Integration Problem

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• March 16th 2010, 03:35 PM
lil_cookie
Integration Problem
Integrate:
$
6(e^{4-3x})^2
$

Is it correct if i do this: $6(e^{8-6x})$ and then solve?

and then let u = 8-6x and du=-6

then the answer comes to $-e^{8-6x} + C$
• March 16th 2010, 03:37 PM
Moo
Hello,

It's all correct (Nod)
• March 16th 2010, 03:47 PM
lil_cookie
thank you, here is another one (I have a test on Friday so I'm trying to do as many problems as I can)

Integrate: $\frac{3}{4}-\frac{1}{\sqrt{6x}}$

I let u = 6x and du = 6dx

the answer I end up with is:
$

-\frac{1}{4}\sqrt{6x}+C
$
• March 16th 2010, 03:52 PM
Moo
Hello,

And what about integrating the constant ? The integral of 1 is x, the integral of 3/4 will be...

But according to your answer, would the integral rather be $\int \frac 34\times \frac{1}{\sqrt{6x}} ~dx$ ? In which case, it's correct - again :p
• March 16th 2010, 04:01 PM
lil_cookie
oops no you were right I did it wrong but thanks for pointing that out :)
• March 16th 2010, 04:10 PM
lil_cookie
the answer should be:

$
\frac{3}{4}x-\frac{1}{2}\sqrt{6x}+C
$

correct?
• March 16th 2010, 04:20 PM
Moo
No, it looks like you forgot du=6 dx :)
Otherwise, it's correct.