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  1. #1
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    average value of function

    Find the average value of the function on the interval?

    H(x)= (cosx)^4sinx [0,pie]
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  2. #2
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    Quote Originally Posted by twilightstr View Post
    Find the average value of the function on the interval?

    H(x)= (cosx)^4sinx [0,pie]
    Average value of the function f(x) over the interval [a, b] is given by \frac{\int_a^b f(x) \, dx}{b - a}.

    For your question the integral is calculated by making the substitution u = \cos x.
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  3. #3
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    Quote Originally Posted by twilightstr View Post
    Find the average value of the function on the interval?

    H(x)= (cosx)^4sinx [0,pie]
    Alternate method of doing the integral (from that of MrF's) is to observe that

    [\cos(x)]^4 \sin(x)=-\frac{1}{5}\frac{d}{dx} [\cos(x)]^5

    and then using the fundamental theorem of calculus.

    CB
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    at the last step, i got -(cos1)^5/5pie + cos(-1)^5/5pie...
    the answer in the book is 2/5pie. how was that was derived?
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    Quote Originally Posted by twilightstr View Post
    at the last step, i got -(cos1)^5/5pie + cos(-1)^5/5pie...
    the answer in the book is 2/5pie. how was that was derived?
    If you post all your working it will be easier to point out the mistake(s) you have have made.
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    favg= -1/pie integral from 1 to -1 u^4du

    =-1/pie[u^5/5] from 1 to -1
    =-1/pie[((cos(1))^5)/5 - ((cos(-1))^5)/5]
    = -((cos(1))^5)/5pie + ((cos(-1)^5)/5pie
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  7. #7
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    Quote Originally Posted by twilightstr View Post
    favg= -1/pie integral from 1 to -1 u^4du

    =-1/pie[u^5/5] from 1 to -1

    Mr F says: What follows is wrong:

    =-1/pie[((cos(1))^5)/5 - ((cos(-1))^5)/5] Mr F says: It should be -1/pi[((cos(0))^5)/5 - ((cos(pi))^5)/5]. But why do this when all you have to do is evaluate the above integral??

    = -((cos(1))^5)/5pie + ((cos(-1)^5)/5pie
    -\frac{1}{\pi} \int_1^{-1} u^4 \, du = \frac{1}{\pi} \int^1_{-1} u^4 \, du = \frac{1}{\pi} \left[ \frac{1}{5} u^5\right]_{-1}^1 = \frac{2}{5 \pi}.
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  8. #8
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    thanks, Mr Fantastic, for pointing out the flaws.
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