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Definite integration problem

  1. Apr 10, 2010 #1
    Can you show me step by step to solve the definite integral of 1/(sqrt(3+x)), lower bound 5 and upper bound 7?
    I'm not that good at calculus, so please explain each step.
    Thanks.
    God bless y'all
     
  2. jcsd
  3. Apr 10, 2010 #2

    HallsofIvy

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    [tex]\int_5^7 \frac{dx}{\sqrt{x+ 3}}= \int_5^7(x+3)^{-\frac{1}{2}} dx[/tex]

    Now you can make the subsitution u= x+ 3 so that du= dx and when x= 5, u= 8, when x= 6, u= 10:
    [tex]\int_8^{10} u^{-\frac{1}{2}} du[/tex]

    Use [tex]\int u^n du= \frac{1}{n+1}u^{n+1}+ C[/tex].
     
  4. Apr 11, 2010 #3
    HallsofIvy: thx for your help, but what you gave, i think, is indefinite integral, rather than a definite integral. from what i know, because definite integral is a limit, it has to end in a certain number, without a variable. thx again for your help though
     
  5. Apr 11, 2010 #4
    To solve for the definite integral, just apply the fundamental theorem of calculus. The hardest part, which HallsofIvy almost completed, was to find the antiderivative.
     
  6. Apr 11, 2010 #5
    oh, i see it...sorry for being an idiot :P
    thx, both of you guys.
     
  7. Apr 12, 2010 #6
    When I apply the Fundamental theorem of calculus in the last step, for LaTeX Code: \\int_8^{10} u^{-\\frac{1}{2}} du , do I use 8 and 10 as my a and b or do I use 5 and 7?
     
  8. Apr 12, 2010 #7
    Use 8 and 10
     
  9. Apr 12, 2010 #8

    Char. Limit

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    Well, if you keep the function in u after integrating, you can just evaluate the bounds for u, and you would use 8 and 10. If you switch u to x+3 after integration, you would evaluate the bounds for x, and you would use 5 and 7.
     
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