Integration Problem

**lil_cookie** · March 16th 2010, 03:35 PM

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 $-e^{8-6x} + C$

**Moo** · March 16th 2010, 03:37 PM

Hello,

It's all correct

**lil_cookie** · March 16th 2010, 03:47 PM

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 $

**Moo** · March 16th 2010, 03:52 PM

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

**lil_cookie** · March 16th 2010, 04:01 PM

oops no you were right I did it wrong but thanks for pointing that out

**lil_cookie** · March 16th 2010, 04:10 PM

the answer should be:

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

correct?

**Moo** · March 16th 2010, 04:20 PM

No, it looks like you forgot du=6 dx

Otherwise, it's correct.

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