First slide

Evaluation of definite integrals

Question

$\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} dx$ is equal to

Easy

A

1/2
B

3/2
C

5/2
D

0

Solution

$\begin{array}{l} \int_{2}^{7} \frac{\sqrt{x} dx}{\sqrt{x} + \sqrt{9 - x}} - - - (i) \\ I = \int_{2}^{7} \frac{\sqrt{9 - x}}{\sqrt{9 - x} + \sqrt{9 - (9 - x)}} dx \\ I = \int_{2}^{7} \frac{\sqrt{9 - x}}{\sqrt{9 - x} + \sqrt{x}} dx - - - - (ii) \end{array}$
On adding Eqs. (i) and (ii), we get
$\begin{array}{l} 2 I = \int_{2}^{7} (\frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} + \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}}) dx \\ = \int_{2}^{7} dx = | x |_{2}^{7} = 5 \\ so, I = \frac{5}{2} \end{array}$

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