Calculating area between curves AP Calculus AB free response 2011 3b
About this video
This video solves part (b) of the 2011 AP Calculus AB free response #3: the area of region R between g(x) = sin(pi x) on top and f(x) = 8x^3 below, from x = 0 to x = 1/2. It finds antiderivatives by sight, -(1/pi)cos(pi x) and 2x^4, and evaluates to get an exact area of 1/pi - 1/8 without a calculator.
You'll be able to
- set up the area between two curves as the integral of (top function - bottom function).
- find the antiderivative of sin(pi x) using the chain rule in reverse.
- evaluate a definite integral exactly, using values like cos(pi/2) = 0 and cos(0) = 1.
Key words
- area between curves
- The integral of the top function minus the bottom function over the interval where they enclose a region.
- antiderivative
- A function whose derivative is the given function.
- chain rule
- The rule for differentiating a function inside another function: multiply by the derivative of the inside.
- u-substitution
- A method for integrating by renaming an inner expression, the reverse of the chain rule.
Transcript
- [Instructor] Now part B, find the area of R. So R is this region right over here, is this region right over here. It's bounded above by G of X, which they've defined as sine of pi X. It's bounded below, it's bounded below by F of X or 8X to the third. So the area is going to be, actually, let me just do it.
I'll scroll down a little bit. I still want to be able to see this graph right over here. Part B, part B, the area of R is going to be equal to the integral from zero, that's this point of intersection right over here, to one half. So let me make it clear. This is zero to one half, zero to one half.
And then the function on the top, so we could just take the area of that, but then we're gonna have to subtract from that the area underneath the function below. Or one way to think about it is the integral from zero to one half of the top function is G of X, which is sine of pi X.
So sine of pi X, sine of pi X. But if we just evaluated this integral, let me just put a DX over here. If we just evaluated this, we would get the area of this entire region. But what we need to do is subtract out the area underneath the second, underneath F of X.
We need to subtract out the area under that. So we just subtract from that F of X. And F of X we already saw is 8X to the third, is 8X to the third power. And now we could just evaluate this. So let me draw a little line here. It's getting a little bit messy. It's getting a little bit messy, I'll just do it down here.
So we need to take the anti-derivative of sine of pi X. Well, the derivative of cosine of X is negative sine X. The derivative of cosine of pi X is negative pi cosine of pi X. So the anti-derivative of sine of pi X is going to be negative one over pi cosine, cosine of Pi X.
And you can verify it for yourself and you might say, wait, so how did you know it was a negative? Well, I put the negative there so that when I take the derivative of the cosine of pi X, I would get a negative sign, but that negative will cancel out the negative to give me a positive here.
And you said, why did you put a one over pi here? Well, when you take the derivative of this thing using the chain rule, you take the derivative of the pi X, you'll get pi that you would multiply everything by, and then you would get negative sine of pi X. And that pi doesn't show up here, so I need something for it to cancel out with.
And that's what this one over pi is going to cancel out with. And you could do U substitution and all the rest if you found something like that useful. But it's in general a good habit, or I guess it's good to be able to do this almost by sight. And you can verify that this derivative is equal to sine of pi X.
So the anti-derivative of sine of pi X is this. The anti-derivative of negative 8X to the third power is negative eight. I'm going to divide it by four. So negative two X to the fourth power. And all I did is I incremented the three to a four and then I divided the eight by the four.
And you can take the derivative of this to verify that it is the same thing as negative 8X to the third power. And we're going to have to evaluate that from zero to one half. When you evaluated one half, so I'm going to get, I'm going to get negative one over pi, cosine of pi over two, cosine of pi over two minus two times one half to the fourth power is one-16th.
So that's it evaluated at one half. And then from that I'm going to subtract negative one over pi, cosine of zero. Let me just write it out. So minus negative one over pi, negative one over pi, cosine of pi times zero, cosine of pi, let me just write cosine of zero pi, I could write or pi times zero, minus two times zero to the fourth.
So that's just going to be minus zero. So let's evaluate this. So to simplify it, we have a cosine of pi over two. Cosine of pi over two is just going to be zero. So this whole thing just becomes zero. And then you have a negative two divided by 16, that's negative one-eighth, negative one-eighth.
And then from that I'm going to subtract this business here. Cosine of zero, this is one. So this is just a negative one over pi, a negative one over pi, and then I have a zero there so I can ignore that. So this is equal to negative one over eight plus plus one over pi, and we are done.
This part of it, you're not allowed to use a calculator. So this is about as far as I would expect them to get to for about as far as I would expect them to expect you to get. And so I'll leave you there. In the next video, we will do part C.
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