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Math Help - Average value of a function problems

  1. #1
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    Average value of a function problems

    Hi,

    I just want to make sure I am doing this right.

    Find the average value of the function n the given interval

    f(x)=sec^2 (x/2) [0, pi/2]

    So I integrated from 0 to pi/2. but I am stuck on what the anti derivative of sec^2 (x/2) is... I know the anti derivative of sec^2 (x) is tan (x)...


    Thanks
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  2. #2
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    \int \sec^2 \frac{x}{a}~dx = a \tan\frac{x}{a}+C ,\forall a \in \mathbb{R}
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  3. #3
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    Hi, thank you!

    I got the answer 4.
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  4. #4
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    Can you help with this problem? I don't even know what it is asking.. It's the same type of problem

    If f ave i[a,b] denotes the average value of f on the interval [a,b] and a<c<b, show that

    f ave[a,b] = (c-a)/(b-a) f ave[a,b]+ (b-c)/(b-a) f ave[c,b]

    ave means average.

    Much thanks! I am stomped on that one.
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  5. #5
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    Do you mean show that \frac{1}{b-a}\int_a^b f(x)~dx=\frac{c-a}{b-a}\int_a^c f(x)~dx +\frac{b-c}{b-a}\int_c^b f(x)~dx ?
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  6. #6
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    I am not sure..... I wish I had a scanner. It basically says this:

    If f ave[a,b] denotes the average value of f on the interval [a,b] and a<c<b, show that

    f ave[a,b] = (c-a)/(b-a) f ave[a,b]+ (b-c)/(b-a) f ave[c,b]

    ave means average.

    So I am guessing i have to proof it?
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  7. #7
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    here is a scanned image of the problem

    Average value of a function problems-problem-24.jpg
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  8. #8
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    Quote Originally Posted by softballchick View Post
    Click image for larger version. 

Name:	problem 24.JPG 
Views:	26 
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ID:	18369
    f_{AVE[a,b]}=\frac{1}{b-a}\int^b_a f(x)dx

    f_{AVE[a,c]}=\frac{1}{c-a}\int^c_a f(x)dx

    f_{AVE[c,b]}=\frac{1}{b-c}\int^b_c f(x)dx

    Now, put all these to you formula...
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  9. #9
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    i'm not sure how to put it in? I think it's obvious that it works, I just don't know how to show it.
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