Mathematics - Area Under Curves Question with Solution | TestHub

Area $=2{\int }_{-2}^0{x}^2\sqrt{x+2}dx$. Put $\sqrt{x+2}=z$, so $x=z^2-2,dx=2zdz:=4{\int }_0^{\sqrt{2}}{\left(z^2-2\right)}^2{z}^{2}dz=$ $\underline{4{\left[\frac{z^7}{7}-\frac{4z^5}{5}+\frac{4z^3}{3}\right]}_0^{\sqrt{2}}}=4\left(\frac{8\sqrt{2}}{7}-\frac{16\sqrt{2}}{5}+\frac{8\sqrt{2}}{3}\right)=\frac{256\sqrt{2}}{105}$.