For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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 subintervals, and then put a rectangle the width of each subinterval 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 subintervals evenly. However, the height of each rectangle can vary depending on the height of the function in that subinterval. Furthermore, the function has different heights throughout the subinterval, so we have to come up with some method to choose an x_{i} in each subinterval to find the height of its associated rectangle: f(x_{i}).
 Here are some of the most common methods for choosing the x_{i} in each subinterval:
 LeftMost Point: We choose x_{i} such that it is the leftmost point in each subinterval. The height of the rectangle is based on its left side.
 RightMost Point: We choose x_{i} such that it is the rightmost point in each subinterval. The height of the rectangle is based on its right side.
 MidPoint: We choose x_{i} such that it is the midpoint in each subinterval. The height of the rectangle is based on its middle.
 Maximum (Upper Sum): We choose x_{i} such that the rectangle is the highest possible for the subinterval. The height of the rectangle is the highest place the function achieves in that subinterval.
 Minimum (Lower Sum): We choose x_{i} such that the rectangle is the lowest possible for the subinterval. The height of the rectangle is the lowest place the function achieves in that subinterval.
 We can evaluate the area approximation by summing up all of the rectangles. This comes out to be
∑
⎛
⎝b−a
⎞
⎠
·f(x_{i}).  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 to b:
lim
∑
⎛
⎝b−a
⎞
⎠
·f(x_{i})  If the above limit exists, it is called the integral from a to b. It is denoted by
⌠
⌡
f(x) dx  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)
 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:
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 "lefthand side of each interval". Since we have a starting location of a=0 and an interval width of 1, our intervals are:Width of each rectangle: b−a n⇒ 3−0 3= 1
The lefthand side of each rectangle are the locations of x=0, 1, 2. We will use these locations to determine the heights.[0, 1] [1, 2] [2, 3]  Compute the height of each rectangle by applying the function to each of the locations we're using:
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:f(0) f(1) f(2) 10 9 6 A_{approx} = 10 ·1 + 9 ·1 + 6 ·1 = 25
 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:
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 "righthand side of each interval". Since we have a starting location of a=0 and an interval width of 1, our intervals are:Width of each rectangle: b−a n⇒ 3−0 3= 1
The righthand side of each rectangle are the locations of x=1, 2, 3. We will use these locations to determine the heights.[0, 1] [1, 2] [2, 3]  Compute the height of each rectangle by applying the function to each of the locations we're using:
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:f(1) f(2) f(3) 9 6 1 A_{approx} = 9 ·1 + 6 ·1 + 1 ·1 = 16
 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:
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 "lefthand side of each interval". Since we have a starting location of a=0 and an interval width of [1/2], our intervals are:Width of each rectangle: b−a n⇒ 3−0 6= 1 2
The lefthand 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.[0, 1 2] [ 1 2, 1] [1, 3 2] [ 3 2, 2] [2, 5 2] [ 5 2, 3]  Compute the height of each rectangle by applying the function to each of the locations we're using:
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:f(0) f( 1 2) f(1) f( 3 2) f(2) f( 5 2) 10 9.75 9 7.75 6 3.75 A_{approx} = 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
 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 xvalues to determine the height of each rectangle:
x= 0, π 2, 3π 2, 2 π  Compute the height of each rectangle by applying the function to each of the locations we're using:
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:g(0) g( π 2) g( 3π 2) g(2π) 2 1 1 2 A_{approx} = 2 · π 2+ 1 · π 2+ 1 · π 2+ 2 · π 2= ⎛
⎝2 + 1 + 1 + 2 ⎞
⎠· π 2= 6 · π 2= 3π
 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 xvalues to determine the height of each rectangle:
x= π 2, π, π, 3π 2  Compute the height of each rectangle by applying the function to each of the locations we're using:
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:g( π 2) g(π) g(π) g( 3π 2) 1 0 0 1 A_{approx} = 1 · π 2+ 0 · π 2+ 0 · π 2+ 1 · π 2= ⎛
⎝1 + 0 + 0 + 1 ⎞
⎠· π 2= 2 · π 2= π
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:




 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 "righthand 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]):
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.]A_{approx} = f ⎛
⎝3 n⎞
⎠· 3 n+ f ⎛
⎝2· 3 n⎞
⎠· 3 n+ f ⎛
⎝3· 3 n⎞
⎠· 3 n+ … + f ⎛
⎝n · 3 n⎞
⎠· 3 n
Make sure that the above sigma notation matches the original, expanded version. It does, so we can continue.A_{approx} = n
∑
i=1f ⎛
⎝i · 3 n⎞
⎠· 3 n  We can now apply the function and simplify to get
A_{approx} = n
∑
i=1f ⎛
⎝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− 9i^{2} n^{2}⎤
⎦· 3 n  Great, we have a formula in summation notation now:
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=1}^{n} ka_{i} = k ∑_{i=1}^{n} a_{i}, where k is a constantA_{approx} = n
∑
i=1⎡
⎣10− 9i^{2} n^{2}⎤
⎦· 3 n
Next, we have two pieces of the summation separated by subtraction, so we can split them into two different summations: Identity: ∑_{i=1}^{n} ( a_{i} ±b_{i} ) = ∑_{i=1}^{n} a_{i} ± ∑_{i=1}^{n} b_{i}n
∑
i=1⎡
⎣10− 9i^{2} n^{2}⎤
⎦· 3 n= 3 n· ⎡
⎣n
∑
i=1⎡
⎣10− 9i^{2} n^{2}⎤
⎦⎤
⎦3 n· ⎡
⎣n
∑
i=1⎡
⎣10− 9i^{2} n^{2}⎤
⎦⎤
⎦= 3 n· ⎡
⎣n
∑
i=110 − n
∑
i=19i^{2} n^{2}⎤
⎦  Continue to break apart the expression by using the identities. Notice that [9/(n^{2})] 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:
Finally, we can use the last two identities from the problem to convert the summations entirely: Identity: ∑_{i=1}^{n} c = cn, where c is a constant3 n· ⎡
⎣n
∑
i=110 − n
∑
i=19i^{2} n^{2}⎤
⎦= 3 n· ⎡
⎣n
∑
i=110 − 9 n^{2}· n
∑
i=1i^{2} ⎤
⎦
Identity: ∑_{i=1}^{n} i^{2} = [(n(n+1)(2n+1))/6]3 n· ⎡
⎣n
∑
i=110 − 9 n^{2}· n
∑
i=1i^{2} ⎤
⎦= 3 n· ⎡
⎣10 ·n − 9 n^{2}· n
∑
i=1i^{2} ⎤
⎦3 n· ⎡
⎣10 ·n − 9 n^{2}· n
∑
i=1i^{2} ⎤
⎦= 3 n· ⎡
⎣10 ·n − 9 n^{2}· n(n+1)(2n+1) 6⎤
⎦  Finally, now that our expression for area has no sum, we can simplify to finish finding our formula:
A_{approx} = 3 n· ⎡
⎣10 ·n − 9 n^{2}· n(n+1)(2n+1) 6⎤
⎦= 30 − 27(n)(n+1)(2n+1) 6n^{3}=30 − 9(n)(2n^{2}+3n+1) 2n^{3}=30 − 9(2n^{3}+3n^{2}+n) 2n^{3}
[Remark: In a previous problem, we figured out the approximate area under f(x) = 10−x^{2} from a=0 to b=3 using three rightside 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:
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(2(3)^{3}+3(3)^{2}+(3)) 2(3)^{3}= 30 − 9(84) 54= 30−14 = 16

 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:
lim
n→ ∞30 − 9(2n^{3}+3n^{2}+n) 2n^{3}  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:
Remember, as n→ ∞, the only "important" parts of the fraction are the n^{3} parts, since they are the fastest growing parts:
lim
n→ ∞30 − 9(2n^{3}+3n^{2}+n) 2n^{3}=
lim
n→ ∞30 − 18n^{3}+27n^{2}+9n 2n^{3}
We can now simplify, canceling out the n^{3} on top and bottom:
lim
n→ ∞30 − 18n^{3}+27n^{2}+9n 2n^{3}=
lim
n→ ∞30 − 18n^{3} 2n^{3}
Finally, since these are now just constants, they are unaffected by the limit:
lim
n→ ∞30 − 18n^{3} 2n^{3}=
lim
n→ ∞30 − 18 2
lim
n→ ∞30 − 18 2= 30 − 18 2= 30 − 9 = 21


 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 x^{2} 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 x^{2} ⇓ Antiderivative ⇓ 10x 1 3x^{3}  Combining these two pieces, we have found our antiderivative:
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.f(x) = 10−x^{2} ⇒ F(x) = 10x − x^{3} 3
[Remark: Don't worry if the above process was a little confusingif 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:
⌠
⌡b
af(x) dx = F(b) − F(a) ⌠
⌡3
010−x^{2} dx = F(3) − F(0) = ⎡
⎣10(3)− (3)^{3} 3⎤
⎦− ⎡
⎣10(0) − (0)^{3} 3⎤
⎦= ⎡
⎣30−3^{2} ⎤
⎦− ⎡
⎣0 ⎤
⎦= 30 − 9
Therefore, by using the fundamental theorem of calculus, we have found that the area underneath the curve of f(x) = 10−x^{2} 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
 Remember what the integral (∫) means: it's a way of talking about the area underneath a curve. In general,
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.⌠
⌡b
af(x) dx ⇔ Area under f(x) from a to b  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
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 usA_{triangle} = h·w 2= 8·4 2= 16
Since the total area is just made up of these two pieces, we add them together to find the area:A_{rectangle} = h·w = 1 ·4 = 4
Since ∫_{−3}^{1} −2x+3 dx just represents this area, we've found its value.A_{total} = 16 + 4 = 20  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
where F(x) is the antiderivative of f(x). Using it, we start by finding the antiderivative of f(x):⌠
⌡b
af(x) dx = F(b) − F(a),
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 = −x^{2} +3x ⎢
⎢1
−3⌠
⌡1
−3−2x+3 dx = −x^{2} +3x ⎢
⎢1
−3= ⎡
⎣−(1)^{2} +3(1) ⎤
⎦− ⎡
⎣−(−3)^{2} +3(−3) ⎤
⎦= ⎡
⎣−1 +3 ⎤
⎦− ⎡
⎣−9 −9 ⎤
⎦= ⎡
⎣2 ⎤
⎦− ⎡
⎣−18 ⎤
⎦
Thus the fundamental theorem of calculus gives the same value for the area, which confirms our previous answer.= 20

 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):
V_{0}, 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:
V_{total, approx} = 1250 + 1134 + 1018 + 902 + 786 + 670 + 554 + 438
Thus the approximate volume in the tank at the end of eight hours is 6752 liters.= 6752  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 equalwidth 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.
*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.
Answer
Area Under a Curve (Integrals)
Lecture Slides are screencaptured 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
 Introduction
 Idea of Area Under a Curve
 Approximation by Rectangles
 Various Methods for Choosing Rectangles
 Rectangle Method  LeftMost Point
 Rectangle Method  RightMost Point
 Rectangle Method  MidPoint
 Rectangle Method  Maximum (Upper Sum)
 Rectangle Method  Minimum
 Evaluating the Area Approximation
 More Rectangles, Better Approximation
 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
 If This Limit Exists, It Is Called the Integral From a to b
 The Process of Finding Integrals is Called Integration
 The Big Reveal
 The Big Reveal  Wait, Why?
