WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP Calculus.
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Today, we are going to be doing some example problems for the fundamental theorem of calculus that we just covered.
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These are going to be extraordinarily straightforward and simple, almost surprisingly.
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Our first example is find the derivative of the following function.
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The function is g(x) and it is defined as the integral from 1 to x, whatever x might be.
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5/ t³ + 9 dt, nice and simple.
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Since the upper limit is x, we would literally just put x where we have the t.
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Our final answer is just 5/ x³ + 9, and we are done.
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That is the beauty of the fundamental theorem.
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When a function is defined as an integral or the upper integral is a variable,
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this particular variable right here xx, we just literally put that in there.
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Let me correct this, a little bit of a notational issue here.
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g’ because we are looking for the derivative.
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g’(x), that is a little more clear, I think.
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We have 5/ x³ + 9.
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Literally, when we take the derivative of this, we are just getting rid of this integral sign
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because differentiation and integration or anti-differentiation are inverse processes.
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That is the whole idea.
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That is it, nice, straightforward, and simple.
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Find the derivative of the following function.
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g(x), upper variable is x, we do the same thing.
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Therefore, the derivative of this g’(x), all we have to do is we put an x where we see a t.
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We get x² sin³ x, that is it, very nice.
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As long as this upper limit of integration is the variable that is inside the function defining it,
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we can just put it straight into the function itself.
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Let us try this one, find the derivative of the following function.
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g(x) = the integral from 1 to x⁴ sin(t) dt.
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Now this is x, this time, we have not x but we have a function of x.
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The only thing we do know is we do the same thing.
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We are just going to be putting in the x here.
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g’(x) is equal to sin(x).
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But because this is not x but it is a function of x, by the chain rule,
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we are just going to multiply by the derivative of this function.
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Let us write that out.
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When the upper limit is not just x but a function of x, some more complicated function of x, we do the following.
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This is the long process, the easy part is just multiply by the derivative of the upper limit, whatever the function happens to be.
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Essentially, what you are doing is you are letting x⁴, the upper limit integration equals some letter u.
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Essentially, you are just doing a substitution.
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And then, du dx is going to equal 4x³.
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And then, you rewrite this g(x) = the integral from 1 to u of sin(t) dt.
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Then, when you do this, when you put this into here, in other words, when you do g’(x) is equal to the sin(u),
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because u is a function of x, by the chain rule, the derivative is sin(u) du dx.
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You are multiplying by du dx.
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What you end up getting is g’(x) is equal to sin(x)⁴ × 4x³.
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This part is exactly the same, you are just putting the upper limit into this variable to get the sin x⁴.
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And then, you are just multiplying by the derivative of that because now it is a function of x, it is not just x itself.
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I hope that makes sense.
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Essentially, what you are doing is you are doing the same thing as you did before.
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If this were just an x, you would put it in and it would be sin(x).
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The derivative of x is just 1, you were technically doing du dx.
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The derivative of x is just 1, which is why it just ends up staying that way.
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But if it is a function, just multiply by its derivative.
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That is your final answer, right there, very straightforward.
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Find the derivative of the following functions, same thing.
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Now π/3, the upper limit is cos(x).
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We are just going to put this into the t's and then multiply everything by the derivative of cos(x).
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g’(x), we would literally just read it off.
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It is going to be 3 × cos(x) + 7 × cos² (x) × the derivative of that function cos(x) which is –sin(x).
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We are done, that is all.
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Just do what the first fundamental theorem of calculus says which is just put the upper limit into those things.
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Then, multiply by ddx of the upper limit.
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That is all, moving along nicely.
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Evaluate the following integral.
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We want to evaluate a definite integral.
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We have a lower limit, an upper limit, they are both numbers.
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We are going to find the antiderivative, the integral.
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And then, we are going to put in 5, we are going to put in 1,
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and we are going to subtract the value of what we get from 1, from the value of the upper limit 5.
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Let us go ahead and do that.
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The integral of this, let us go ahead and actually multiply this out first because we have x³ - 4x².
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This is going to be the integral from 1 to 5 of x⁶ – 4x × that is going to be 4x⁴, then another 4x⁴.
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It is -8x⁴ + 16x², I just multiply this out, dx.
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And that is going to equal x⁷/7 - 8x⁵/5 + 16x³/3.
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I'm going to evaluate this from 1 to 5.
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In other words, I’m going to put 5 in for all of these and get a number, then I'm going to subtract by 1.
