JEE Challenger
More from Application of Integrals
MathMedAdv 2008 • Paper 1

Area Bounded by Curve and Coordinate Axes

The area of the region bounded by the curve y=f(x)y = f(x), the xx-axis, and the lines x=ax = a and x=bx = b, where −∞<a<b<−2-\infty < a < b < -2, is

Options

A

∫abx3((f(x))2−1)dx+bf(b)−af(a)\int_a^b \frac{x}{3\left((f(x))^2 - 1\right)} dx + bf(b) - af(a)

Correct
B

−∫abx3((f(x))2−1)dx+bf(b)−af(a)-\int_a^b \frac{x}{3\left((f(x))^2 - 1\right)} dx + bf(b) - af(a)

C

∫abx3((f(x))2−1)dx−bf(b)+af(a)\int_a^b \frac{x}{3\left((f(x))^2 - 1\right)} dx - bf(b) + af(a)

D

−∫abx3((f(x))2−1)dx−bf(b)+af(a)-\int_a^b \frac{x}{3\left((f(x))^2 - 1\right)} dx - bf(b) + af(a)

Topics & Concepts

Application of IntegralsArea Under CurvesDefinite Integration

Step-by-Step Solution

To find the area of the region bounded by the curve y=f(x)y = f(x), the xx-axis, and the vertical lines x=ax = a and x=bx = b where −∞<a<b<−2-\infty < a < b < -2, we start from the fundamental definition of the area under a curve:

Area=∫aby dx=∫abf(x) dx\text{Area} = \int_a^b y \, dx = \int_a^b f(x) \, dx

Using integration by parts on ∫abf(x) dx\int_a^b f(x) \, dx:
∫abf(x) dx=[xf(x)]ab−∫abxf′(x) dx\int_a^b f(x) \, dx = \left[ x f(x) \right]_a^b - \int_a^b x f'(x) \, dx
=bf(b)−af(a)−∫abxf′(x) dx= b f(b) - a f(a) - \int_a^b x f'(x) \, dx

From the given function's defining relation, we have:
f′(x)=−13((f(x))2−1)f'(x) = -\frac{1}{3\left((f(x))^2 - 1\right)}

Substituting this expression for f′(x)f'(x) into the integral:
−∫abxf′(x) dx=−∫abx(−13((f(x))2−1))dx=∫abx3((f(x))2−1) dx-\int_a^b x f'(x) \, dx = -\int_a^b x \left( -\frac{1}{3\left((f(x))^2 - 1\right)} \right) dx = \int_a^b \frac{x}{3\left((f(x))^2 - 1\right)} \, dx

Thus, the area simplifies to:
Area=∫abx3((f(x))2−1) dx+bf(b)−af(a)\text{Area} = \int_a^b \frac{x}{3\left((f(x))^2 - 1\right)} \, dx + bf(b) - af(a)

This corresponds to Option (A).

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