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Integral Calculus

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Question : 71 of 73

Marks: +1, -0

\int\limits_{0}^{2} \; \frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}} dx

is equal to

$-1$
$0$
$2$
$1$

Solution:

Concept:

Use the substitution

x \to 2 - x

property of definite integrals.

Explanation:

Let

I = \int\limits_{0}^{2} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}} \, dx

.

Using the property

\int_0^a f(x) dx = \int_0^a f(a-x) dx

, we get:

I = \int_0^2 \frac{\sqrt{2-x}}{\sqrt{2-x} + \sqrt{x}} \, dx

.

Add the two expressions for

I

:

2I = \int_0^2 \frac{\sqrt{x} + \sqrt{2-x}}{\sqrt{x} + \sqrt{2-x}} \, dx = \int_0^2 1 \, dx = 2

.

Thus,

I = 1

.

Answer:

The integral equals

1

, which corresponds to option D.

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