 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
 Example 2
 Example 3
 Example 4
 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  LeftMost Point 5:12
 The LeftMost Point
 Rectangle Method  RightMost Point 5:58
 The RightMost Point
 Rectangle Method  MidPoint 6:42
 Horizontal MidPoint
 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 SubIntervals
 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
Precalculus with Limits Online Course
Transcription: Area Under a Curve (Integrals)
Hiwelcome back to Educator.com.0000
Today, we are going to talk about area under a curvealso 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 functionwe can take the integral of that, and we can get location out of it.0039
Why is that? Well, velocity tells us motionhow 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 termif we put all of the motion togetherif 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 stufflet'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 xaxis on the interval going from a to b?"0079
So, we can fill that in; and the shadedin 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 perfectlywhat 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 subintervals.0146
In this case, we have broken it into n = 4; here are 1, 2, 3, 4 equalwidth subintervals.0150
So, we go from a to b, and we break it into n equalwidth chunks.0158
In each one of these subintervals, we will base the height of each rectangle off of some x_{i} in each subinterval.0164
That is to say, we will choose some horizontal location in the subinterval; and then we will see what height that horizontal location goes to.0170
That horizontal location, as a pointwhat height is that point at?0178
And then, that is going to set the height of that rectangle for that subinterval.0183
So, for the i^{th} subinterval, we will use x_{i} to determine that height.0189
The i^{th} rectangle has a height of f evaluated at x_{i};0195
so x_{i} is the horizontal reference location that we use to determine the height for the i^{th} rectangle.0200
For example, in our third subinterval, we might look at x_{3}, the horizontal reference location for our third subinterval.0207
We check, and we see what height that ends up going to; and so, that would be f evaluated at x_{3}.0215
And then, that determines the height for our third rectangle.0222
We use x_{i} to determine the horizontal reference location for our i^{th} subinterval,0226
which then determines the height for our i^{th} rectangle.0232
We use this x_{i} to determine height by getting f(x_{i}) to tell us what that rectangle's height is.0238
Now, there is a variety of ways to choose our x_{i}.0244
In any given subinterval, 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 x_{i}, for each one of these subintervals:0268
because the x_{i}, that horizontal reference location, determines the height for that rectangle,0272
it is going to affect the height of the i^{th} rectangle,0278
because height is just f evaluated at whatever horizontal reference location we chose, our x_{i}.0281
By using a different method to choose our x_{i}, 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 x_{i} 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 subinterval?0320
We can see an example of this right here: what we do for our first subinterval0323
we look: where would the leftmost point of that subinterval 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 subinterval, 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 subinterval, 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 subinterval, what is the height that the rightmost horizontal location gets mapped to?"0361
For our first subinterval, 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 subinterval, 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 subinterval, we look at the rightmost location of that subinterval; 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 subinterval and the far right side of the subinterval.0404
But we could also ask what happens to the one in the middle.0409
For our first subinterval, 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 subinterval, we look at the midpoint of that subinterval.0422
We go up, and that tells us the height for our rectangle here.0426
For our third subinterval, 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 subinterval, and we see which one of these points gets mapped to the highest possible location.0457
We look over the entire subinterval, and we see which one is highest in this portion.0463
So, in this case, the highest location that we get for this subinterval is here; so that determines the height of the entire rectangle.0468
For our second subinterval, the highest possible location we reach in this subinterval is this location right here.0474
So, that determines the height of that entire second rectangle.0480
For our third subinterval, we see that the highest possible height we reach in that subinterval 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 subintervals, 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 subintervals.0533
We can look and see, for our first subinterval, 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 subintervals.0547
We can have our second subinterval; the lowest possible height in that second subinterval is right here,0551
so we end up evaluating that as the height of our second rectangle.0558
For our third subinterval, the lowest possible height is right here; so that gives us the height of our third rectangle.0561
And for our fourth and final subinterval, the lowest possible height over that subinterval 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 undercutting 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 subintervals.0615
That is how we did this, right from the beginning; we split it into n equalwidth subintervals.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 equalwidth subintervals, 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 i^{th} rectangle's height0651
