

So, our width is equal to b - a, divided by n; that is the width of one of our rectangles, b - a over n.0641

To figure out the height for each rectangle...well, the ith rectangle's height0651

depends on the specific xi we chose for it, with the different methods that we were just talking about.0654

There are various different ways that we can choose that xi.0659

But whatever xi we end up choosing will tell us the height; so just by definition,0661

our height is equal to f evaluated at the various xi that we chose.0665

So, with these two things in mind, we now have our width, and we have our height.0670

The area for the ith rectangle, any given rectangle, is going to be that rectangle's width,0675

which is b - a divided by n, times that rectangle's height, which was determined by the xi we chose, so our height right here.0680

So, the area of our ith rectangle is equal to the width, b - a over n, times the height of that ith rectangle,0689

which is going to be f evaluated at xi--whatever horizontal reference location we chose.0695

If we want to figure out what the total approximation is, we need to add up each one of our rectangles.0700

We aren't concerned with just one of our rectangles; we want to add up all of the rectangles in our approximation.0705

So, we sum them all up; and we can use sigma notation for that.0710

So, we have i = 1, our first rectangle, to n, the nth rectangle, which is our last rectangle,0713

since we are using a total of n sub-intervals; and then we just end up adding up the area from each one of these.0720

So, it is going to be b - a over n, the width of each one of these, times the height of each one of these, f(xi).0726

Notice that whatever method we choose to determine the height of the rectangles, the more rectangles we use, the better our approximation becomes.0734

Over here, we have n = 4; but over here, we have n = 8.0742

And notice: in this second one, we end up having a closer approximation.0747

We have less missing chunks; the more rectangles we use for our approximation,0752

the closer our approximation becomes to actually giving us the area underneath the curve.0757

Thus, our approximation becomes ever more accurate as the number of rectangles, n, goes off to infinity.0763

So, as we increase our n-value higher and higher and higher and higher, our approximation becomes better and better and better and better.0769

As our n slides off to infinity, we will be able to get, effectively, what the perfect value is underneath that curve.0776

With this realization in mind, remember: the approximation we just figured out for the area with n rectangles0784

was the sum of i = 1 up until n of our width for each rectangle, (b - a)/n,0789

times the height of each rectangle, f evaluated at its specific xi.0795

And this approximation becomes more accurate as our n goes off to infinity,0799

as the number of rectangles we have becomes more and more and more and more.0803

We get a finer and finer sense of the area underneath the curve.0806

With that idea in mind, we take the limit at infinity to find the area underneath the curve.0809

The area under our function f(x) from a to b, in that interval from a to b, is the limit as n goes to infinity,0815

as our number of rectangles slides off to infinity, of our approximation formula, the sum from i = 1 to n of the width, (b - a)/n, times the height, f(xi).0822

It is the limit as our number of rectangles slides off to infinity of our normal approximation formula.0834

If this limit exists (and it might not for some weird functions, but for most of the functions we are used to,0841

it will end up existing), what that limit is: it is called the integral from a to b.0846

It is denoted with the integral from a to b...that is the integral sign there,0853

that new sign that you probably haven't seen before: integral from a to b of f(x)dx.0856

The process of finding integrals is called integration.0861

The integral from a to b tells us the area underneath the curve, f(x), from a to b (that interval a to b)--really cool stuff here.0865

Here is the really amazing part: the integral from a to b of f(x)dx is based on the anti-derivative of f(x)--0875

that is, the derivative process done in reverse on f(x).0885

So, we talked about the idea of taking the derivative of some function, of being able to see some function,0889

and then turn it into another function that talks about the rate of change of that.0894

For that, we had f(x) become, through the derivative, f'(x).0898

But we can also talk about if we did the reverse of this process.0905

Instead of taking a derivative, we did the anti-derivative, where we worked our way up the chain.0910

And we symbolized that with the capital F(x); F(x) is the one that you can take the derivative of, and it becomes f(x).0914

So, F(x), the anti-derivative, is the thing where, if you take its derivative, you just get f(x),0925

the function that we are starting with, what is inside of our integral.0932

We can take the derivative process and reverse it, and we are able to start talking about the area underneath the curve.0935

