Area of region between graph

**alny** · February 19th 2010, 10:14 PM

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...

**drumist** · February 19th 2010, 10:51 PM

Originally Posted by alny

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:

$A=\int_0^{1/3} x(1+3x^2)^3 \, dx$

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 $u=1+3x^2$ .

**Prove It** · February 19th 2010, 10:51 PM

Originally Posted by alny

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...

You need to evaluate $\int_0^{\frac{1}{3}}{x(1 + 3x^2)^3\,dx}$ .

$\int_0^{\frac{1}{3}}{x(1 + 3x^2)^3\,dx} = \frac{1}{6}\int_0^{\frac{1}{3}}{6x(1 + 3x^2)^3\,dx}$ .

Now let $u = 1 + 3x^2$ so that $\frac{du}{dx} = 6x$ .

Also note that when $x = 0, u = 1$ and when $x = \frac{1}{3}, u = \frac{4}{3}$ .

The integral becomes

$\frac{1}{6}\int_0^{\frac{1}{3}}{6x(1 + 3x^2)^3\,dx} = \frac{1}{6}\int_1^{\frac{4}{3}}{u^3\,du}$

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