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- anonymous

evaluate the indefinite integral
integral of (6)/(e^(8x))dx

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- anonymous

evaluate the indefinite integral
integral of (6)/(e^(8x))dx

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- anonymous

\[\int\limits \frac{ 6 }{ e ^{8x} }dx\]

- anonymous

can i use u substitution?

- anonymous

so if i make u=e^(8x) my du would be 1/8du = e^(8x)dx

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- anonymous

so then i would have 1/8 integral of 6/u ?

- anonymous

not sure if i am right so far

- anonymous

is it possible to take out the u from the bottom?

- anonymous

is it true that if i take out u from the bottom of the fraction it would become u^(-1) ?

- Empty

You're on the right track you'll need a substitution I think! I'll help walk you through this, first let's take out the coefficient 6 and use exponent rules to rewrite it like this:
\[6 \int e^{-8x}dx\]
From here you can use this substitution:
\[u=-8x \]
Take the derivative of this to get:
\[\frac{du}{dx} = -8\]
multiply \(dx\) by both sides:
\[du=-8dx\]
divide -8 from both sides:
\[\frac{-1}{8} du = dx\]
Now you can plug these into your for \(-8x=u\) and \(\frac{-1}{8}du = dx\)
\[6 \int e^u * \frac{-1}{8} du\]
Pull this constant out as well to get
\[\frac{-6}{8} \int e^udu\]
Now see if you can take it from here.

- anonymous

thank you i will try it

- anonymous

so the integral will be e^(u+1)/(u+1) ?

- anonymous

and i plug in u back in now which is -8x

- anonymous

so my answer is (e^(-8x+1))/((-8x+1)+1)

- anonymous

hmm i think i did it wrong that answer is not correct

- Empty

So what you're doing is the rule for polynomials:
\[\int x^n dx = \frac{x^{n+1}}{n+1}\]
but the type of integral you're looking at is:
\[\int e^x dx = e^x\]
Which you can check by taking the deriativatives of the right hand sides of the equations since integrals are antiderivatives

- anonymous

oh i see, so the integral of e^u is just e^u... and all i would have to do is plug in my u and multiply it by -6/8 ...

- anonymous

so my answer is -3/4e^(-8x)

- anonymous

thank you!

- Empty

Awesome glad I could help!

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