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anonymous

  • 5 years ago

Evaluate aniti-derivitive 5x^4 e^x^5 dx I know I need to get the u and du, but what about this problem tells me to do that?

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  1. anonymous
    • 5 years ago
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    the fact that the derivative of x^5 is present

  2. anonymous
    • 5 years ago
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    This one is a bit less obvious, but what part of this equation seems like a good candidate for a u sub?

  3. anonymous
    • 5 years ago
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    5x^4

  4. anonymous
    • 5 years ago
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    not quite. But that does stand out.

  5. anonymous
    • 5 years ago
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    I'm looking at the \(e^{x^5}\) as being particularly ugly, and I'd like to have it be something nice like \(e^u\)

  6. anonymous
    • 5 years ago
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    You'll find that in the beginning, you are mostly just guessing at things to pick for u and seeing if they work out nicely. As you get more practice you'll be able to spot things better.

  7. anonymous
    • 5 years ago
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    So you can certainly try working with the x^4 as u, but then you'll have \(e^{ux}\)

  8. anonymous
    • 5 years ago
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    which isn't as nice.

  9. anonymous
    • 5 years ago
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    And you'll have \(5u/x^3\) in front because your du will be \(4x^3 dx\)

  10. anonymous
    • 5 years ago
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    So that's a lot of mixed x's and u's.

  11. anonymous
    • 5 years ago
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    Which often (especially in the beginning) means you're on the wrong track.

  12. anonymous
    • 5 years ago
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    du = 5x^4 ??

  13. anonymous
    • 5 years ago
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    When you are first learning u substitution picking a good u, should simplify the problem a lot. What did you pick for u?

  14. anonymous
    • 5 years ago
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    u = x^5

  15. anonymous
    • 5 years ago
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    ok good. Yes. So what is dx in terms of du ?

  16. anonymous
    • 5 years ago
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    du = \(5x^4 dx \implies dx = ?\)

  17. anonymous
    • 5 years ago
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    not sure I want to think I take derivative of 5x^4 ??? not sure

  18. anonymous
    • 5 years ago
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    No, just divide. \[du = 5x^4 dx \implies dx = \frac{1}{5x^4} du\]

  19. anonymous
    • 5 years ago
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    So now we have something we can plug in for dx and it'll cancel nicely with the product of 5x^4 out front.

  20. anonymous
    • 5 years ago
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    so will I put e^u * 1/5x^4

  21. anonymous
    • 5 years ago
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    Don't forget the 5x^4 you have in front of the e^u from the initial equation.

  22. anonymous
    • 5 years ago
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    or reverse it 1/5x^4 first

  23. anonymous
    • 5 years ago
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    Neither.. Let \(u = x^5 \implies du = 5x^4 dx \implies dx = \frac{1}{5x^4}du\) \[\int 5x^4e^{x^5}dx = \int(5x^4e^u )\frac{1}{5x^4}du\]

  24. anonymous
    • 5 years ago
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    Do you follow that and understand where each piece came from?

  25. anonymous
    • 5 years ago
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    I don't really understand why 5x^4 stayed in front

  26. anonymous
    • 5 years ago
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    Where should it have gone? It's part of the equation, I can't make it evaporate ;)

  27. anonymous
    • 5 years ago
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    All I did was substitute u for x^5 and replaced dx with my expression with du.

  28. anonymous
    • 5 years ago
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    But I can't do anything to the 5x^4 yet, because that's not x^5 = u.

  29. anonymous
    • 5 years ago
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    Does that make sense?

  30. anonymous
    • 5 years ago
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    yes it does

  31. anonymous
    • 5 years ago
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    But when we do that, we get something nice for our new version that should be easier to take the anti-derivative of.

  32. anonymous
    • 5 years ago
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    ?? (5x^4 * 1/u *e^u) * 1/5x^4 du

  33. anonymous
    • 5 years ago
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    never mind th 1/u should be just 1/1 shouldn't it

  34. anonymous
    • 5 years ago
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    Umm.. close \[\int (5x^4e^u)\frac{1}{5x^4}du = \int e^u\frac{5x^4}{5x^4}du = \int e^u du\]

  35. anonymous
    • 5 years ago
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    I can see that because it is all multipilcation no + or -

  36. anonymous
    • 5 years ago
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    Right.

  37. anonymous
    • 5 years ago
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    so now will I replace u with x^5

  38. anonymous
    • 5 years ago
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    After you integrate then you replace it back.

  39. anonymous
    • 5 years ago
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    err take the anti-derivative.

  40. anonymous
    • 5 years ago
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    Sorry, later on they're going to tell you that anti-derivatives are called integrals. ;p

  41. anonymous
    • 5 years ago
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    so will the answer be e^x^5 + c

  42. anonymous
    • 5 years ago
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    Yep.

  43. anonymous
    • 5 years ago
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    good!!!!! Thanks gotta go now

  44. anonymous
    • 5 years ago
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    Though you should slap some parens in there for readability.

  45. anonymous
    • 5 years ago
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    when you type it that is. I'm sure it's written right on your paper.

  46. anonymous
    • 5 years ago
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    thanks

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