Mathematics - Area Under Curves Question with Solution | TestHub

The given curves are $y^2=4x$ and $y=|x|$.

Since $y^2=4x$ is defined only for $x\ge 0$, we consider the line $y=|x|=x$ for $x\ge 0$.

To find the points of intersection, substitute $y=x$ into $y^2=4x$:

$x^2=4x$

$x^2-4x=0$

$x(x-4)=0$

This gives $x=0$ and $x=4$.

In the interval $[0,4]$, the parabola $y=2\sqrt{x}$ lies above the line $y=x$.

The required area $A$ is given by:

$A={\int }_0^4(2\sqrt{x}-x)dx$

$A={\left[2\left(\frac{x^{3/2}}{3/2}\right)-\frac{x^2}{2}\right]}_0^4$

$A={\left[\frac{4}{3}{x}^{3/2}-\frac{x^2}{2}\right]}_0^4$

Substituting the limits:

$A=\left(\frac{4}{3}(4{)}^{3/2}-\frac{4^2}{2}\right)-0$

$A=\frac{4}{3}(8)-\frac{16}{2}$

$A=\frac{32}{3}-8$

$A=\frac{32-24}{3}=\frac{8}{3}$ sq. units

Answer: $\frac{8}{3}$ sq. units