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anonymous

  • 5 years ago

Evaluate: anti derivative Square root (3x+9x) dx would the answer be square root (3/2x^2+9x) + c

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  1. amistre64
    • 5 years ago
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    sqrt(12x) .... is that correct?

  2. anonymous
    • 5 years ago
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    \[\sqrt{3x+9x}\]

  3. anonymous
    • 5 years ago
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    Is that what you have?

  4. anonymous
    • 5 years ago
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    \[\sqrt{3x + 9} dx\]

  5. anonymous
    • 5 years ago
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    Ah. Have you done u substitutions yet?

  6. amistre64
    • 5 years ago
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    (3x + 9)^1/2 you need a 3, so multiply it by 3/3 and draw out the (1/3)

  7. amistre64
    • 5 years ago
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    (1/3) [S] 3 (3x+9)^1/2

  8. amistre64
    • 5 years ago
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    1/3 [S] u^1/2 du is what it becomes

  9. anonymous
    • 5 years ago
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    no I have not I am having trouble with this section. Not sure when I need to substitute

  10. amistre64
    • 5 years ago
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    2sqrt(u^3)/9 if I did it right

  11. amistre64
    • 5 years ago
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    u = 3x+9 :)

  12. amistre64
    • 5 years ago
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    ..+C...

  13. amistre64
    • 5 years ago
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    after that last "bout" my brain is acting like a neanderthal

  14. radar
    • 5 years ago
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    amistre64 you've the patient of Job!

  15. amistre64
    • 5 years ago
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    lol....thanx ;)

  16. anonymous
    • 5 years ago
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    As a u substitution you would just say \[u = 3x+9 \implies du = 3dx \implies dx = \frac{du}{3}\] Then \[\int \sqrt{3x+9}\ dx = \int \sqrt{u} * \frac{1}{3}du\] Then take the anti derivative of the u version and when you are done substitute back in 3x+9 for u.

  17. anonymous
    • 5 years ago
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    ok but why will I use substitution. What about this problem tells me to substitute?

  18. amistre64
    • 5 years ago
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    that "3x+9" under the radical is the clue. in order to derive down to this state, we had to use the chain rule. And that part is now missing from this equation; we essentially need to get it back.

  19. amistre64
    • 5 years ago
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    Dx(3x+9) = 3 ..... so we need to account for a "missing" 3

  20. amistre64
    • 5 years ago
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    if u = 3x+9 ; then du = 3 dx and dx= du/3

  21. amistre64
    • 5 years ago
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    sqrt(u) du -------- is what we can work out easily 3

  22. anonymous
    • 5 years ago
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    When you are finding anti-derivatives of a function composed of another function \[\int f(g(x)) dx\] You will want to use a u substitution. In this case \[f(a) = \sqrt{a}\] and \[g(x) = 3x+9\] And you're finding the anti-derivative f(g(x) dx \[f(g(x)) = f(3x+9) = \sqrt{3x+9}\]

  23. anonymous
    • 5 years ago
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    is the answer (3x+9)^2/3 / 9 +C

  24. anonymous
    • 5 years ago
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    Not quite. What did you get for the anti derivative of \[\frac{1}{3}\sqrt{u}\ du?\]

  25. anonymous
    • 5 years ago
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    Actually I see where your mistake was. You need to bring down the reciprocal of the exponent _after_ you add 1 to it.

  26. anonymous
    • 5 years ago
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    Actually, it looks like you did that. I think it was just a book keeping mistake.

  27. anonymous
    • 5 years ago
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    You're missing a factor of 2 on the front. And your exponent is flipped the wrong way. It should be: \[\int \frac{\sqrt{u}}{3}du \] \[= \int \frac{u^{1/2}}{3} du\] \[= \frac{2}{3}*\frac{u^{3/2}}{3} + C\] \[= \frac{2(3x+9)^{3/2}}{9} + C\]

  28. anonymous
    • 5 years ago
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    I see what you are doing but it just isn't clicking for me, but I am working on it. The 2/3 * u^3/2/3 The 2/3 is just the recp of 3/2???

  29. anonymous
    • 5 years ago
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    Yes. Recall that the anti-derivative of \[x^a \implies \frac{1}{a+1}*x^{a+1} + C\] We can confirm this is true, by taking the derivative of the right side and verifying that we get the left side.

  30. anonymous
    • 5 years ago
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    Where a and C are constants.

  31. anonymous
    • 5 years ago
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    I am going to start a new question

  32. anonymous
    • 5 years ago
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    Ok!

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