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Math Help - Mean value theorem for integrals?

  1. #1
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    Mean value theorem for integrals?

    Find the value of c guaranteed by the mean value theorem for integrals for the function over the indicated interval

    f(x)= x-2 square root(x) [0,2]

    I am not really sure how to do this as my textbook does not really provide an example. Can anyone show me how to solve it?
    Last edited by homeylova223; August 14th 2011 at 06:49 PM.
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    MHF Contributor Siron's Avatar
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    Re: Mean value theorem for integrals?

    Are you familiar with this proposition:
    If a function f is continuous in [a,b] and differentiable in ]a,b[ then their exists an element  c \in ]a,b[ wherefore:
    \int_{a}^{b} f(x)dx = (b-a)\cdot f(c)

    You've given a,b so you can integrate the function and calculate c.
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  3. #3
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    Re: Mean value theorem for integrals?

    I have integrated the function to get

    (1/2)x^2-(4/3)x^(3/2)+c
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    Re: Mean value theorem for integrals?

    That's indeed the primitive function. But now we have a definite integral so:
    \int_{1}^{2} (x-2\sqrt{x})dx =\int_{1}^{2}xdx - 2\int_{1}^{2}x^{\frac{1}{2}}dx=\left[\frac{x^2}{2}-\frac{4x^{\frac{3}{2}}}{3}\right]_{1}^{2}=...?

    If you've calculated this definite integral then you've:
    \int_{1}^{2} (x-2\sqrt{x})dx=(2-1)\cdot f(c)
    If f(x)=x-2\sqrt{x} then f(c)=c-2\sqrt{c}

    That means:
    \int_{1}^{2} (x-2\sqrt{x})dx=c-2\sqrt{c}

    Solve this equation for c
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    Re: Mean value theorem for integrals?

    I got -1.771 when plugging in [0,2] but I am still confused on finding c. I made I typo also I mean between [0,2].
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    Re: Mean value theorem for integrals?

    If the interval is [0,2] then:
    \int_{0}^{2} (x-2\sqrt{x})dx=2-\frac{8\sqrt{2}}{3} which is approximately -1,771...
    So:
    -1,77124=2\cdot(c-2\sqrt{c})

    Solve this equation for c.
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    Re: Mean value theorem for integrals?

    So I have

    -1.771=2(c-2 square root(c))

    -1.771=2c-4c

    -1.771=-2c
    c=.8855 ?

    Yet the back of my book says .4380 as the answer.
    Last edited by homeylova223; August 14th 2011 at 05:38 PM.
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    Re: Mean value theorem for integrals?

    Quote Originally Posted by homeylova223 View Post
    So I have

    -1.771=2(c-2 square root(c))

    -1.771=2c-4c

    -1.771=-2c
    c=.8855 ?

    Yet the back of my book says .4380 as the answer.
    ...
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    Re: Mean value theorem for integrals?

    Hmm what would 2 times 2 square root(c) be?
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    Re: Mean value theorem for integrals?

    Quote Originally Posted by homeylova223 View Post
    Hmm what would 2 times 2 square root(c) be?
    4 square root(c)
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    Re: Mean value theorem for integrals?

    Alright I have

    2c-4 square root(c)=-1.771 I am actually having an problem solving this do I have to square the square square root c and -1.771

    How would I go about solving this?
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  12. #12
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    Re: Mean value theorem for integrals?

    Quote Originally Posted by homeylova223 View Post
    Alright I have

    2c-4 square root(c)=-1.771 I am actually having an problem solving this do I have to square the square square root c and -1.771

    How would I go about solving this?

    2-\frac{8}{3}\sqrt{2}=2(c-2\sqrt{c})

    1-\frac{4}{3}\sqrt{2}=c-2\sqrt{c}

    1-\frac{4}{3}\sqrt{2}-c=-2\sqrt{c}

    (1-\frac{4}{3}\sqrt{2}-c)^2=(-2\sqrt{c})^2

    1+\frac{32}{9}+c^2-\frac{8}{3}\sqrt{2}-2c+c \frac{8}{3}\sqrt{2}=4c

    \frac{41}{9}+c^2-\frac{8}{3}\sqrt{2}-2c+c \frac{8}{3}\sqrt{2}=4c

    c^2+c(\frac{8}{3}\sqrt{2}-4-2)+\frac{41}{9}-\frac{8}{3}\sqrt{2}=0

    c^2+c(\frac{8}{3}\sqrt{2}-6)+\frac{41}{9}-\frac{8}{3}\sqrt{2}=0

    Solve this for c.
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