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Integral Calculus

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Question : 23 of 73

Marks: +1, -0

Value of

\int\limits_{2}^{5} \; \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} dx

is

$\frac{5}{2}$
$\frac{7}{2}$
$\frac{3}{2}$
none

Solution:

Concept:

Use the property

\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx

to simplify the integral.

Explanation:

Let

I = \int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} \, dx

.

Replace

x

by

7-x

(since

a+b=7

):

I = \int_{2}^{5} \frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}} \, dx

.

Add the two expressions for

I

:

2I = \int_{2}^{5} \left( \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} + \frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}} \right) dx = \int_{2}^{5} 1 \, dx = 5-2 = 3

.

Thus

I = \frac{3}{2}

.

Answer:

\frac{3}{2}

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