anonymous
  • anonymous
evaluate the indefinite integral integral of (6)/(e^(8x))dx
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[\int\limits \frac{ 6 }{ e ^{8x} }dx\]
anonymous
  • anonymous
can i use u substitution?
anonymous
  • anonymous
so if i make u=e^(8x) my du would be 1/8du = e^(8x)dx

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so then i would have 1/8 integral of 6/u ?
anonymous
  • anonymous
not sure if i am right so far
anonymous
  • anonymous
is it possible to take out the u from the bottom?
anonymous
  • anonymous
is it true that if i take out u from the bottom of the fraction it would become u^(-1) ?
Empty
  • 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
  • anonymous
thank you i will try it
anonymous
  • anonymous
so the integral will be e^(u+1)/(u+1) ?
anonymous
  • anonymous
and i plug in u back in now which is -8x
anonymous
  • anonymous
so my answer is (e^(-8x+1))/((-8x+1)+1)
anonymous
  • anonymous
hmm i think i did it wrong that answer is not correct
Empty
  • 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
  • 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
  • anonymous
so my answer is -3/4e^(-8x)
anonymous
  • anonymous
thank you!
Empty
  • Empty
Awesome glad I could help!

Looking for something else?

Not the answer you are looking for? Search for more explanations.