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My question is from Apostol's Vol. 1: One-variable calculus with introduction to linear algebra textbook.

Page 94. Exercise 17. We have defined $\pi$ to be the area of a unit circular disk. In Example 3 of Section 2.3, we proved that $\pi=2\int_{-1}^1\sqrt{1-x^2}\mathrm dx$. Use properties of the integral to compute the following in terms of $\pi$:

$(a)\int_{-3}^3\sqrt{9-x^2}\mathrm dx;\qquad$ $(b)\int_0^2\sqrt{1-\frac{1}{4}x^2}\mathrm dx;\qquad$ $(c)\int_{-2}^2(x-3)\sqrt{4-x^2}\mathrm dx.$

My attempt at a solution. I solved (a) and (b), the answers were $\frac{9}{2}\pi$ and $\frac{\pi}{2}$. But I have trouble solving the third one. for (c) we have $$\int_{-2}^2(x-3)\sqrt{4-x^2}\mathrm dx=2\int_{-1}^1(2x-3)\sqrt{4-4x^2}\mathrm dx$$ $$=2\int_{-1}^12(2x-3)\sqrt{1-x^2}\mathrm dx=4\int_{-1}^1(2x-3)\sqrt{1-x^2}\mathrm dx.$$ Now as I understand, this is definite integral of multiple of two functions, which can be solved by integration by parts. But since I haven't learned that yet, I want to ask how I can compute this in terms of $\pi$?

Community
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George Apriashvili
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3 Answers3

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Your last integral

$$4\int_{-1}^{1} (2x-3)\sqrt{1-x^2}\;dx$$

Can be split into the two integrals

$$4\int_{-1}^{1} 2x\sqrt{1-x^2}\;dx + 4\int_{-1}^{1}(-3)\sqrt{1-x^2}\;dx$$

The second integral you can compute in terms of $\pi$, and the first integral can be computed via $u$-substitution.

  • I did that before, and that's where I stopped. I will read on $u$-substitution now to understand how to proceed. Thanks. – George Apriashvili Jan 31 '15 at 22:08
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    Actually, looking over the first integral again, you can actually completely avoid $u$-substitution: $x\sqrt{1-x^2}$ is an odd function, so $$\int_{-1}^{0} x\sqrt{1-x^2}; dx = -\int_{0}^{1} x\sqrt{1-x^2}; dx$$ and $$\int_{-1}^{1} x\sqrt{1-x^2}; dx =\int_{-1}^{0} x\sqrt{1-x^2}; dx + \int_{0}^{1} x\sqrt{1-x^2}; dx = 0$$

    (Of course, it's perfectly valid to use $u$-substitution, but I think the above argument is cleaner for the particular integral you want to evaluate).

     –  Jan 31 '15 at 22:12
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    That's so much better, yes. So we have $$\int_{-1}^{1} x\sqrt{1-x^2}\mathrm dx =\int_{-1}^{0} x\sqrt{1-x^2}\mathrm dx + \int_{0}^{1} x\sqrt{1-x^2}\mathrm dx = 0.$$ And main integral would be $$8\int_{-1}^{1} x\sqrt{1-x^2}\mathrm dx+4\int_{-1}^{1}(-3)\sqrt{1-x^2}\mathrm dx=-6*2\int_{-1}^1\sqrt{1-x^2}\mathrm dx=-6\pi.$$ – George Apriashvili Jan 31 '15 at 22:21
  • @GeorgeApriashvili Yep, looks good to me! –  Jan 31 '15 at 22:23
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$$\int_{-1}^1(2x-3)\sqrt{1-x^2}\mathrm dx=\int_{-1}^1 2x\sqrt{1-x^2}\mathrm dx-3\int_{-1}^1\sqrt{1-x^2}\mathrm dx=\int_{-1}^1 2x\sqrt{1-x^2}\mathrm dx-\frac{3 \pi}{2}$$

Liza
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The only new integral here is $$ \int\limits_{-1}^1 x \sqrt{1-x^2} dx = \left[-\frac{1}{3}(1-x^2)^{3/2}\right]_{-1}^1 = 0 $$ otherwise the known integral for $\pi$ shows up.

mvw
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