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Integral Calculus

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Question : 70 of 73

Marks: +1, -0

\int\limits_{0}^{2a} \; \frac{f(x)}{f(x)+f(2a-x)} dx

a
$-a$
0
$2a$

Solution:

Concept:

Use the property

\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx

to find the value of the given integral.

Explanation:

Let

I = \int_{0}^{2a} \frac{f(x)}{f(x) + f(2a - x)} dx

.

Substitute

x = 2a - t

, then

dx = -dt

, limits:

x=0 \to t=2a

,

x=2a \to t=0

.

Thus

I = \int_{2a}^{0} \frac{f(2a - t)}{f(2a - t) + f(t)} (-dt) = \int_{0}^{2a} \frac{f(2a - t)}{f(t) + f(2a - t)} dt

.

Since the variable is dummy,

I = \int_{0}^{2a} \frac{f(2a - x)}{f(x) + f(2a - x)} dx

.

Add the two expressions for

I

:

2I = \int_{0}^{2a} \frac{f(x) + f(2a - x)}{f(x) + f(2a - x)} dx = \int_{0}^{2a} 1 \, dx = 2a

.

Hence

I = a

.

Answer:

a

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