JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Definite Integrals

Let

f : [- 1, 2] \to [0, \infty)

be a continuous function such that

f (x) = f (1 - x)

for all

x \in [- 1, 2]

. Let

R_{1} = \int_{- 1}^{2} x f (x) d x

, and

R_{2}

be the area of the region bounded by

y = f (x), x = - 1, x = 2

, and the

x

-axis. Then

Answer: (C)

Solution

Using the property

\int_{a}^{b} g (x) d x = \int_{a}^{b} g (a + b - x) d x

R_{1} = \int_{- 1}^{2} x f (x) d x = \int_{- 1}^{2} (1 - x) f (1 - x) d x = \int_{- 1}^{2} (1 - x) f (x) d x

Adding these two expressions:

2 R_{1} = \int_{- 1}^{2} [x + (1 - x)] f (x) d x = \int_{- 1}^{2} f (x) d x = R_{2}

Therefore,

2 R_{1} = R_{2}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option