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Putting them into all the x here.
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When I put 5 into this, I get 78,000, 5⁷ is 78,125/7.
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When I put 5 into here, multiply by 8 and divide by 5.
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I get - 2500/5, I hope you are going to be confirming my arithmetic.
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I’m notorious for arithmetic mistakes.
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+ 2000/3 that takes care of the 5.
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I’m going to subtract by putting 1 in.
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This is going to be 1/7 - 8/5 + 16/3.
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Find the antiderivative of the integrand, and then evaluate.
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Put the upper number into this, you are going to get a value.
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Put the lower number into the x values, you are going to get this and then you subtract.
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The rest is just arithmetic.
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I'm just going to go ahead and leave it like this.
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You can go ahead and do the arithmetic, if you want, to get a single answer.
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Evaluate the following integral, once again, we are using the second fundamental theorem of calculus.
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The integrals of f from 2 to 8 is going to be the f(8) - f(2), F(8) – F(2), that is what the second fundamental theorem says.
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Remember, it is this one, if from a to b of f(x) dx is just equal to f(b) – f(a), where F is the antiderivative of f.
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This is going to be the integral from 2 to 8, I’m going to separate these out.
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I’m going to write this as x²/3, it is going to be x/ x²/3 - 3/ x²/3.
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x/ x²/3 dx - the integral from 2 to 8 of 3/ x²/3 dx.
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That is going to equal, the integral from 2 to 8 of x¹/3 dx - the integral from 2 to 8 of 3 × x⁻²/3 dx.
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I can go ahead and integrate these, not a problem.
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1/3 + 1 is 4/3, I get x⁴/3/ 4/3 - 3 × x – 2/3 + 1 is 1/3/ 1/3.
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That is going to equal ¾ x⁴/3 - 9x¹/3.
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I'm going to evaluate that from 2 to 8.
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When I put 8 in for here, I'm going to get 12.
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When I put 8 into here, it is going to be -18, that is the first one.
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-, when I put 2 into here, I end up with 1.89.
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And then, - when I put 2 into here, I end up with 11.34.
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There is your answer.
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Once you have the antiderivative, you put the upper limit into the x value to evaluate, that gives you the first term.
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You put the lower limit into the x, that gives you the second term.
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You subtract the second term from the first term, f(b) - f(a).
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Nice and straightforward.
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Another good example, the integral from π/6 to 2π/ 3 of cosθ dθ.
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This is going to equal where the integral of cosθ dθ is going to be sin(θ).
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We are going to evaluate sin(θ) from π/6 to π/3.
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Not a problem, that is going to be the sin(π/3) - sin(π/6).
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We are just putting in upper and lower limit into the actual variable of integration.
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The sin(π/3) is √3/2 – sin(π/6) which is ½.
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The integral from 1 to 3 of 5/ x² + 1.
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You should recognize that the integral of 1/ x² + 1 is equal to the inv tan.
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Remember, the integral or the antiderivative, whatever you want to call it,
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of 1 + x² dx that was equal to the inv tan(x).
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Therefore, this is, that is fine, I will go ahead and separate it out.
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This is just a constant, it is 5 integral from 1 to 3, 1/ x² + 1 dx = 5 × inv tan(x), evaluated from 1 to 3.
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That is it, that is all it is.
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This is just 5 × inv tan(3) – inv tan(1).
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That is all, you can go ahead and multiply the 5 through.
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You can do 5 or you can evaluate this from here to here.
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This - this and then multiply the constant through.
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The order actually does not matter.
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I’m just somebody who tends to like to see the constant on the outside, personal choice.
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g(x) is equal to the integral from 0 to x of f(t) dt.
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f being the function whose graph is given below.
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This graph right here, this is a function of f.
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Answer the following questions.
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We take a look at this graph, it looks like some sort of a modified sin or cos graph, definitely periodic.
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We are taking a look real quickly, we notice that we have about a 3. something, then we have a 6.2 something.
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It turns out that these are actually just π/2, π.
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I’m going to go ahead and label them that way.
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I left the labels just as is, but it is nice to recognize the numbers for what it is that they are.
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Especially, since you are talking about a periodic function, then more than likely the graphs that you are going to be dealing with
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are going to pass through the major points, π/2, π, π/6, π/3, π/4, things like that.
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In this case, let me mark these off.
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3.14, this is π.
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This over here looks like it is π/2.
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Over here, this looks like 2π and this looks like 3π/ 2, 4π/ 2.
