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Definite integration problem

  • Thread starter Moonflower
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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].
 
  • #3
21
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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
 
  • #4
614
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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.
 
  • #5
21
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oh, i see it...sorry for being an idiot :P
thx, both of you guys.
 
  • #6
21
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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?
 
  • #7
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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?
Use 8 and 10
 
  • #8
Char. Limit
Gold Member
1,204
14
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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