Find the area of the region between the graph of y = x(1+3x^2)^3 and the x-axis, from x=0 to x=1/3
This one has also got me torn...
The region between x=0 and x=1/3 is positive everywhere, so we can just calculate:
$\displaystyle A=\int_0^{1/3} x(1+3x^2)^3 \, dx$
It is possible to solve this integral either by expanding out the expression term by term or you can avoid most of the algebra by using a substitution with $\displaystyle u=1+3x^2$.
You need to evaluate $\displaystyle \int_0^{\frac{1}{3}}{x(1 + 3x^2)^3\,dx}$.
$\displaystyle \int_0^{\frac{1}{3}}{x(1 + 3x^2)^3\,dx} = \frac{1}{6}\int_0^{\frac{1}{3}}{6x(1 + 3x^2)^3\,dx}$.
Now let $\displaystyle u = 1 + 3x^2$ so that $\displaystyle \frac{du}{dx} = 6x$.
Also note that when $\displaystyle x = 0, u = 1$ and when $\displaystyle x = \frac{1}{3}, u = \frac{4}{3}$.
The integral becomes
$\displaystyle \frac{1}{6}\int_0^{\frac{1}{3}}{6x(1 + 3x^2)^3\,dx} = \frac{1}{6}\int_1^{\frac{4}{3}}{u^3\,du}$