This is really cool stuff; armed with this notation of F(x), it turns out that the integral of a to b of f(x)dx--0941

that is, the area underneath the curve from a to b of some f(x) function--0950

is equal to the anti-derivative of f, evaluated at b, minus the anti-derivative of f, evaluated at a.0955

Wow, that is amazing; it is just so incredibly elegant that there is this way to talk about the area underneath the curve0963

with this thing that we just talked about that had rate of change.0970

And it seems really shocking that these things are connected at all.0972

But it is absolutely amazing that it allows us to really easily find area for something that would be otherwise very difficult to work through,0975

that infinite limit that we were just talking about.0982

So, you might be wondering why this happens; why is there this connection between the area underneath the curve,0985

and this anti-derivative, where we are talking about what the height is...the anti-derivative of the height of it0991

gives us the area underneath the curve...it seems really surprising at first.0996

In short, what we can do is think of the area underneath the curve as being a special function, a(x).1000

Now, notice: we have this area underneath the curve, the shaded portion, as being a(x).1007

Notice that the rate of change for the area is based on the height of the function at any given location.1013

For example, if we consider this x location right here, how fast our area function is changing is based on how tall our function is in that moment.1020

And how tall is a function? Well, that is just what you get when you evaluate the function, f(x).1029

Notice: if we had gone to some other horizontal location, where we had a different height,1034

we would end up having a very different speed that our area was growing at.1039

If the height of the function is small, our area grows at a slow rate.1044

If the height of the function is high, our area grows at a fast rate.1048

So, the height of the function changes how fast our area grows.1052

How fast does something grow? Well, that is what we are talking about: rate of change.1056

That means that since height is connected to area through rate of change, since height gives us the rate of change of area,1060

well, rate of change was a derivative that we were just talking about.1069

So, height is the derivative of area, because height is the rate of change of our area.1073

That means that the reverse works as well; we can look at the symmetric version of this.1078

Area is based on the anti-derivative of height; since height is the derivative of area,1083

that means that area must be the anti-derivative of height.1088

Since we go in one direction, if we go in the opposite direction, doing the opposite thing, we get the other version.1091

Since a'(x) = f(x)--that is to say, the derivative of x equals little f(x), if we do the anti-derivative of f,1095

we end up getting the anti-derivative of a', which is just our area function that we started with.1104

If this a little bit confusing to you now, don't worry; it very well might be, and it would be perfectly reasonable.1109

Just think about it when you end up getting exposed to this new idea when you actually take a calculus class.1114

This ends up being a fair bit of a way into calculus class, but I think it is a really cool idea.1119

And now that you understand what is going on intuitively, or at least have the seed ready to blossom at a future date,1123

when you see it later, you will think, "Oh, now I understand what that guy was talking about--1130

it is now starting to make sense!" and there is really cool stuff here.1135

I really, really love this stuff; and I hope that you are getting some sense of just how amazingly beautiful all of this stuff in math is.1138

All right, let's start looking at some examples.1144

Find an approximation to the area under f(x) = x2 from a = 0 to b = 3 by using three equal-width rectangles--1147

that is, n = 3--on the left-hand point of each interval.1155

The first thing: let's just get a quick sketch here, so that we can see what is going on.1159

f(x) = x2: we will just, really quickly, sketch what we are looking at here.1163

And so, let's say here is a = 0; here is b = 3.1167

So, if we are going to use three equal-width rectangles, n = 3, we are breaking it up into three chunks.1173

And we are going to evaluate each one of these rectangles on the left-hand point of each interval.1179

Three equal chunks here...if it is the left-hand point, then that first one there, this one here, and this one here give us the area for each one of these.1183

Notice that that first rectangle won't have any area at all, because we are evaluating the left-hand point of each interval.1196

First, let's check and see what the width is: the width is going to be...1203

since it is equal width for each one of these...how long is our interval?1208

That is b - a, divided by...how many sub-intervals do we break it into? n.1212

So, we have 3 for our ending location, minus 0 for our starting location, divided by 3 (is the total number of sub-intervals).1216

