Question

If y = u bisects the area bounded by region given by x2 < y < 2x then:

Answer 1

Consider the inequality,

$\longrightarrow x^2<2x$

$\longrightarrow x^2-2x<0$

$\longrightarrow x(x-2)<0$

$\Longrightarrow x\in(0,\ 2)$

The equality is satisfied at $x\in\{0,\ 2\}.$

Then the area is equal to,

$\displaystyle\longrightarrow A=\int\limits_0^2(2x-x^2)\ dx$

$\displaystyle\longrightarrow A=\left[x^2-\dfrac{x^3}{3}\right]_0^2$

$\displaystyle\longrightarrow A=\dfrac{4}{3}$

Point of intersection of $y=u$ and $y=x^2$ is $(\sqrt u,\ u).$

Point of intersection of $y=u$ and $y=2x$ is $\left(\dfrac{u}{2},\ u\right).$

The line $y=u$ bisects the area, then,

$\displaystyle\longrightarrow \dfrac{A}{2}=\int\limits_0^{\frac{u}{2}}2x\ dx+\int\limits_{\frac{u}{2}}^{\sqrt u}u\ dx-\int\limits_0^{\sqrt u}x^2\ dx$

$\displaystyle\longrightarrow \dfrac{2}{3}=\Big[x^2\Big]_0^{\frac{u}{2}}+\Big[ux\Big]_{\frac{u}{2}}^{\sqrt u}-\left[\dfrac{x^3}{3}\right]_0^{\sqrt u}$

$\displaystyle\longrightarrow \dfrac{2}{3}=\dfrac{u^2}{4}+u\sqrt u-\dfrac{u^2}{2}-\dfrac{u\sqrt u}{3}$

$\displaystyle\longrightarrow \dfrac{2}{3}=-\dfrac{u^2}{4}+\dfrac{2u\sqrt u}{3}$

$\displaystyle\longrightarrow\underline{\underline{3u^2-8u\sqrt u+8=0}}$