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onegirl

  • one year ago

Determine whether the integral converges or diverges. Find the value of the integral if it converges.

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  1. onegirl
    • one year ago
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  2. onegirl
    • one year ago
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    @Reaper534 can u help?

  3. terenzreignz
    • one year ago
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    Let's just focus on the integral first, keeping in mind that... \[\Large \int\limits_1^\infty x^{-\frac43}= \lim_{b\rightarrow\infty}\int\limits_1^bx^{-\frac43}\]

  4. onegirl
    • one year ago
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    okay

  5. terenzreignz
    • one year ago
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    Well, can you integrate \[\Large \int x ^{-\frac43}=\color{red}?\]

  6. onegirl
    • one year ago
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    yes i get -3/3sqrt(x)

  7. terenzreignz
    • one year ago
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    This \[\Large \frac{-3}{3\sqrt{x}}\]?

  8. onegirl
    • one year ago
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    yes!

  9. terenzreignz
    • one year ago
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    Or this \[\Large \frac{-3}{\sqrt[3]x}\]

  10. onegirl
    • one year ago
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    the second one

  11. terenzreignz
    • one year ago
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    lol... it's cube root, not 3sqrt because \(\Large3\sqrt x\) means something entirely different, ok?

  12. onegirl
    • one year ago
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    Okay

  13. terenzreignz
    • one year ago
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    Okay, so, we have... \[\Large \int\limits_1^b x^{-\frac43}=\left.\frac{-3}{\sqrt[3]x}\right]_1^b\]

  14. terenzreignz
    • one year ago
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    Can you evaluate this bit?

  15. onegirl
    • one year ago
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    i think

  16. terenzreignz
    • one year ago
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    Well then, what do you get?

  17. onegirl
    • one year ago
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    hmm i'm working on it

  18. onegirl
    • one year ago
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    eh...i'm getting a wrong answer :/

  19. terenzreignz
    • one year ago
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    Fundamental theorem of Calculus? \[\Large \int\limits_a^b f'(x)dx = \left.f(x)\right]_a^b = f(b)-f(a)\]

  20. onegirl
    • one year ago
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    ohhhh okay

  21. terenzreignz
    • one year ago
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    So... \[\Large \int\limits_1^b x^{-\frac43} \ dx=\left.\frac{-3}{\sqrt[3]x}\right]_1^b=\color{red}?\]

  22. onegirl
    • one year ago
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    so i plug in b and 1 into that then subtract?

  23. terenzreignz
    • one year ago
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    Yes... but that's not the end yet, just plug in for now, and tell me what you get :)

  24. onegirl
    • one year ago
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    okay so -3/3sqrt(b) - -3/3sqrt(1)

  25. terenzreignz
    • one year ago
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    I'm assuming by 3sqrt you mean cube root :D Okay, that being the case, you're right :) \[\Large \int\limits_1^b x^{-\frac43}=\frac{-3}{\sqrt[3]b}+\frac3{\sqrt[3]1}= -\frac3{\sqrt[3]b}+3\] Catch me so far?

  26. onegirl
    • one year ago
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    yes

  27. terenzreignz
    • one year ago
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    Now, we're supposed to take the improper integral to infinity, right? Remember this... \[\Large \int\limits_1^\infty x^{-\frac43} \ dx = \color{red}{\lim_{b\rightarrow\infty}}\int\limits_1^bx^{-\frac43} \ dx\] Now is the time to apply that limit (which we haven't done yet) \[\Large \color{red}{\lim_{b\rightarrow\infty}}\left(-\frac3{\sqrt[3]b}+3\right) \]

  28. onegirl
    • one year ago
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    okay

  29. terenzreignz
    • one year ago
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    So... evaluating the limit...?

  30. onegirl
    • one year ago
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    umm i get 3? :/

  31. terenzreignz
    • one year ago
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    That's right!

  32. onegirl
    • one year ago
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    ohh okay lol

  33. onegirl
    • one year ago
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    so thats the final answer?

  34. terenzreignz
    • one year ago
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    Well, you technically have two questions, but since there was an answer, then the integral converges, and it converges to 3 ^.^

  35. onegirl
    • one year ago
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    okay thanks

  36. terenzreignz
    • one year ago
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    No problem :)

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