3/3 gets us 1, so we have a width of 1 for each one of these.1223

So, at this point, we can go in and see where we have our first location; the sub-interval will go from 0 to 1;1228

the next one will go from 1 to 2; the next one, the last one, goes from 2 to 3.1235

All right, with all of this in mind, we can now see about evaluating each one of these rectangles.1240

Our first rectangle will be i = 1; our second rectangle, i = 2; and our last, final rectangle, i = 3.1244

Where will we evaluate our first rectangle? Well, that is the left-hand point.1253

If our first rectangle...remember: it is going from 0 to 1; 0 to 1 is the sub-interval it is evaluating.1257

The left-hand point is 0; so it is going to have a height of f at 0.1266

What is the width? The width is 1, so 1 times f(0).1271

In general, notice that that is also just the same thing as saying the width, (b - a)/n, times our function evaluated at...however we determined our xi.1275

In this case, we are determining based on left-hand points.1287

So, 1 times f at 0...our f(x) equals x2, so we have 1(0)2,1290

which comes out to be just 0 for the area of our first rectangle, this right here.1296

And we can see that it is going to have to be completely flat, not really a rectangle,1301

just a chunk of line, because it doesn't have any height, because we are evaluating at the left-hand side.1305

For our second rectangle, once again, there is a width of 1, times the height (will be evaluated at the left-hand side here); it will be 1.1309

So, f evaluated at 1...1 times 12 gets us just 1; so there is an area of 1 for our second rectangle.1316

And our third and final rectangle: 1 times f evaluated at 2, because the left-hand from 2 to 3 is going to be 2;1324

1 times f(2) is 1 times 22, when we evaluate that function; and that comes out to 4.1332

So, the total amount of area that we get for our approximation, the total approximation we get,1338

is going to be equal to each of these added up together--the first rectangle, 0,1343

plus the second rectangle, 1, plus the third, final rectangle (4 there)--each of our areas.1349

0 + 1 + 4 gets us a total area of 5 for our entire approximation of a = 0 to b = 3 with three sub-intervals and the left-hand point for each one.1356

Our second example is very similar to our previous example.1368

We are finding an approximation to the area under f(x) = x2 from a = 0 to b = 31371

by using four equal-width rectangles, n = 4, on the maximum point of each interval.1378

The only difference here is that we are now using four rectangles, and we are doing it based on the maximum point,1383

the highest location for each one of these intervals.1390

We draw in our curve...x2...we have something like this; we are going, once again, from a = 0 to b = 3.1392

So, there is our interval; and we are going to be looking for the maximum point of each interval.1400

If it is four equal-width rectangles (let's draw that in really quickly: 1, 2, 3, 4),1404

notice: the maximum point of each interval--where is that going to be here?--well, that is going to be here.1409

Where is that going to be for this one?--that will be here.1414

Where will that be for this one?--that will be here; and where for this one?--that will be here.1416

The maximum point for this one is basically the same thing as saying the right-hand side.1420

Now, I want to point out that this is not always true.1428

As we saw when we were working through this lesson, right-hand point and maximum can give us totally different rectangle pictures.1430

However, for something like f(x) = x2, where it is just constantly growing, constantly growing, constantly growing,1436

since it is always growing as it goes off to the right, that means that for any sub-interval,1442

the right-most point will be at the highest height; so the right-most point1446

is the same thing as maximum for the specific case of the function x2.1449

With a different function, you might end up having different things; so it is something that you have to think about.1454

But in this specific case, it will be the same thing as just evaluating at the right-hand side.1458

What is the width of each one of these rectangles?1462

Well, the width is (b - a)/n; in that case, we have a width of 3 total, divided by 4 for each one.1464

And let's put that in decimal for ease, because we are going to end up using a calculator to crunch these numbers,1470

because we will have a lot of decimals showing up otherwise.1474

Our very first horizontal location is 0, because we start at 0.1477

The next one will be 0.75; we have a width of 0.75.1482

The next one will be at 1.5, because we are another 0.75 step ahead of that.1486

The next one will be at 2.25, another 0.75 step ahead of that; and finally, we finish at 3, which makes sense.1491

We have to start at 0 and end at 3; and we work by step after step of our width, 0.75, to make it up each time.1498

