Question

In what ratio does the $x-$axis divide the area of the region bounded by the parabolas $y=4x-x^2$ and $y=x^2-x$ ?

• A. $4:121$
• B. $4:131$
• C. $121:4$
• D. None of these

Answer 1

The correct option is C $121:4$

Given parabolas are $y=4x-x^2$

and $y=-(x-2{)}^2+4$

or $(x-2{)}^2=-(y-4)$

Therefore, it is a vertically downward parabola with vertex at $(2,4)$ and its axis is $x=2$

And $y=x^2-x$

$\Rightarrow y={(x-\frac{1}{2})}^2-\frac{1}{4}$

$\Rightarrow {(x-\frac{1}{2})}^2=y+\frac{1}{4}$

This is a opening upward parabola having its vertex at $(\frac{1}{2},-\frac{1}{4})$
and its axis is $x=\frac{1}{2}$

The points of intersection of given curves are

$4x-x^2=x^2-x\Rightarrow 2x^2=5x\Rightarrow x(2-5x)=0\Rightarrow x=0,\frac{5}{2}$

Also, $y=x^2-x$, meets $x$-axis at $(0,0)$ and $(1,0)$

$\therefore \text{Area},A_1={\int }_0^{5/2}[(4x-x^2)-(x^2-x)]dx={\int }_0^{5/2}(5x-2x^2)dx={[\frac{5}{2}{x}^2-\frac{2}{3}{x}^3]}_0^{5/2}=\frac{5}{2}{\left(\frac{5}{2}\right)}^2-\frac{2}{3}\cdot {\left(\frac{5}{2}\right)}^3=\frac{5}{2}\cdot \frac{25}{4}-\frac{2}{3}\cdot \frac{125}{8}$

$=\frac{125}{8}(1-\frac{2}{3})=\frac{125}{24}$

This area is considering above and below $x-$ axis both.

Now, for area below $x-$axis separately. We consider

$A_2=-{\int }_0^1(x^2-x)dx={[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

Therefore, net area above the $x-$axis

$A_1-A_2=\frac{125-4}{24}=\frac{121}{24}$

Hence, ratio of area above the $x-$axis and area below $x-$axis is

$=\frac{121}{24}:\frac{1}{6}=121:4$