Mathematics - Area Under Curves Question with Solution | TestHub

To find the points of intersection of the parabola with the x-axis, substitute $y=0$:

$4x-x^2=0$

$x(4-x)=0\Rightarrow x=0,x=4$

The required area is the integral of the curve from $x=0$ to $x=4$:

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

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

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

$A=32-\frac{64}{3}$

$A=\frac{96-64}{3}=\frac{32}{3}$ sq. units.

Answer: $\frac{32}{3}$