Mathematics · Area under the Curves

JEE Main 2024 — 31 January, Shift 2 — Question 11

The area of the region enclosed by the parabola $y=4 x-x^{2}$ and $3 y=(x-4)^{2}$ is equal to :

Answer: C

Area $=\left|\int_{1}^{4}\left[\left(4 x-x^{2}\right)-\frac{(x-4)^{2}}{3}\right]\right| d x$

Area $=\left|\frac{4 x^{2}}{2}-\frac{x^{3}}{3}-\frac{(x-4)^{3}}{9}\right|_{1}^{4}$

$=\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right|$

$\Rightarrow(27-21)=6$

Answer key and solution verified before publishing.

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