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Quick help with average value function problem please

  1. Dec 15, 2004 #1
    I am in a time crunch and I am stumped........

    I need to find the average value function of f(x) = (3x+5)^2 on [1, 2]


    help please..
     
  2. jcsd
  3. Dec 15, 2004 #2

    arildno

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    What would you suppose to be the method of finding it out?
     
  4. Dec 15, 2004 #3
    .

    ok..i typed the problem wrong....its (3x+5)^5


    I know the formula is 1/b-a int from 1 to 2 f(X)dx

    but We havent gone over this along with 4 other different problems I have left....I don't know how to handle the 5th power?
     
  5. Dec 15, 2004 #4

    arildno

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    You'd know how to handle the integral [tex]\int_{1}^{2}t^{5}dt[/tex]

    Right?
     
  6. Dec 15, 2004 #5
    i think I could figure it out...we haven't gone over this material because we ran out of time...but I have figured out problems similar to that.....
     
  7. Dec 15, 2004 #6
    is this a substitution problem?
     
  8. Dec 15, 2004 #7

    arildno

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    Allright:
    If I asked you to find:
    [tex]\frac{d}{dt}\frac{1}{6}t^{6}[/tex]
    I hope you agree with me that we have:
    [tex]\frac{d}{dt}\frac{1}{6}t^{6}=t^{5}[/tex]

    Hence, to solve the problem:
    1) Make a variable substitution (you've done this type of stuff, right?): t=3x+5
    2) since t=8 when x=1 and t=11 when x=2, we get the integral:
    [tex]\frac{1}{3}\int_{8}^{11}t^{5}dt[/tex]
    which is the average value you're seeking.
     
  9. Dec 15, 2004 #8
    no....I haven't done this stuff, but with what you just posted, I might be able to figure it out....thanks, I'll try it again.........
     
  10. Dec 15, 2004 #9
    so I don't use the formula I posted above?..I am just looking in the book, its a different type of problem(polynomial), but it gets a constant as the answer and it uses a formula to get it....
     
    Last edited: Dec 15, 2004
  11. Dec 15, 2004 #10

    arildno

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    This is certainly the formula I've used:
    Since 2-1=1, we have:
    [tex]\hat{f}=\frac{1}{1}\int_{1}^{2}(3x+5)^{5}dx=\int_{1}^{2}(3x+5)^{5}dx=\int_{8}^{11}\frac{1}{3}t^{5}dt[/tex]
     
  12. Dec 15, 2004 #11
    thats what I got after you explained it before, but does it simplify from there or is that the answer? ..never mind...i see...sub back in the value of t and solve....thanks, I got it...
     
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