Our first rectangle...figure out the area for that one; our second rectangle; our third rectangle; and our fourth rectangle.1506

OK, so our first rectangle: remember, the right-hand point is the same thing as the maximum point in the specific case of the function x2.1514

So, what is the width here? The width is 3/4; remember, in general, the area of any rectangle is the width times the height.1523

So, in this case, it will be (b - a)/n, because that is always going to be the width if we have n equal sub-intervals going from a to b.1530

b - a is the total length; divide by n for the width.1537

Multiply that times the height of each one of them, our f evaluated at xi.1541

And xi will depend on how we are choosing the point to look at.1545

In this case, we are looking at the maximum point, which ends up being the right-hand point;1548

so we will always end up looking at the right side of each one of these sub-intervals.1551

OK, back to our first one: 3/4, our width, times the height--where does the height end up getting evaluated?1555

Well, if we are going from 0 to 0.75 (that is our first sub-interval), we are going to end up looking at the most right-hand part, which is 0.75.1561

So, f...plug in 0.75; 3/4 times f(0.75) is the same thing as 0.75 times 0.752.1570

Since f is just x2, just a squaring function, it is 0.75 times 0.752.1583

We work that out with a calculator, and we get 0.422; great.1589

The second rectangle: once again, we evaluate 3/4, the width, times f evaluated now at 1.5,1594

because the right side of this one, the right side of our second rectangle, will be 1.5.1601

Here was our first rectangle; now we are on our second rectangle; its right side is 1.5.1607

So, 3/4 times 1.52: work that out with a calculator; we get 1.688.1612

The third rectangle is this one right here: the right side of that rectangle, the maximum height, is 2.25.1623

It is still the same width; the width will never end up changing, because we set the width to be equal for all of them.1629

So, f...plug in 2.25 from the right side (in this specific case);3/4...our function is squared, so 2.252;1634

that comes out to be 2.797; and our fourth and final rectangle...width times the height that it evaluates at...1641

that is going to be 3, because the far right-hand side of our final interval is 3.1650

3/4 times f(3)...remember, the right-hand side, in this specific case, happens to be the maximum.1654

If you worked with a different function, you would have to think about what was going to be the maximum location there.1661

3/4 times 32 comes out to be 6.75.1666

If we want to know what the total area approximation is, we add up all of these numbers.1671

So, it is going to be area equals 0.422 + 1.688 + 3.797 + 6.75.1676

We add up all four of the rectangles, and that tells us the total approximation we got by using this specific method.1686

And that comes out to be 12.657; so that is the total approximation we end up getting here.1693

Notice: at this point, we now have the lower sum and the upper sum.1701

In our first example, we chose minimum, because we chose the left-hand side; and we ended up getting 5 out of that.1705

At this point, we now have just chosen the maximum, so we got an upper sum,1711

something that is the most area it could possibly be; and we ended up getting 12.657.1715

So, whatever the actual area underneath that curve is, we know that it has to be between 51721

(because that was the lower sum in our first example) and 12.657 (because we just got an upper sum in this, our second example).1725

So, whatever the actual value of the area underneath that curve between 0 and 3 is, it has to be somewhere between 5 and 12.657.1731

Now, in our third example, we are going to actually figure out what it is precisely by taking an infinite limit.1739

For our third example, find the precise area under f(x) = x2 from a = 0 to b = 3,1745

using the limit as the number of rectangles, n, goes off to infinity.1751

To do this, we will also end up needing this specific identity, this sum of i = 1 to n of i2 = n(n + 1)(2n + 1), all divided by 6.1755

But we won't actually end up using that until we get about halfway through this thing.1766

So, first, how do we set this thing up?1769

Once again, here is just a quick picture to help us clarify things.1772

We have x2; we are working our way from a = 0 to b = 3.1777

All right, and we are going to end up cutting it into some number of sub-intervals.1785

We are going to end up cutting it into n sub-intervals.1789

And then, from there, we will let n go off to infinity.1790

But we have to start by figuring out what the area would be if we had just some actual value for n, and then we let n run off to infinity.1793

First, what is our width going to end up being?1801

Our width will end up being (b - a)/n, because that is always the case: (b - a)/n.1803

