Question

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

• A. $4:23$
• B. $4:27$
• C. $4:19$
• D. none of these

Answer 1

Answer: (A)

We have, $y=x^2-4x\dots \left(i\right)$
and $y=2x-x^2\dots \left(ii\right)$
Both are parabolas.
Solving $(i)$ and $(ii)$, we get
$x^2-4x=2x-x^2$
$\Rightarrow x=0$ or $x=3$

$A_1=\underset{0}{\overset{2}{\int }}\left(2x-x^2\right)dx=\frac{4}{3}$
$A_2=\underset{0}{\overset{2}{\int }}\left(x^2-4x\right)dx+\underset{2}{\overset{3}{\int }}\left[\left(2x-x^2\right)-\left(x^2-4x\right)\right]dx$
$=\frac{x^3}{3}-\frac{4x^2}{2}{|}_0^2+\underset{2}{\overset{3}{\int }}\left(6x-2x^2\right)dx$
$=\left|\frac{8-24}{3}\right|+{\left|\frac{6x^2}{2}-\frac{2x^3}{3}\right|}_2^3$

$=\left|\frac{-16}{3}\right|+\left[\left(27-18\right)-\left(12-\frac{16}{3}\right)\right]=\frac{16}{3}+\frac{7}{3}=\frac{23}{3}$

$\therefore$ Required ratio$=\frac{A_1}{A_2}=\frac{4}{3}:\frac{23}{3}$
$=4:23$