First slide

Evaluation of definite integrals

Question

$\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9 - x} + \sqrt{x}} d x =$

Easy

A

$\frac{1}{2}$
B

$\frac{3}{2}$
C

3
D

1

Solution

Let $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9 - x} + \sqrt{x}} d x______(1)$
$I = \int_{3}^{6} \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} d x_____(2)$

$(1) + (2) \Rightarrow 2 I = \int_{3}^{6} 1 d x = 3$

$\Rightarrow I = \frac{3}{2}$

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