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Math Help - integration proof using MVT

  1. #1
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    integration proof using MVT

    I have been asked to prove the following:

    Suppose that f is continuous on [a,b]. Prove that there exists a point c such that the integral from a to be is f(c)(b-a).

    Proof:
    Let F(x)= integral from a to x of f(t)dt, then by the Mean Value Theorem there exists a value f(b) such that a \leqb \leqx. Furthermore, the integral from a to b of f(t)dt can be calculated to be f(c)(b-a).

    This is all I have so far, I realize there are holes. Can you please help me with this?
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  2. #2
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    Re: integration proof using MVT

    Quote Originally Posted by CountingPenguins View Post
    Suppose that f is continuous on [a,b]. Prove that there exists a point c such that the integral from a to be is f(c)(b-a).
    Define F(x) = \int_a^x {f(t)dt} , then the function F is continuous on [a,b].

    Moreover F'(x) = f(x). So apply MVT to F(x) on [a,b].
    Last edited by Plato; July 26th 2011 at 08:33 AM.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: integration proof using MVT

    I would prove it this way, by distinguishing tree cases:
    a<b
    a=b
    a>b
    If you find a<b then you'll find the other two, you may consider the proof for a=b as trivial.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: integration proof using MVT

    Quote Originally Posted by CountingPenguins View Post
    I have been asked to prove the following:

    Suppose that f is continuous on [a,b]. Prove that there exists a point c such that the integral from a to be is f(c)(b-a).

    Proof:
    Let F(x)= integral from a to x of f(t)dt, then by the Mean Value Theorem there exists a value f(b) such that a \leqb \leqx. Furthermore, the integral from a to b of f(t)dt can be calculated to be f(c)(b-a).

    This is all I have so far, I realize there are holes. Can you please help me with this?


    A different approach...

    Let f(x) be integrable function on [a,b], then we can easily prove the following:

    i)If f(x)\geq m for all x\in[a,b] then \int_{a}^{b}f(x) \;dx \geq m(b-a)

    Proof:

    For every partition P of [a,b] and for chosen points, t_i\in\Delta x_i we eill have:

    \sigma _P(t_1,...,t_n)=\sum_{i=1}^{n}f(t_i)\Delta x_i\geq \sum_{i=1}^{n}m\Delta x_i=m(b-a).

    \int_{a}^{b}f(x) \;dx =\lim_{\Delta(P)\to 0}\sigma _P(t_1,...,t_n)\geq m(b-a).

    ii) If f(x)\leq M for all x\in[a,b] then \int_{a}^{b}f(x) \;dx \leq M(b-a)

    The proof is identical to proof in i).


    Now, by combining i) and ii) we will get the next result:

    If f(x) is integrable on [a,b] , and let M be the supremum of f(x) on [a,b] and m is the infinum of f(x) on [a,b]. Then:

    m(b-a)\leq \int_{a}^{b}f(x) \;dx\leq M(b-a) (1)


    Now, to your question:

    By dividing (1) by (b-a), we will get:


    m\leq \frac{1}{b-a}\int_{a}^{b}f(x) \;dx\leq M

    Now, using MVT we conclude the existence of c\in[a,b] such that:

    f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)
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