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bettyboop8904

Can someone give me a quick recap of integrals? I have an example and there are 5 problems that are done similarly. Example is in the post with equation converter helper.

  • one year ago
  • one year ago

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  1. bettyboop8904
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    \[\int\limits_{0}^{1} (9x ^{e}+e ^{x})dx\] Do I use u substitution?

    • one year ago
  2. CanadianAsian
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    I can't tell what the power of x is but it looks like an e? Either way, you can break this integral into two parts. \[\int\limits_{0}^{1}9x^e + \int\limits_{0}^{1}e^x\] Then just take integrate the individual integrals.

    • one year ago
  3. CanadianAsian
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    The rule of integration for X^n where n is a constant is \[x^{n+1}/(n+1)\] And the integral of e^x is just e^x

    • one year ago
  4. bettyboop8904
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    Wait I'm confused on the last post. \[\frac{ x ^{n+1} }{ n+1 }\] Isn't that the anti-derivative?

    • one year ago
  5. bettyboop8904
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    is that what an integral is?

    • one year ago
  6. CanadianAsian
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    Exactly! An integral is the anti-derivative.

    • one year ago
  7. hartnn
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    recap of integrals ? try this.... http://openstudy.com/users/hartnn#/updates/50960518e4b0d0275a3ccfba

    • one year ago
  8. CanadianAsian
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    Well not exactly, but a key component of taking the integral of something is finding the anti-derivative. If there are no bounds, then an integral is essentially the anti-derivative.

    • one year ago
  9. bettyboop8904
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    Oh ok = ) So that's how you solve an integral. you use the rule that you can split the integrals up and then solve each one by finding the anti-derivative. Oh yeah bc you have to incorporate the integral 0 to 1 right?

    • one year ago
  10. bettyboop8904
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    so does that mean you would have: \[\frac{ 9x ^{e+1} }{ e+1 }\] and \[\frac{ e ^{x+1} }{ x+1 }\] is that right?

    • one year ago
  11. hartnn
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    the first part is correct, for e^x \(\int e^xdx=e^x+c\)

    • one year ago
  12. CanadianAsian
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    But since this is a definite integral (it has bounds) you can ignore the c, which will cancel out anyway.

    • one year ago
  13. bettyboop8904
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    oh yeah that's right because the integral of e^x is itself right?

    • one year ago
  14. CanadianAsian
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    Yup

    • one year ago
  15. bettyboop8904
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    Ok so what's next? or is that all I have to do with it?

    • one year ago
  16. bettyboop8904
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    It just says evaluate the integral

    • one year ago
  17. CanadianAsian
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    Once you've taken the anti-derivative of it you plug in your bounds for x So for the integral of e^x for example: you'd get \[e^1 - e^0\] Which simplifies into e-1 But you have to remember that you're adding this to the integral of the first half.

    • one year ago
  18. bettyboop8904
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    Oh wait I just realized I typed the problem out. There is no 9 in front of the x^e

    • one year ago
  19. bettyboop8904
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    *typed it out wrong lol

    • one year ago
  20. CanadianAsian
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    The answer you posted above was correct for the anti-derivative of x^e, just take out the 9 then.

    • one year ago
  21. bettyboop8904
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    stupid question how do you get e^-1 from e^1-e^0?

    • one year ago
  22. CanadianAsian
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    so e^1 = e e^0 = 1 So e^1 - e^0 = e - 1

    • one year ago
  23. bettyboop8904
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    oh ok I thought the -1 was the exponent of e from the rule of two bases being the same the exponents divide you know what I'm trying to say? lol

    • one year ago
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