Evaluating the integral:

Printable View

Aug 10th 2010, 07:51 PM
bobsanchez

Evaluating the integral:

10x(5x^2 - 3)^4 dx

I got it to [10x (5x^2)^5]/5

First off, is this correct? And if so, should it be simplified?
Aug 10th 2010, 07:57 PM
eumyang

Not quite.

$\displaystyle \int 10x(5x^2 - 3)^4 \,dx$

$\begin{aligned} u &= 5x^2 - 3 \\ du &= 10x \,dx \end{aligned}$

$\begin{aligned} \displaystyle \int u^4 \,du &= \dfrac{u^5}{5} + C \\ &= \dfrac{1}{5}(5x^2 - 3)^5 + C \end{aligned}$
Aug 10th 2010, 08:32 PM
bobsanchez

Okay, so if the 10x is already there you don't have to put it back in the problem?
Aug 10th 2010, 08:57 PM
eumyang

That's right, you don't.
Aug 10th 2010, 08:59 PM
bobsanchez

Okay, thanks guys. That probably explains some of my other problems. I really appreciate it.

All times are GMT -8. The time now is 09:24 AM.