Question

Area lying in the first quadrant and bounded by the curve $y=x^3$ and the line $y=4x$ is

• A. 2
• B. 3
• C. 4
• D. 8

Answer 1

Answer: (D)

Solving the equation of the given curves for x, we get x = 0, 2, - 2
The required area is symmetrical about the origin.
So, Reqd. area $=2{\int }_0^2(4x-x^3)dx$
$=2{\left[\frac{4x^2}{2}-\frac{x^4}{4}\right]}_0^2=2\left[\frac{16}{2}-\frac{16}{4}\right]=8$