# Help on another Definite Integral problem

• Apr 12th 2009, 08:47 PM
Jim Marnell
Help 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 PM
TheEmptySet
Quote:

Originally Posted by Jim Marnell
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!

for $\displaystyle 6e^{3x}$ use the sub u=3x

The 2nd one needs to be memorized. Here is the hint

$\displaystyle \frac{d}{dx}\ln(x)=\frac{1}{x}$ so

$\displaystyle \int \frac{1}{x}dx=...$
• Apr 12th 2009, 08:55 PM
mollymcf2009
Quote:

Originally Posted by Jim Marnell
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!

Just integrate term by term and factor out your constants. I'll get you started:

$\displaystyle 6 \int^{2}_{1} e^{3x} dx - \int^{2}_{1} \frac{1}{x} dx$

* For the first one, do a u-substitution with u = 3x

• Apr 12th 2009, 09:02 PM
Jim Marnell
yep, thanks for getting me started!
• Apr 12th 2009, 09:15 PM
Jim 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 PM
Jhevon
Quote:

Originally Posted by Jim 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

Consider $\displaystyle \int 6e^{3x}~dx$

Let $\displaystyle u = 3x$

$\displaystyle \Rightarrow du = 3~dx$

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

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

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