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anonymous

  • 5 years ago

the indefinite integral i am given is (e^x - 6x^2)/6... i need to evaluate it. please help!

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  1. nowhereman
    • 5 years ago
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    Do you know the integral of e^x and the power rule?

  2. anonymous
    • 5 years ago
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    i assume the integral is e^x

  3. anonymous
    • 5 years ago
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    i need to use substitution, but i dont know what to substitute... i've tried it a lot already

  4. nowhereman
    • 5 years ago
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    You don't need to use any elevated integration techniques. If you want to prove that your result is right, just differentiate it and if you get the original integrand, you're fine.

  5. anonymous
    • 5 years ago
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    what about the 6 in the denominator? how do i do that? is it just 6x?

  6. nowhereman
    • 5 years ago
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    No, that is just a factor of 1/6 which you can pull-out.

  7. anonymous
    • 5 years ago
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    wow.. i got it! thanks! i have another one tho if u are willing to help?

  8. nowhereman
    • 5 years ago
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    Sure :-)

  9. anonymous
    • 5 years ago
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    (5-2xe^x)/x

  10. nowhereman
    • 5 years ago
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    Pull that division into the difference and you should have two known integrals.

  11. anonymous
    • 5 years ago
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    like (5/x)-((2xe^x)/x)?

  12. nowhereman
    • 5 years ago
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    Yes, and reduce that x in the second term.

  13. anonymous
    • 5 years ago
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    ok. then i would need substitution right?

  14. nowhereman
    • 5 years ago
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    For what? You should know the integral of 1/x and e^x and that's all there is.

  15. anonymous
    • 5 years ago
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    i dont know the 1/x one. is it natural log?

  16. nowhereman
    • 5 years ago
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    Exactly, sometimes it is defined this way \[\ln x := \int_1^x\frac 1 y \mathrm d y\] but if you have defined it as the inverse function of e^x you could use the inverse-differentiation rule.

  17. anonymous
    • 5 years ago
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    i got that one! thanks a lot. this is my last one... (x-2)/(x^2-4x+4)^3..... that requires substitution correct?

  18. nowhereman
    • 5 years ago
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    Yes, a little one :-) but you should first observe that \[x^2 - 4x + 4 = (x-2)^2\]

  19. anonymous
    • 5 years ago
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    so i can cancel one of them right away?

  20. nowhereman
    • 5 years ago
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    yes

  21. anonymous
    • 5 years ago
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    would it turn into (x-2)^6?

  22. anonymous
    • 5 years ago
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    in the denominator

  23. nowhereman
    • 5 years ago
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    Sure, that is just one of the exponentiation rules.

  24. anonymous
    • 5 years ago
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    then after the cancellation i am left with 1/(x-2)^5?

  25. nowhereman
    • 5 years ago
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    Right.

  26. anonymous
    • 5 years ago
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    now i substitute u=x-2

  27. nowhereman
    • 5 years ago
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    Yes, but as du/dx = 1 if you are a little more experienced you could just do it in your head.

  28. anonymous
    • 5 years ago
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    i think it would be 1/((1/6)*(x-2)^6)

  29. nowhereman
    • 5 years ago
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    Not quite, maybe you should write it is \[(x-2)^{-5}\] instead to see it better.

  30. anonymous
    • 5 years ago
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    (-1/4)*(x-2)^-4?

  31. nowhereman
    • 5 years ago
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    Yeah, good.

  32. anonymous
    • 5 years ago
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    is this website free to use?

  33. nowhereman
    • 5 years ago
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    It costs both of us our time, but you won't have to pay for it ;-)

  34. anonymous
    • 5 years ago
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    so y do ppl like u go around and answer questions?

  35. nowhereman
    • 5 years ago
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    I love math and helping other people.

  36. anonymous
    • 5 years ago
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    it seems strange that there is no charge for this service. or breech of security

  37. nowhereman
    • 5 years ago
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    I don't know what you mean by breech of security; but the idea of being is just that everyone can ask questions and give answers here. So if I had a question regarding my study, I could ask them too. Only that it seems there are a lot of high-school students here lately :-D

  38. anonymous
    • 5 years ago
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    hmm... thats cool. i guess i'm a skeptic. but with quick responses like the ones u gave me i am becoming a believer

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