Mathematics - Area Under Curves Question with Solution | TestHub

$\frac{dy}{dx}-\frac{y}{x+1}=x+1-\frac{3}{x+1}$. With integrating factor $\frac{1}{x+1}:\frac{d}{dx}\left(\frac{y}{x+1}\right)=1-\frac{3}{(x+1{)}^2}$, so $\frac{y}{x+1}=x+\frac{3}{x+1}+c$, i.e. $y=(x+1)(x+c)+3$. Since $(2,0)$ lies on it, $0=3(2+c)+3$, so $c=-3$ and $y=x^2-2x$. Area $={\int }_{-3}^3\left|x^2-2x\right|dx={\int }_{-3}^0\left(x^2-2x\right)dx+{\int }_0^2\left(2x-x^2\right)dx+{\int }_2^3\left(x^2-2x\right)dx=18+\frac{4}{3}+\frac{4}{3}=\frac{62}{3}$.