Mathematics - Area Under Curves Question with Solution | TestHub

We have, $y^2=2x-1 ...\left(1\right)$ and $y^2=4x-3 ...\left(2\right)$

We first find the points of intersection of the given parabolas by solving the equations $\left(1\right) \& \left(2\right)$simultaneously.

This gives, $2x-1=4 x-3$

$\Rightarrow 2x=2$  $\Rightarrow x=1$

Now, put $x=1$ in equation $\left(1\right)$, we get $y=\pm 1$.

Now, the equations $\left(1\right) \& \left(2\right)$ can be written as

$y^2=2\left(x-\frac{1}{2}\right)$

$y^2=4\left(x-\frac{3}{4}\right)$ respectively.

The graphs of these two curves are shown in figure.

We have to find the area of the shaded region.

Now, the equations $\left(1\right) \& \left(2\right)$ can be written as

$x=\frac{y^2+1}{2}$ & $x=\frac{y^2+3}{4}$ respectively.

It is clear from the figure, required area $A=2{\int }_0^1\left(x_2-x_1\right)dy$  ($\because$ Area is symmetrical about $x-$ axis)

where $x_1=\frac{y^2+1}{2}$ and  $x_2=\frac{y^2+3}{4}$

$\Rightarrow A=2{\int }_0^1\left(\frac{y^2+3}{4}-\frac{y^2+1}{2}\right)dy$ $=\frac{2}{4}{\int }_0^1\left(1-y^2\right)dy$

$=\frac{1}{2}{\left[y-\frac{y^3}{3}\right]}_0^1$   $=\frac{1}{2}\left[1-\frac{1}{3}\right]$

$=\frac{1}{2}\left(\frac{2}{3}\right)=\frac{1}{3}$ sq.units.