First slide

Introduction to definite integration

Question

$\int_{0}^{2} \sqrt{\frac{2 + x}{2 - x}} d x$ is equal to

Moderate

A

$\frac{π}{2} + 1$
B

$π + \frac{3}{2}$
C

$π + 1$
D

None of these

Solution

$\int_{0}^{2} \sqrt{\frac{2 + x}{2 - x}} d x = \int_{0}^{2} \frac{2 + x}{\sqrt{4 - x^{2}}} d x$
$= 2 \int_{0}^{2} \frac{1}{\sqrt{4 - x^{2}}} d x + \int_{0}^{2} \frac{x d x}{\sqrt{4 - x^{2}}}$

$= 2 {[\sin^{- 1} (\frac{x}{2})]}_{0}^{2} - \frac{1}{2} \int_{0}^{2} \frac{- 2 x}{\sqrt{4 - x^{2}}} d x$

$= 2 (\frac{π}{2}) - \frac{1}{2} {[\frac{\sqrt{4 - x^{2}}}{1 / 2}]}_{0}^{2}$

$= π - (0 - 2) = π + 2$

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