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anonymous

  • one year ago

how do i solve this? thanks!!

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  1. anonymous
    • one year ago
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    \[\int\limits_{0}^{1} \frac{ 1 }{ 1+\sqrt[3]{x} }dx\]

  2. anonymous
    • one year ago
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    what do you know so far?

  3. anonymous
    • one year ago
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    would u = 1 + 3sqrtx and du = dx?

  4. Astrophysics
    • one year ago
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    This requires a u sub, what is the obvious u sub here?

  5. anonymous
    • one year ago
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    you seem to know enough to tackle the problem

  6. Astrophysics
    • one year ago
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    Try it out and see what you get!

  7. Astrophysics
    • one year ago
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    You can also try \[u=\sqrt[3]{x}\] but then you probably will end up doing two substitutions, mess around with it and see what you get

  8. anonymous
    • one year ago
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    ohh okay, err so would this be on the right track? u = 1 + 3sqrtx and du = 3udu du = 3sqrtx

  9. anonymous
    • one year ago
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    I'm still quite confused :( @Astrophysics :(

  10. anonymous
    • one year ago
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    oops *dx=3sqrtx ? :/

  11. Jhannybean
    • one year ago
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    \[\int \frac{dx}{1+\sqrt[3]{x}}\]\[u=x~,~ du=dx\]\[\int\frac{du}{1+u^{1/3}}\]

  12. Jhannybean
    • one year ago
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    I left out the numbers

  13. anonymous
    • one year ago
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    okay!! and so the next step would be the solve this integral? :/

  14. ganeshie8
    • one year ago
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    you really want to sub \(u=x\) ?

  15. anonymous
    • one year ago
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    wait, so we don't sub u=x? :/

  16. ganeshie8
    • one year ago
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    what good does that do

  17. anonymous
    • one year ago
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    im not sure :/ what would i make the sub be then? :/

  18. ganeshie8
    • one year ago
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    unless u hate the letter `x` ..

  19. zepdrix
    • one year ago
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    \(\large\rm u=1+\sqrt[3]{x}\) or \(\large\rm u=\sqrt[3]{x}\) I think both of these subs will work out just fine :) Err no, go with the first one. I dunno bout that other guy, he gave me a funny look.

  20. zepdrix
    • one year ago
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    Where'd you leave off food man? :O Where you stuck at?

  21. anonymous
    • one year ago
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    lol im confused:/ so u = x ?

  22. Jhannybean
    • one year ago
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    Oh what about changing this to arctan function..

  23. zepdrix
    • one year ago
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    \[\large\rm u=1+\sqrt[3]{x},\qquad\qquad du=\frac{1}{3x^{2/3}}dx\]Do you understand how to differentiate that u? :o

  24. anonymous
    • one year ago
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    i think so :) what is next?

  25. zepdrix
    • one year ago
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    Now you have to apply a bunch of sneaky little tricks!

  26. anonymous
    • one year ago
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    how do i do that?? :O

  27. anonymous
    • one year ago
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    just differentiate what you have and try to finish it

  28. zepdrix
    • one year ago
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    Isolate the dx, that's one of the things we're substituting stuff in for after all.\[\large\rm 3x^{2/3}du=dx\]I'm gonna use rules of exponents to write it like this:\[\large\rm 3(x^{1/3})^2du=dx\]

  29. anonymous
    • one year ago
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    that means \(\large x^\frac{1}{3} \)

  30. zepdrix
    • one year ago
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    We can sub something in for that x^(1/3) by using our equation involving u.

  31. zepdrix
    • one year ago
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    \(\large\rm u=1+x^{1/3}\quad\to\quad x^{1/3}=u-1\)

  32. zepdrix
    • one year ago
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    So there is our dx,\[\large\rm 3(u-1)^2du=dx\]

  33. zepdrix
    • one year ago
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    Was that super confusing? :o

  34. Jhannybean
    • one year ago
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    Oh why did I miss this lol D'oh!

  35. anonymous
    • one year ago
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    As I have said earlier, you have sufficient knowledge to tackle this problem. Just see to it that you finish what you started so you can learn different techniques along the way.

  36. anonymous
    • one year ago
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    okay, haha yes a bit confusing but i think i follow.. sort of…. :P what do i do next?

  37. zepdrix
    • one year ago
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    Substitute in your pieces,\[\large\rm \color{orangered}{1+\sqrt[3]{x}=u},\qquad\qquad \color{royalblue}{dx=3(u-1)^2du}\]\[\large\rm \int\limits \frac{1}{\color{orangered}{1+\sqrt[3]{x}}}\color{royalblue}{dx}=?\]

  38. zepdrix
    • one year ago
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    And from there, it shouldn't be too bad :) You have to expand out the square in the numerator, and you can divide each term by u, and integrate term by term.

  39. anonymous
    • one year ago
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    best coaching ever

  40. anonymous
    • one year ago
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    okay, er so do i get this? \[\frac{ 3x ^{2/3} }{ 2 } - 3\sqrt[3]{x} + 3\log(\sqrt[3]{x}+1) + C \]

  41. anonymous
    • one year ago
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    @zepdrix ?

  42. zepdrix
    • one year ago
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    sec, doing some calculations :)

  43. anonymous
    • one year ago
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    okie :)

  44. zepdrix
    • one year ago
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    What did you get after integrating in u? I feel like you're missing a 2 in the middle maybe.\[\large\rm =3\left(\frac{1}{2}u^2-2u+\ln u\right)\]Something similar or no? :o We can substitute back in a sec, I'm just curious if yours looks the same up to this point.

  45. anonymous
    • one year ago
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    yes, i have that :

  46. anonymous
    • one year ago
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    :)

  47. zepdrix
    • one year ago
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    When you get to this point, you normally have two options: ~find new boundary values for your integral, in terms of u, instead of x. ~or undo your substitution and use the original values for integrating. I have a feeling... that with this monster it's going to be easier to get new boundaries for u as opposed to subbing back in those weird x's.\[\large\rm x=0 \qquad\to\qquad u=?\]\[\large\rm x=1\qquad\to\qquad u=?\]

  48. anonymous
    • one year ago
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    ohh okay :) u=1 , u=2 ?

  49. zepdrix
    • one year ago
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    \[\large\rm =3\left(\frac{1}{2}u^2-2u+\ln u\right)_1^2\]Mmm sounds good!

  50. anonymous
    • one year ago
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    so now i plug in? and i get 3((2-4+ln(2) -(1/2-2+ln(1)) ?

  51. anonymous
    • one year ago
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    so 3(-2+ln(2) - 1/2 +2 -ln(1) = 3(-1/2 +ln(1) ?

  52. zepdrix
    • one year ago
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    Your ln(2) disappeared, uh oh

  53. anonymous
    • one year ago
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    oh oops so it should be 3(-1/2 + ln(2) + ln(1)) ?

  54. zepdrix
    • one year ago
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    You could go a tad further with it if you wanted, ln(1)=0. \[\large\rm =ln(2^3)-\frac{3}{2}\] but whatever, yay good job \c:/

  55. anonymous
    • one year ago
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    ooh okay, so would that be my solution?

  56. zepdrix
    • one year ago
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    Yes, that or a decimal value, depending on which your teacher prefers.

  57. anonymous
    • one year ago
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    ooh yay!! thanks so much!! to all of you!!:)

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