Mathematics - Area Under Curves Question with Solution | TestHub

The intersection points of $y=x^2-2x$ and $y=x$ are given by $x^2-2x=x$, which simplifies to $x^2-3x=0$. Thus, $x(x-3)=0$, so $x=0$ and $x=3$. For $x\in (0,3)$, the line $y=x$ is above the parabola $y=x^2-2x$. The area $A$ is given by the integral:$A={\int }_0^3(x-(x^2-2x))dx={\int }_0^3(3x-x^2)dx$ $A={\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]}_0^3=\left(\frac{3(3^2)}{2}-\frac{3^3}{3}\right)-\left(\frac{3(0{)}^2}{2}-\frac{0^3}{3}\right)=\frac{27}{2}-9=\frac{27-18}{2}=\frac{9}{2}$ The area is $4.5$.