Find the area bounded by $x=2-5y-3y^2$ and $x=\mathbf{0}$

Given parabola is $x=2-5y-3y^2$

And the line $\text{ is }x=0$

$\therefore 2-5y-3y^2=0$

$3y^2+5y-2=0$

$3y^2+6y-y-2=0$

$3y(y+2)-1(y+2)=0$

$(3y-1)(y+2)=0$

$y=-2,y=\frac{1}{3}$

$\text{ Required area }={\int }_{-2}^{\frac{1}{3}} \left(2-5y-3y^2\right)dy$

$={\left[2y-\frac{5y^2}{2}-y^3\right]}_{-2}^{1/3}$

$=\left(\frac{2}{3}-\frac{5}{18}-\frac{1}{27}\right)-(-4-10+8)$

$=\frac{19}{54}+6=\frac{343}{54}\text{ sq.units }$