Answer to Question #261347 in Calculus for Artika

Question

Find the area of the region enclosed by the graphs of f(x) = x^3 and and g(x) = x^2+2x.

Answer 1

Points of intersections of f(x) and g(x):

$x^3=x^2+2x$

$x(x^2-x-2)=0$

$x=-1, x=0, x=2$

Area of the region enclosed the graphs:

$A=\int_{-1}^0(x^3-x^2-2x)dx+\int_0^2(x^2+2x-x^3)dx=$

$=(\frac{x^4}{4}-\frac{x^3}{3}-x^2)|_{x=-1}^{x=0}+(\frac{x^3}{3}+x^2-\frac{x^4}{4})|_{x=0}^{x=2}=$

$=\frac{5}{12}+\frac{8}{3}=\frac{37}{3}.$