depends on the specific x_{i} 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 x_{i}.0659
But whatever x_{i} we end up choosing will tell us the height; so just by definition,0661
our height is equal to f evaluated at the various x_{i} 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 i^{th} 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 x_{i} we chose, so our height right here.0680
So, the area of our i^{th} rectangle is equal to the width, b  a over n, times the height of that i^{th} rectangle,0689
which is going to be f evaluated at x_{i}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 n^{th} rectangle, which is our last rectangle,0713
since we are using a total of n subintervals; 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(x_{i}).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 nvalue 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 x_{i}.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(x_{i}).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 antiderivative 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 antiderivative, 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 antiderivative, 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)dx0941
that is, the area underneath the curve from a to b of some f(x) function0950
is equal to the antiderivative of f, evaluated at b, minus the antiderivative 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 antiderivative, where we are talking about what the height is...the antiderivative 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 antiderivative of height; since height is the derivative of area,1083
that means that area must be the antiderivative 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 antiderivative of f,1095
we end up getting the antiderivative 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 about1130
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) = x^{2} from a = 0 to b = 3 by using three equalwidth rectangles1147
that is, n = 3on the lefthand 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) = x^{2}: 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 equalwidth 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 lefthand point of each interval.1179
Three equal chunks here...if it is the lefthand 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 lefthand 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 subintervals 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 subintervals).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 subinterval 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 lefthand point.1253
If our first rectangle...remember: it is going from 0 to 1; 0 to 1 is the subinterval it is evaluating.1257
The lefthand 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 x_{i}.1275
In this case, we are determining based on lefthand points.1287
So, 1 times f at 0...our f(x) equals x^{2}, 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 lefthand side.1305
For our second rectangle, once again, there is a width of 1, times the height (will be evaluated at the lefthand side here); it will be 1.1309
So, f evaluated at 1...1 times 1^{2} 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 lefthand from 2 to 3 is going to be 2;1324
1 times f(2) is 1 times 2^{2}, 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 togetherthe 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 subintervals and the lefthand 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) = x^{2} from a = 0 to b = 31371
by using four equalwidth 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...x^{2}...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 equalwidth rectangles (let's draw that in really quickly: 1, 2, 3, 4),1404
notice: the maximum point of each intervalwhere 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 righthand 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, righthand point and maximum can give us totally different rectangle pictures.1430
However, for something like f(x) = x^{2}, 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 subinterval,1442
the rightmost point will be at the highest height; so the rightmost point1446
is the same thing as maximum for the specific case of the function x^{2}.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 righthand 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 righthand point is the same thing as the maximum point in the specific case of the function x^{2}.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 subintervals 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 x_{i}.1541
And x_{i} 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 righthand point;1548
so we will always end up looking at the right side of each one of these subintervals.1551
OK, back to our first one: 3/4, our width, times the heightwhere does the height end up getting evaluated?1555
Well, if we are going from 0 to 0.75 (that is our first subinterval), we are going to end up looking at the most righthand 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 0.75^{2}.1570
Since f is just x^{2}, just a squaring function, it is 0.75 times 0.75^{2}.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 1.5^{2}: 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 2.25^{2};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 righthand side of our final interval is 3.1650
3/4 times f(3)...remember, the righthand 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 3^{2} 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 lefthand 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) = x^{2} 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 i^{2} = 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 x^{2}; 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 subintervals.1785
