Mathematics - Area Under Curves Question with Solution | TestHub

The given curve is $x=2-y-y^2$.

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

$2-y-y^2=0$

$y^2+y-2=0$

$(y+2)(y-1)=0$

$y=-2,y=1$

The required area is given by:

$A={\int }_{-2}^{1}xdy$

$A={\int }_{-2}^1(2-y-y^2)dy$

$A={\left[2y-\frac{y^2}{2}-\frac{y^3}{3}\right]}_{-2}^1$

$A=\left(2(1)-\frac{1}{2}-\frac{1}{3}\right)-\left(2(-2)-\frac{(-2{)}^2}{2}-\frac{(-2{)}^3}{3}\right)$

$A=\left(2-\frac{1}{2}-\frac{1}{3}\right)-\left(-4-2+\frac{8}{3}\right)$

$A=\frac{7}{6}-\left(-6+\frac{8}{3}\right)$

$A=\frac{7}{6}-\left(\frac{-10}{3}\right)$

$A=\frac{7}{6}+\frac{20}{6}=\frac{27}{6}=\frac{9}{2}$ sq. units.

Answer: $\frac{9}{2}$ sq. units