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mariomintchev Group Title

i need help with #17

  • 2 years ago
  • 2 years ago

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  1. mariomintchev Group Title
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    • 2 years ago
  2. mariomintchev Group Title
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    i know the chain rule is used

    • 2 years ago
  3. TuringTest Group Title
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    not really, it should just be a u-sub\[u=t+4\]\[du=dt\]

    • 2 years ago
  4. mariomintchev Group Title
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    i need more help than that

    • 2 years ago
  5. TuringTest Group Title
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    well if you use that sub what does this integral become\[\int_{0}^{x}\sqrt{t+4}dt\]?write it in terms of u

    • 2 years ago
  6. mariomintchev Group Title
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    yeah im stuck at 1/2(t+4)^(-1/2)

    • 2 years ago
  7. TuringTest Group Title
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    you took the derivative or something...

    • 2 years ago
  8. TuringTest Group Title
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    we are integrating here, that's the opposite

    • 2 years ago
  9. mariomintchev Group Title
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    that response was to whoever said use the power rule... they deleted their comment

    • 2 years ago
  10. TuringTest Group Title
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    they just didn't quite understand your question I think \[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]now rewrite the integral in terms of u

    • 2 years ago
  11. mariomintchev Group Title
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    im not sure how to do that

    • 2 years ago
  12. TuringTest Group Title
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    \[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu\]which you should be able to integrate

    • 2 years ago
  13. TuringTest Group Title
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    notice we put the \(u\) where the \(t+4\) was and the \(du\)0 where the \(dt\) was. that's the essence of how a u-sub in integration works

    • 2 years ago
  14. mariomintchev Group Title
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    the only part im not sure how to get is the - 16 / 3

    • 2 years ago
  15. TuringTest Group Title
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    what do you get after integrating?

    • 2 years ago
  16. TuringTest Group Title
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    after integrating you get ________ evaluated from __ to __ fill in the blanks

    • 2 years ago
  17. mariomintchev Group Title
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    t+4 0 x

    • 2 years ago
  18. TuringTest Group Title
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    not quite... would it help you if I reminded you that we can rewrite this as\[\int u^{1/2}du\]??

    • 2 years ago
  19. TuringTest Group Title
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    then you can use the power rule for \(integration\) (basically the opposite of the power rule for differentiation)

    • 2 years ago
  20. TuringTest Group Title
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    which is\[\int x^ndx=\frac{x^{n+1}}{n+1}\]

    • 2 years ago
  21. mariomintchev Group Title
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    can you solve the problem so i can see how its done? ill understand it better that way.

    • 2 years ago
  22. TuringTest Group Title
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    I will try to guide you a little farther than I normally would, but I hope you don't ask the exact same question again later!

    • 2 years ago
  23. mariomintchev Group Title
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    i try not to do that unless i forget a step or something

    • 2 years ago
  24. TuringTest Group Title
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    \[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}\]now put it back in terms of t and put in the evaluation\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}\]evaluate this \(carefully\). Now do you see where the last term comes from?

    • 2 years ago
  25. mariomintchev Group Title
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    not really lol

    • 2 years ago
  26. TuringTest Group Title
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    ok full solution:

    • 2 years ago
  27. TuringTest Group Title
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    \[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}\]now put it back in terms of t and put in the evaluation\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}=\frac23(x+4)-\frac23(0+4)^{3/2}\]\[=\frac23(x+4)-\frac23(4)^{3/2}=\frac23(x+4)-\frac23(8)=\frac23(x+4)-\frac{16}3\]please don't ask me to give the final answer again, it's very buch not my style... I hope this makes sense to you after looking it over. Good luck!

    • 2 years ago
  28. TuringTest Group Title
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    very much*

    • 2 years ago
  29. mariomintchev Group Title
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    i wont. thanks for your time/help!

    • 2 years ago
  30. TuringTest Group Title
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    *typo above (I dropped the ^3/2 in the last few steps)\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}\]now put it back in terms of t and put in the evaluation\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}=\frac23(x+4)^{3/2}-\frac23(0+4)^{3/2}\]\[=\frac23(x+4)^{3/2}-\frac23(4)^{3/2}=\frac23(x+4)^{3/2}-\frac23(8)=\frac23(x+4)^{3/2}-\frac{16}3\]

    • 2 years ago
  31. mariomintchev Group Title
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    ok i see what was done. thanks. :D

    • 2 years ago
  32. TuringTest Group Title
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    very welcome :)

    • 2 years ago
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