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

$\begin{array}{1 1} \large\frac{2}{3} \\ \large\frac{4}{3} \\ \large\frac{1}{3} \\ \large\frac{2}{5} \end{array}$

Toolbox:
• whenever the area bounded by a curve and a line is divided into two equal parts,then the limits of the curve can also be split into two.
Here it is given in the area between $x=y^2$ and x=4 is divided into two equal parts by the line x=a.
From the fig we can see that the area is symmetrical about x-axis.
hence the required area can be divided as,$y_1$ and $y_2$.

Given $y_1=y_2$.

Area of $y_1=\int_0^a \sqrt x\;dx$--------(1)
Integrating this we get,
Area of $y_1=\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_0^9$

$\;\;\;\;\quad\qquad=\frac{2}{3}[x^\frac{3}{2}]$
on applying limits we get,
Area of $y_1=\frac{2}{3}[a^\frac{3}{2}]$--------(2)

Area of $y_2=\int_a^4\sqrt x\;dx$----------(3)
on integrating we get,
Area of $y_2=\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_a^4$

$\;\;\;\;\quad\qquad=\frac{2}{3}[x^\frac{3}{2}]_a^4$
on applying limits we get,
Area of $y_2=\frac{2}{3}[4^\frac{3}{2}-a^\frac{3}{2}]$

$\;\;\;\;\quad\qquad=\frac{2}{3}[8-a^\frac{3}{2}]$---------(4)
Combining (3) and (4) gives the required area.
But since they are equal in area let us equate both the equations.
$\quad\frac{2}{3}a^\frac{3}{2}=\frac{2}{3}[8-a^\frac{3}{2}]$

$\;\;\Rightarrow a^\frac{3}{2}=[8-a^\frac{3}{2}]$

$\;\Rightarrow 2a^\frac{3}{2}=8$

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

The value of a is $(4)^\frac{2}{3}$.