In this case, a = 0; b = 3; so our width is going to be 3, divided by the number of rectangles we choose to use, 3 divided by n.1809

That is the width of each one of these.1816

Now, we need to decide how we are going to determine where we choose our xi.1818

It is going to end up actually making our notation just a little bit easier if we choose right-most; so we are just going to arbitrarily choose right-most.1823

I want to point out to you, though, that it doesn't actually matter which one we choose: right-most, left-most, midpoint, upper sum, lower sum...1831

If it does eventually converge to a single value, if our limit does exist as n goes off to infinity, then we will have found the area underneath it.1838

And no matter how we choose to set up our cuts and our heights, as the number of rectangles goes farther and farther off to infinity,1846

our area has to get closer and closer to the actual thing,1854

no matter how we set up the heights, if it can get to an actual value under the area.1856

So, it doesn't matter which one we choose, specifically.1861

Right-most is nice, because it is easy to do it notation-wise; so if you end up having to do a similar problem,1863

you can basically just copy the method I am doing here to set up.1867

And it will end up working for you in notation, as well.1870

All right, let's get back to this: we will set up our right-most xi for each one.1873

So, what will xi end up being? Well, our first xi will be...1878

a is the far left side, so what would be the next one?1886

Well, that would be a plus...what is our width? Our width is 3/n...3/n for the first xi.1888

So, in general, to get out to the ith xi, we start at a.1896

And then, if we are at the ith xi, we will be...3/n is our width each time we go forward a step.1900

How many steps did we take forward? i steps.1905

If we start at a, and then we want to get to the xi, which is the far right side of any one,1908

well, the first one would show up at +1(3/n); the second one would show up at +2(3/n); and the third one would show up at +3(3/n).1912

It is the number of steps, times our width each time, 3/n.1924

So, the number of steps that we have taken, if we are at the right-most of each side, is just going to be i, whatever sub-interval we are at.1928

If we are in our first sub-interval, we have taken one width-step to get to the right side.1935

If we are in our tenth sub-interval, we have taken ten width-steps to get to the right side of that tenth sub-interval.1938

In general, xi is equal to a, our starting location, plus 3/n, times i.1944

In the specific case of this problem, we can plug in what our a equals, 0.1950

We have xi = 0 + 3/n(i); so 3 over n times i gives us our xi for this specific problem.1955

But if you were doing just any problem in general, you would want to use that first part, a plus...1964

and also, you wouldn't use 3/n; you would end up using whatever your specific width ended up being, which would be (b - a)/n.1969

It is not necessarily 3/n; that is this specific problem that we are working right here.1975

All right, if we have xi for each one, what will end up being the height for each one of our rectangles?1980

The height of the ith rectangle is going to be f evaluated at xi, which to say f evaluated at 3/n times i.1985

So, in general, what we do next is set up our limit: the limit as n goes off to infinity of our sum.2001

The approximation for n rectangles will be to start at our first rectangle, i = 1, and go out to our last rectangle, n.2009

And it is going to be the width of each one of these rectangles, (b - a)/n, times the height of each one of these rectangles, f(xi).2015

Now, this formula here will always end up working for any problem that you have set up.2023

So, that is a useful thing to work with.2028

Now, we are going to start using what we have in our specific problem--we will start plugging things in.2030

Our limit is still the same; our summation is still the same; we are going from the first to the last.2034

Our (b - a)/n...we figured out that that was a width of 3, divided by the number of rectangles we chose, 3 divided by n,2040

times f evaluated at xi...well, what is f evaluated at xi?2046

Well, f(x) = x2; so if our xi is 3/n times i (that was what we figured out that our xi has to be),2050

then f evaluated at 3/n, times i, is 3/n times i squared.2061

And we have that right here; we now plug that in: 3/n times i, the whole thing squared.2070

At this point, we can now simplify that just a little bit, and we have the limit as n goes off to infinity of the sum, i = 1 to n,2077

our first rectangle to our last rectangle; the square distributes--we have 32/n2 times i2.2086

We also have that 3/n there; so we simplify that to 27, over n3, times i2.2093

All right, at this point, we are now ready to move on to the second half of this.2103