We are going to end up cutting it into n subintervals.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 x_{i}.1818
It is going to end up actually making our notation just a little bit easier if we choose rightmost; so we are just going to arbitrarily choose rightmost.1823
I want to point out to you, though, that it doesn't actually matter which one we choose: rightmost, leftmost, 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
Rightmost is nice, because it is easy to do it notationwise; 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 rightmost x_{i} for each one.1873
So, what will x_{i} end up being? Well, our first x_{i} 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 x_{i}.1888
So, in general, to get out to the i^{th} x_{i}, we start at a.1896
And then, if we are at the i^{th} x_{i}, 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 x_{i}, 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 rightmost of each side, is just going to be i, whatever subinterval we are at.1928
If we are in our first subinterval, we have taken one widthstep to get to the right side.1935
If we are in our tenth subinterval, we have taken ten widthsteps to get to the right side of that tenth subinterval.1938
In general, x_{i} 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 x_{i} = 0 + 3/n(i); so 3 over n times i gives us our x_{i} 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 x_{i} for each one, what will end up being the height for each one of our rectangles?1980
The height of the i^{th} rectangle is going to be f evaluated at x_{i}, 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(x_{i}).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 problemwe 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 x_{i}...well, what is f evaluated at x_{i}?2046
Well, f(x) = x^{2}; so if our x_{i} is 3/n times i (that was what we figured out that our x_{i} 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 distributeswe have 3^{2}/n^{2} times i^{2}.2086
We also have that 3/n there; so we simplify that to 27, over n^{3}, times i^{2}.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/n^{3} times i^{2}, 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/n^{3} 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 notationthe 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 changethe limit has to do something2158
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/n^{3}, 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 i^{2}...at this point, look: we have the sum as i = 1 goes off to n of i^{2}.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/n^{3};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/n^{3} 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 n^{2} + n, times 2n + 1, all over n^{3} times 6; great.2234
Limit as n goes off to infinity...we can finish expanding our factors there.2249
We have 27, times n^{2} times 2n (becomes 2n^{3}); n^{2} + 1...n^{2} times +1; n times 2n gets 2n^{2}...2253
So, we have a total of +3n^{2}; and n times 1 gets us + n, all over 6n^{3}.2264
Limit as n goes off to infinity...we have 27 times 2n^{3}; that comes out to be 54n^{3};2274
plus 3 times n^{2}...that comes out to be 81n^{2}; plus 27n, all divided by 6n^{3}.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 54n^{3}, over 6n^{3}, plus 81n^{2}, plus 27n, all over 6n^{3}.2310
Now, this portion right here, the n^{3} and the n^{3}, 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 n^{3} on the bottom; that is a 3 degree on the bottom; but on the top, we only have n^{2} and n.2343
So, that is a 2 degree and a 1 degree; those can't compete with an n^{3} 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 9pretty 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 antiderivative 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 antiderivative of f evaluated at b,2413
minus the antiderivative 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) = x^{2} 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 antiderivative of f(x).2441
So, the very first step that we need to figure out here is what the antiderivative is to f(x).2452
f(x) = x^{2}; 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) = x^{5}, 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 5x^{4}.2481
There is this nice, easy rule for taking derivatives with the power rule.2489
If we have x^{2}, 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 antiderivative 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 x^{3}; if we took the derivative of this,2534
we would bring the 3 down, and we would get 3 times x^{3  1}, or 3x^{2}.2539
But there is this problem here: it is not 3x^{2}; it is just plain x^{2}.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 antiderivative to it.2555
Let's check and make sure that that works: let's check: if F(x) is equal to x^{3}/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 x^{2}, which is what we started with; that checks outgreat.2584
So now, we have figured out what our antiderivative 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 antiderivative of our function evaluated at the far end, minus the antiderivative at our starting location.2608
The ending location, minus the starting locationwhat is our thing?2616
It is x^{3}/3; that is what our F(x) is; so we have 3^{3}/3  0^{3}/3.2619
0^{3}/3 just disappears; so we have 3^{3}/3; 3 cancels this into a 2;2630
the 3 on the bottom cancels our top exponent down by 1; we have just 3^{2}; 3^{2} is 9nice.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 ishow 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 doinggoodbye!2723
1 answer
Last reply by: Professor SelhorstJones
Sun Mar 8, 2015 9:10 PM
Post by Jamal Tischler on March 8, 2015
If 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 answer
Last reply by: Professor SelhorstJones
Mon Aug 26, 2013 11:09 PM
Post by Richard Gregory on August 22, 2013
Thank 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.