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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    \[\int\limits \frac{ 6 }{ e ^{8x} }dx\]

  2. anonymous
    • one year ago
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    can i use u substitution?

  3. anonymous
    • one year ago
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    so if i make u=e^(8x) my du would be 1/8du = e^(8x)dx

  4. anonymous
    • one year ago
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    so then i would have 1/8 integral of 6/u ?

  5. anonymous
    • one year ago
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    not sure if i am right so far

  6. anonymous
    • one year ago
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    is it possible to take out the u from the bottom?

  7. anonymous
    • one year ago
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    is it true that if i take out u from the bottom of the fraction it would become u^(-1) ?

  8. Empty
    • one year ago
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    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
    • one year ago
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    thank you i will try it

  10. anonymous
    • one year ago
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    so the integral will be e^(u+1)/(u+1) ?

  11. anonymous
    • one year ago
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    and i plug in u back in now which is -8x

  12. anonymous
    • one year ago
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    so my answer is (e^(-8x+1))/((-8x+1)+1)

  13. anonymous
    • one year ago
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    hmm i think i did it wrong that answer is not correct

  14. Empty
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    so my answer is -3/4e^(-8x)

  17. anonymous
    • one year ago
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    thank you!

  18. Empty
    • one year ago
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    Awesome glad I could help!

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