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Evaluate the Integral sqrt(1+7s)ds

  1. May 12, 2012 #1
    1.The question is Evaluate the Integral [itex]\sqrt{1+7s}[/itex]ds



    2. Ok, so i've tried this problem but I'm not coming up with the correct solution. Could you help me.



    3. This is how i'm handling the problem.

    Evaluate: (1+7s) ^1/2
    = [itex]\frac{2(1+7s)^(3/2)}{3}[/itex] +C



    The answer is supposed to be 2/21 (1+7s) ^(3/2) + C Could someone please "spell out" how to get this. I dont understand! THank you!
     
  2. jcsd
  3. May 12, 2012 #2
    take the derivative of [itex]\frac{2}{3}(1+7s)^{3/2}[/itex] and then you'll probably see what's going wrong. You're forgetting the chain rule.

    EDIT: let [itex]f(s)=s^{3/2}[/itex] and [itex]g(s)=1+7s[/itex]. Then [itex]\frac{2}{3}(1+7s)^{3/2}=\frac{2}{3}f(g(s))[/itex] and by the chain rule [itex]\frac{d}{ds}(\frac{2}{3}(1+7s)^{3/2})=\frac{2}{3}f'(g(s))*g'(s)[/itex]. Can you tell me what f'(g(s)) and g'(s) are?
     
    Last edited: May 12, 2012
  4. May 12, 2012 #3
    What you did is apply,

    [tex]\int \sqrt{x} dx = \frac{2x^{\frac{3}{2}}}{3} + C[/tex]

    What happens when your given integral is

    [itex]\int \sqrt{ax+b}\ dx[/itex]

    Hint: Use substitution.
     
  5. May 12, 2012 #4
    AHHHH!! I got it!! THank you guys so much!!! I truly appreciate it!
     
  6. May 13, 2012 #5
    What happens when you are taking the integral of something like this.....



    ∫[itex]\frac{6r^2}{\sqrt{6-r^3}}[/itex]

    I tried:
    u = 6-r^3
    du = 3r^2

    but i cannot get the correct answer. Help would be great!
     
  7. May 13, 2012 #6

    HallsofIvy

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    With [itex]u= 6- r^3[/itex], [itex]du= -3r^2 dr[/itex]. You seem to have left out the negative sign.
     
  8. May 13, 2012 #7

    sharks

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    Your integral in post #5 is in this general form:
    [tex]\int \frac{f'(x)}{\sqrt{f(x)}}\,dx=2\sqrt{f(x)}+C[/tex]
    In that case, you just rearrange the real constant coefficient of the numerator to match that of f'(x).
     
    Last edited: May 13, 2012
  9. May 13, 2012 #8
    Great.. So if I use that equation then I am getting


    -2 (6-r^3)^(1/2) + C

    But the book says it should be -4(6-r^3)^(1/2) +C
    What am i missing?
    (thankyou so much for the help!!)
     
  10. May 13, 2012 #9

    sharks

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    You wrote your integral wrong. It should be:
    [tex]\int \frac{6r^2}{\sqrt{6-r^3}}\,dr[/tex]
     
  11. May 13, 2012 #10
    You should write out the steps that you've tried.

    You let [itex]u=6-r^{3}[/itex] and [itex]du=-3r^{2}dr[/itex].

    Then [itex]3r^2dr=-du[/itex]

    However, notice that the numerator is actually [itex]6r^{2}[/itex], not [itex]3r^{2}[/itex]

    Also, can you tell me what is [itex]\int\frac{1}{\sqrt{x}}dx[/itex]?
     
    Last edited: May 13, 2012
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