First slide

Evaluation of definite integrals

Question

$\int_{a}^{b} \sqrt{(x - a) (b - x)} d x (b > a)$ is equal to

Moderate

A

$π (b - a)^{2} / 8$
B

$π (b + a)^{2} / 8$
C

$π (b - a)^{2}$
D

$π (b + a)^{2}$

Solution

Let $t = \frac{x - a + x - b}{2} = x - \frac{1}{2} (a + b)$
and $α = \frac{b - a}{2}$ , sot that
$\begin{array}{l} \int_{a}^{b} \sqrt{(x - a) (b - x)} d x = \int_{- α}^{α} \sqrt{(t + α) (α - t)} d t \\ = 2 \int_{0}^{α} \sqrt{α^{2} - t^{2}} d t \\ {= t \sqrt{α^{2} - t^{2}} + α^{2} \sin^{- 1} (\frac{t}{α})]}_{0}^{α} \\ = \frac{π α^{2}}{2} = \frac{π (b - a)^{2}}{8} . \end{array}$

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