We can now start working to figure out what this infinite series ends up coming out to be.2107

All right, continuing on with our example: we have that the area is equal to the limit as n goes off to infinity from i = 1 to n2113

of 27/n3 times i2, whatever that ends up happening to be.2121

And at a later point, we will end up putting this identity into the problem so that we can actually solve this thing out.2126

All right, the first thing to notice is that the 27/n3 part doesn't actually do anything inside of the sum.2131

The n, as far as the sum is concerned, is actually a fixed value.2140

The n here is just a constant; remember when we were working with sigma notation--the number on top was just some number.2144

It didn't change around during the course of doing things.2149

So, since the n isn't changing inside of the sigma (it will change over here, because n will go off to infinity,2152

but as far as the sigma is concerned, it doesn't change--the limit has to do something--2158

so since the sigma doesn't have anything happening), we can actually pull it outside of the sigma.2162

The first step here is to see that what we really have is: we can write this as limit as n goes off to infinity.2168

We pull out the 27/n3, because it is not affected if it does it on the inside or the outside.2173

It is just a scalar constant, as far as our series here is concerned.2179

So, i = 1 up to n of i2...at this point, look: we have the sum as i = 1 goes off to n of i2.2184

So, we can now swap it out for this portion right here.2195

So, we swap it out for that portion right here; we have the limit as n goes off to infinity of 27/n3;2199

and we swap out n times n + 1 times 2n + 1, all over 6.2207

Great; let's work to simplify things out a bit: the limit as n goes off to infinity...2217

we will keep our 27/n3 off to the side for just a moment.2222

Well, let's actually put it into the thing, but we will deal with this part first.2228

n times n + 1 times 2n + 1...let's expand this a little bit.2231

We do a little bit of expanding; we have n2 + n, times 2n + 1, all over n3 times 6; great.2234

Limit as n goes off to infinity...we can finish expanding our factors there.2249

We have 27, times n2 times 2n (becomes 2n3); n2 + 1...n2 times +1; n times 2n gets 2n2...2253

So, we have a total of +3n2; and n times 1 gets us + n, all over 6n3.2264

Limit as n goes off to infinity...we have 27 times 2n3; that comes out to be 54n3;2274

plus 3 times n2...that comes out to be 81n2; plus 27n, all divided by 6n3.2285

All right, at this point, let's move this all up here, so that we can keep working on it.2300

Notice that, if we want, we can break this into two separate fractions.2306

We have the limit as n goes to infinity of 54n3, over 6n3, plus 81n2, plus 27n, all over 6n3.2310

Now, this portion right here, the n3 and the n3, will end up canceling out.2331

We have 54 divided by 6; but the limit as n goes off to infinity for this portion...2336

well, we have n3 on the bottom; that is a 3 degree on the bottom; but on the top, we only have n2 and n.2343

So, that is a 2 degree and a 1 degree; those can't compete with an n3 on the bottom.2350

A degree of 3 on the bottom is going to end up crushing that in the long term.2355

As n rides off to infinity, our denominator is going to get so much bigger than our numerator that this whole part here just crushes down to 0.2359

That means that we are left, as our n goes off to infinity, with simply having 54 divided by 6.2366

And 54 divided by 6 gets us 9; so the total area underneath that curve is actually precisely equal to 9--pretty cool.2372

Notice that it does take a little bit of challenging effort to work through this limit as n goes to infinity.2381

We can work through it slowly, but it is not easy.2387

We had to pull up this kind of arcane formula.2389

It is not a very difficult summation formula, but it is not one that we probably know immediately.2392

So, this stuff isn't super easy to work with (limit as n goes off to infinity), if we are trying to do this infinite cut method.2396

And that is why that fundamental theorem of calculus, that integral with anti-derivative stuff, is so useful.2403

Let's see just how powerful that is now in our final example of the course.2409

Using the amazing fact that the integral of a to b of f(x)dx is equal to the anti-derivative of f evaluated at b,2413

minus the anti-derivative of f evaluated at a (and remember: the integral is just the area underneath the curve, from a to b,2421

for some curve defined by f(x)), find the area under f(x) = x2 from a = 0 to b = 3.2428

