JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Definite Integrals

Let

f : [0, 2] \to R

be a function which is continuous on

[0, 2]

and is differentiable on ( 0,2 ) with

f (0) = 1

. Let

F (x) = \int_{0}^{x^{2}} f (t) d t

for

x \in [0, 2]

. If

F^{'} (x) = f^{'} (x)

for all

x \in (0, 2)

, then

F (2)

equals

Answer: (B)

Solution

By Leibniz rule:

F^{'} (x) = f (x^{2}) \cdot 2 x = f (x) \cdot 2 x

.

Given

F^{'} (x) = f^{'} (x)

, we have

2 x f (x) = f^{'} (x)

.

Separating variables:

\frac{f ^{'} ( x )}{f ( x )} = 2 x

.

Integrating:

ln f (x) = x^{2} + K

.

Using

f (0) = 1

K = 0

, so

f (x) = e^{x^{2}}

.

Therefore:

F (x) = \int_{0}^{x^{2}} e^{t} d t = e^{x^{2}} - 1

Thus

F (2) = e^{4} - 1

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option