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Compute some integrals in terms of π

Recall we have defined \pi as the area of the unit disk. We have proved (Apostol, Section 2.3, Example 3) that

    \[ \pi = 2 \int_{-1}^1 \sqrt{1-x^2} \, dx. \]

Use this and the theorems on the property of integrals to compute the following in terms of \pi:

  1. \displaystyle{\int_{-3}^3 \sqrt{9-x^2} \, dx}.
  2. \displaystyle{\int_0^2 \sqrt{1-\frac{1}{4}x^2} \, dx}.
  3. \displaystyle{\int_{-2}^2 (x-3) \sqrt{4-x^2} \, dx}.

  1. Using the expansion/contraction property of the integral (Apostol, Theorem 1.19) we compute:

        \begin{align*}   \int_{-3}^3 \sqrt{9-x^2} \, dx &= 3 \int_{-1}^1 \sqrt{9 - (3x)^2} \, dx & (\text{Exp/Cont property}) \\   &= 9 \int_{-1}^1 \sqrt{1-x^2} \, dx & (\text{Factoring out a constant}) \\   &= \frac{9}{2} \pi &(\text{Expression for } \pi \text{ given}).  \end{align*}

  2. Using the expansion/contraction property of the integeral (Apostol, Theorem 1.19) we compute:

        \begin{align*}   \int_0^2 \sqrt{1 - \frac{1}{4}x^2} \, dx &= 2 \int_0^1 \sqrt{1 - x^2} \, dx \\   &= \int_{-1}^1 \sqrt{1-x^2} \, dx \\   &= \frac{\pi}{2}  \end{align*}

  3. Using the linearity properties of the integral, as well as, expansion/contraction and the formula for \pi given above. We also use this exercise (Apostol, Section 1.25, Exercise #26 part (b)) which establishes that the integral from -b to b of an odd function is 0.

        \begin{align*}  \int_{-2}^2 (x-3) \sqrt{4 - x^2} \, dx &= \int_{-2}^2 x \sqrt{4-x^2} \, dx - 3 \int_{-2}^2 \sqrt{4-x^2} \, dx \\  &= - 6 \int_{-1}^1 \sqrt{4-4x^2} \, dx & (x \sqrt{4-x^2} \text{ is an odd function})\\   &= -12 \int_{-1}^1 \sqrt{1-x^2} \, dx \\  &= -6 \pi. \end{align*}

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