This part right here is the area that we are looking for.2435

So, what we want to do is figure out what F(b) - F(a) is; F(x) is the anti-derivative of f(x).2441

So, the very first step that we need to figure out here is what the anti-derivative is to f(x).2452

f(x) = x2; when we worked through the derivative examples, we talked about the power rule,2457

where you take the exponent, and you move it down.2465

So, for example, if we have f(x) = x5, the derivative of x is equal to...move that exponent, the 5, down...2467

we have 5 times x, and then we subtract 1 from the exponent; so - 1 from it at that point...it will be 5x4.2481

There is this nice, easy rule for taking derivatives with the power rule.2489

If we have x2, and we want to reverse the process, well, since it drops down by 12493

each time we end up taking a derivative, that means that the anti-derivative must push our exponent up by 1.2498

Since it will drop down one when we take the derivative, when we go to the derivative of F(x), remember:2506

since we can take the derivative of F(x) to just get f(x), we have to have this relationship2513

of our exponent dropping down when we take the derivative of F(x).2520

The derivative of F(x), that 3, must drop down to a square, since 3 - 1 will go down to 2.2525

So, we know that the exponent that must start there must be 3.2531

However, if we were to work this out, we would end up getting x3; if we took the derivative of this,2534

we would bring the 3 down, and we would get 3 times x3 - 1, or 3x2.2539

But there is this problem here: it is not 3x2; it is just plain x2.2545

So, how can we get rid of this 3 coefficient at the front? We just divide the whole thing by 3.2549

So, we divide the whole thing by 3, and now we have the anti-derivative to it.2555

Let's check and make sure that that works: let's check: if F(x) is equal to x3/3, then what is the derivative of F(x)?2560

Well, it should turn out to be f(x); we have this cubed exponent; we bring that down and out to the front.2570

That is going to be equal to 3 times what we started with; 3 - 1 becomes a 2; it is still divided by 3;2576

3 and 3 on the bottom cancel out, and we have x2, which is what we started with; that checks out--great.2584

So now, we have figured out what our anti-derivative is.2591

At this point, we can now actually use this portion of the formula:2594

F evaluated at...what is our b?...3, minus F evaluated at...what is our a?...0:2598

so, F(3) - F(0)...the anti-derivative of our function evaluated at the far end, minus the anti-derivative at our starting location.2608

The ending location, minus the starting location--what is our thing?2616

It is x3/3; that is what our F(x) is; so we have 33/3 - 03/3.2619

03/3 just disappears; so we have 33/3; 3 cancels this into a 2;2630

the 3 on the bottom cancels our top exponent down by 1; we have just 32; 32 is 9--nice.2637

And that is the exact same thing that we got by working through that long infinite series method,2645

where we had the limit as n goes to infinity of this approximation.2650

And notice how much faster this was than that previous method.2653

And this was with me carefully explaining a bunch of ideas that we have just seen for the very first time.2656

If you were used to doing this...when you get used to this method, you can fly through it.2661

You can do it so much faster than trying to work through the limits.2664

That precise, formal limit method is something that you learn at the very beginning, just so we can introduce this really, really amazing fact.2667

This is the fundamental theorem of calculus; this idea is so important that it gets the name "fundamental theorem of calculus."2674

It is really cool stuff here.2679

All right, that finishes this course; it has been a pleasure teaching you.2681

I hope that you have not just learned how these things work and how to use them,2684

but that you have also started to gain some idea of how math works on a deeper, more intuitive level.2688

Things are starting to make more sense to you than they were at the beginning of the course.2693

Beyond just the specific material that you have worked with, you are starting to get a better sense of just how math works in a personal way.2696

And even if you don't decide to continue on with math, I hope you have started to develop2702

a little bit more of an appreciation for just how cool it is--how many uses it ends up having.2705

This stuff is really great, and it forms the basis for so much of the society that we have.2709

The basis of technology, in many ways, is mathematics; plus, I think that this stuff is just beautiful and cool for itself, just to study it.2714

All right, we will see you at Educator.com later.2721

And I wish you luck in whatever you end up doing--goodbye!2723