Integrate:

$\displaystyle

6(e^{4-3x})^2

$

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

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

then the answer comes to $\displaystyle -e^{8-6x} + C$

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- Mar 16th 2010, 03:35 PMlil_cookieIntegration Problem
Integrate:

$\displaystyle

6(e^{4-3x})^2

$

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

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

then the answer comes to $\displaystyle -e^{8-6x} + C$ - Mar 16th 2010, 03:37 PMMoo
Hello,

It's all correct (Nod) - Mar 16th 2010, 03:47 PMlil_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: $\displaystyle \frac{3}{4}-\frac{1}{\sqrt{6x}}$

I let u = 6x and du = 6dx

the answer I end up with is:

$\displaystyle

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

$ - Mar 16th 2010, 03:52 PMMoo
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 $\displaystyle \int \frac 34\times \frac{1}{\sqrt{6x}} ~dx$ ? In which case, it's correct - again :p - Mar 16th 2010, 04:01 PMlil_cookie
oops no you were right I did it wrong but thanks for pointing that out :)

- Mar 16th 2010, 04:10 PMlil_cookie
the answer should be:

$\displaystyle

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

$

correct? - Mar 16th 2010, 04:20 PMMoo
No, it looks like you forgot du=6 dx :)

Otherwise, it's correct.