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Q.
Area lying in the first quadrant and bounded by the curve $y = x^3$ and the line $y = 4x$ is
Application of Integrals
Solution:
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^{2}_{0} (4x - x^3) dx$
$ = 2 \left[\frac{4x^{2}}{2} - \frac{x^{4}}{4}\right]^{2}_{0} = 2\left[\frac{16}{2} - \frac{16}{4}\right] =8$