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Calculating the definite integral using FTC pt 2

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Sorry that I am not up on latex yet, but will describe the problem the best I can.
    On the interval of a=1 to b= 4 for X. ∫√5/√x.

    2. Relevant equations


    3. The attempt at a solution
    My text indicates the answer is 2√5. I have taken my anti derivative and plugged in b and subtracted plugged in a. Perhaps I am having an issue with the exponents. Thanks for any advice.
     
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  3. Nov 23, 2016 #2

    stevendaryl

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    The general rule for the integral of powers of [itex]x[/itex] is:

    [itex]\int_{a}^{b} x^n dx = \frac{x^{n+1}}{n+1}|_a^b = \frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1}[/itex]

    What is the value of [itex]n[/itex] for your problem?
     
  4. Nov 23, 2016 #3
    I have ∫5^(1/2) * x^(-1/2). then get [10^(3/2)/3]*[2x^(1/2)]. Not sure what you mean by n value for this problem. My interval a,b is 1,4?
     
  5. Nov 23, 2016 #4

    Ray Vickson

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    Ca
    Can you write ##1/\sqrt{x}## as ##x^n## for some ##n##?
     
  6. Nov 23, 2016 #5

    stevendaryl

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    You already got it: [itex]n=\frac{-1}{2}[/itex].

    You seem to be getting confused about the factor [itex]5^\frac{1}{2}[/itex]. Why don't you first figure out the answer to the simpler problem:

    [itex]\int_1^4 x^{-\frac{1}{2}} dx[/itex]
     
  7. Nov 23, 2016 #6
    Answering that problem I get the anti-derivative of 2x^(1/2). subtracting a from b I get 3.
     
  8. Nov 23, 2016 #7

    Ray Vickson

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    OK, so now put back the ##\sqrt{5}## in the correct way.
     
  9. Nov 24, 2016 #8

    stevendaryl

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    Well, that's not correct. Could you post your calculation?
     
  10. Nov 24, 2016 #9
    My bad, I re-did it and got 2. I still get the anti derivative of 2x^(1/2).
     

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  11. Nov 24, 2016 #10

    stevendaryl

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    Now, multiply by [itex]\sqrt{5}[/itex], and what do you get?
     
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