# Evaluating the integral:

• August 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?
• August 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}
• August 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?
• August 10th 2010, 08:57 PM
eumyang
That's right, you don't.
• August 10th 2010, 08:59 PM
bobsanchez
Okay, thanks guys. That probably explains some of my other problems. I really appreciate it.