Question

The area between $x=y^2$ and $x=4$ is divided into two equal parts by the line $x=a$, find the value of $a$.

Answer 1

Given curve
$y^2=x$
Since the line $x=a$ divides the region into the equal parts
$\therefore$ Area of OBAO = Area of ABCDA

$\Rightarrow 2\times \text{(Area of OEBO)}=2\times \text{(Area of EFCBE)}$

$\Rightarrow 2\times {\int }_0^{a}ydx=2\times {\int }_a^{4}ydx$

$\Rightarrow {\int }_0^{a}ydx={\int }_a^{4}ydx$ ..(i)

Now,

$y^2=x$

$\Rightarrow y=\pm \sqrt{x}$

Since, we are calculating the area in first quadrant

$\therefore y=\sqrt{x}$

${\int }_0^{a}ydx={\int }_a^{4}ydx$

$\Rightarrow {\int }_0^a\sqrt{x}dx={\int }_a^4\sqrt{x}dx$

$\Rightarrow {\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^a={\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_a^4$

$\Rightarrow \frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_0^a=\frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_a^4$

$\Rightarrow (a{)}^{\frac{3}{2}}-0=(4{)}^{\frac{3}{2}}-(a{)}^{\frac{3}{2}}$

$\Rightarrow 2(a{)}^{\frac{3}{2}}=2^3\Rightarrow (a{)}^{\frac{3}{2}}=2^2$

$\Rightarrow a=(2^2{)}^{\frac{2}{3}}=2^{\frac{4}{3}}=4^{\frac{2}{3}}$

Final answer:
Therefore, value of a is $(4{)}^{\frac{2}{3}}$