Area Under Curves — JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 Question with Solution

Consider the functions defined implicitly by the equation

y^{3} - 3 y + x = 0

on various intervals in the real line.
If

x \in (- \infty, - 2) \cup (2, \infty)

, the equation implicitly defines a unique real valued differentiable function

y = f (x)

.
If

x \in (- 2, 2)

, the equation implicitly defines a unique real valued differentiable function

y = g (x)

satisfying

g (0) = 0

.

The area of the region bounded by the curve

y = f (x)

, the

x

-axis, and the lines

x = a

and

x = b

, where

- \infty < a < b < - 2

, is

(A) $\int_{a}^{b} \frac{x}{3 ( ( f ( x ) ) ^{2} - 1 )} d x + b f (b) - a f (a)$ ✓ Correct
(B) $- \int_{a}^{b} \frac{x}{3 ( ( f ( x ) ) ^{2} - 1 )} d x + b f (b) - a f (a)$
(C) $\int_{a}^{b} \frac{x}{3 ( ( f ( x ) ) ^{2} - 1 )} d x - b f (b) + a f (a)$
(D) $- \int_{a}^{b} \frac{x}{3 ( ( f ( x ) ) ^{2} - 1 )} d x - b f (b) + a f (a)$

Answer: (A)

Solution

The area is:

A = \int_{a}^{b} f (x) d x

Using integration by parts with

u = f (x)

d v = d x

A = [x f (x)]_{a}^{b} - \int_{a}^{b} x f^{'} (x) d x = b f (b) - a f (a) - \int_{a}^{b} x f^{'} (x) d x

From the implicit differentiation:

f^{'} (x) = \frac{- 1}{3 ( f ( x ) ^{2} - 1 )}

Therefore:

\int_{a}^{b} x f^{'} (x) d x = \int_{a}^{b} x \cdot \frac{- 1}{3 ( f ( x ) ^{2} - 1 )} d x = - \int_{a}^{b} \frac{x}{3 ( f ( x ) ^{2} - 1 )} d x

Substituting back:

A = b f (b) - a f (a) - (- \int_{a}^{b} \frac{x}{3 ( f ( x ) ^{2} - 1 )} d x)

A = b f (b) - a f (a) + \int_{a}^{b} \frac{x}{3 ( f ( x ) ^{2} - 1 )} d x

Answer: (A)

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option