## anonymous one year ago evaluate the indefinite integral integral of (6)/(e^(8x))dx

1. anonymous

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

2. anonymous

can i use u substitution?

3. anonymous

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

4. anonymous

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

5. anonymous

not sure if i am right so far

6. anonymous

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

7. anonymous

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

8. 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.

9. anonymous

thank you i will try it

10. anonymous

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

11. anonymous

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

12. anonymous

13. anonymous

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

14. 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

15. 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 ...

16. anonymous

17. anonymous

thank you!

18. Empty