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Q.
The ratio in which the $x$-axis divides the area of the region bounded by the curves $y = x^{2} - 4x$ and $y= 2x - x^{2}$ is
Application of Integrals
Solution:
We have, $y=x^{2}-4x \quad\ldots\left(i\right)$
and $y = 2x - x^{2}\quad\ldots\left(ii\right)$
Both are parabolas.
Solving $(i)$ and $(ii)$, we get
$x^{2}-4x=2x-x^{2}\,$
$\Rightarrow \quad x = 0$ or $x = 3$
$A_{1}=\int\limits_{0}^{2}\left(2x-x^{2}\right)dx=\frac{4}{3}$
$A_{2}=\int\limits_{0}^{2}\left(x^{2}-4x\right)dx+\int\limits_{2}^{3}\left[\left(2x-x^{2}\right)-\left(x^{2}-4x\right)\right]dx$
$=\frac{x^{3}}{3}-\frac{4x^{2}}{2}\bigg|_{0}^{2}+\int\limits_{2}^{3}\left(6x-2x^{2}\right)dx$
$=\left|\frac{8-24}{3}\right|+\left|\frac{6x^{2}}{2}-\frac{2x^{3}}{3}\right|_{2}^{3}$
