I'm lost on this one, just need a starting point.

$\displaystyle \int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$

Thanks for any help!

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- Apr 12th 2009, 08:47 PMJim MarnellHelp on another Definite Integral problem
I'm lost on this one, just need a starting point.

$\displaystyle \int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$

Thanks for any help! - Apr 12th 2009, 08:55 PMTheEmptySet
- Apr 12th 2009, 08:55 PMmollymcf2009
- Apr 12th 2009, 09:02 PMJim Marnell
yep, thanks for getting me started!

- Apr 12th 2009, 09:15 PMJim Marnell
for the 1st term, would it equal, $\displaystyle u=3x,,,, du=3,,, then,,, 6e^u(3)?$

I think thats the term thats confusing me the most - Apr 12th 2009, 09:22 PMJhevon
Consider $\displaystyle \int 6e^{3x}~dx$

Let $\displaystyle u = 3x$

$\displaystyle \Rightarrow du = 3~dx$

$\displaystyle \Rightarrow \frac 13 ~du = dx$

$\displaystyle \int 6e^{{\color{red}3x}}~{\color{blue}dx} = \int 6e^{\color{red}u} \cdot {\color{blue}\frac 13~du}$

$\displaystyle = 2 \int e^u~du$