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It looks like we have got 5π/ 2.
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At least that way we have some numbers that we are accustomed to seeing,
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instead of just referring them as 3.1, 1.5, things like that.
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Give the values of x for which g(x) achieves its local maxes and mins.
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You have to be very careful here, this is a graph of f.
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g is the integral of f from 0 to x.
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Therefore, g is actually the area under the graph.
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The graph actually falls below the x axis, the area is going to become negative.
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Give the values of x for which g(x) achieves its local maxes and mins.
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As we start moving from 0 x, moving forward g(x), as x gets bigger and bigger,
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this value is just the area under the graph itself.
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It is going to hit a high point, when I hit π/2.
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And then, at π/2, because the graph itself falls below the axis, as I keep going,
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as x keeps getting bigger and bigger, the integral itself is going to be negative.
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The area is going to hit a maximum and then the integral is going to start to go negative.
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The area is going to diminish.
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At this point, the area is going to go down.
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At some point, it is going to hit 0.
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The area is going to rise to a certain number, that is up to here.
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We are going to start subtracting from the area, at some point it is going to hit 0.
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Over here, it is going to hit a minimum because now this area under the x axis is a lot bigger than the area above the x axis.
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The integral from 0 to π is going to be a negative number.
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It is actually going to hit a maximum.
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One of the maximums that it is going to hit is right there.
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And then, it is going to hit a minimum right there.
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After that, it is going to climb again, the area under the graph.
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That is what we are doing, g(x) is the area of f, the area under the graph of f because this is the graph of f.
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The area is going to get bigger again.
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At this point, it is going to hit another maximum.
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And then at this point, the area is going to be negative.
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We are going to hit another minimum, 2π.
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We are going to increase the area again.
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It is going to pass this and it is going to hit another maximum.
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That is what is going to happen.
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Give the values of x for which g(x) achieves its local maxes and mins.
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It is going to achieve its local maxes and mins where the area under the graph hits a maximum and a minimum.
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It is a maximum at π/2, a minimum at π.
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Maximum at 3π/ 2, a local minimum at 2π, local max and min at 5π/ 2.
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We have π/2, π, 3π/ 2, 2π, and 5π/ 2.
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The reason is because g(x) is the net area under the graph of f.
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This red is the graph of f.
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Where does g(x) achieves its absolute max, between 0 and 8, this point right here.
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Let us go ahead and draw this out.
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Basically, what we are going to have is this.
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Let me do this in black.
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g(x) is going to look like this, it is going to rise.
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It is going to hit a maximum here and it is going to hit a minimum.
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It is going to start rising again.
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It is going to hit a maximum and it is going to hit a minimum, and it is going to hit a maximum.
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It is going to start coming down to there.
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The area keeps getting bigger and bigger.
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The absolute value, the area keeps getting bigger and bigger.
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Our highest point is going to be this point right here, which is going to be our 5π/ 2.
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Again, because g(x), that is what we want, where does g(x) achieve its absolute max.
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g(x) is the net area under the graph.
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Positive, if it is above the x axis, negative if it is below.
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This is the graph of g.
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On what intervals is g(x) concave up or concave down?
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For concave up or concave down, we want to see the sin(g”).
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g’ is equal to f.
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By the fundamental theorem of calculus, if I take the derivative of this, I get rid of this.
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I just have f(x), this is g’ that mean g“ is equal to f’.
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f’ is just the derivative of the red graph.
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The derivative is the slope.
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It is going to be concave up where the slope is positive.
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It is going to be concave down where the slope is negative.
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If the slope is positive, that is f’.
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f’ happens to be g”.
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Therefore, we have from here to here, is concave up.
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From here to here, it is concave down.
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From here to here, concave up.
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From here to here, concave down.
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From here to here, concave up.
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And from here to 8, we got concave down which is exactly what we see.
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Concave up graph, this is g that we are looking at.
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Now it is concave down, it starts to be concave up again up to here.
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From here to here, it is concave down, it matches.
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I hope that makes sense.
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You are given a graph f, if g is defined as the integral from 0 to x of f.
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It is the net area under that particular graph.
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I hope that made sense.
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Let us take a look at what it actually looks like.
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The original function f was this one right here.
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This one, this is the actual g(x).
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This was our f(x).
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It is a maximum here, it is a minimum here.
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Local, hits a local max here, it is right there.
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It is a local min here, it is right there.
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Hits a local max here, that is right there.
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It is exactly what we said before.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.