First slide

Evaluation of definite integrals

Question

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

Easy

A

2
B

1
C

$\frac{1}{2}$
D

$\frac{3}{2}$

Solution

$\begin{array}{l} I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9 - x} + \sqrt{x}} d x - - - - (1) \\ I = \int_{3}^{6} \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} d x - - - - (2) \\ (1) + (2) \Rightarrow 2 I = \int_{3}^{6} d x = (x)_{3}^{6} = 6 - 3 = 3 \\ \Rightarrow I = \frac{3}{2} \end{array}$

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