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Q.
The area bounded by the curves $f(x)=x^3-3 x$ and $g(x)=2 x^2$ in the second quadrant is
Application of Integrals
Solution:
$y_1$ and $y_2$, we get $x=0,-1,3$
$\text { Required area }= \int\limits_{-1}^0\left(y_1-y_2\right) d x=\int\limits_{-1}^0\left(x^3-3 x-2 x^2\right) d x $
$=\left(\frac{x^4}{4}-\frac{3 x^2}{2}-\frac{2 x^3}{3}\right)_{-1}^0 $
$=0-\left(\frac{1}{4}-\frac{3}{2}+\frac{2}{3}\right)=-\left(\frac{3-18+8}{12}\right)=\frac{7}{12} .$
