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Thread: 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$\displaystyle \leq$b$\displaystyle \leq$x. 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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    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 $\displaystyle F(x) = \int_a^x {f(t)dt} $, then the function $\displaystyle F$ is continuous on $\displaystyle [a,b]$.

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

    I would prove it this way, by distinguishing tree cases:
    $\displaystyle a<b$
    $\displaystyle a=b$
    $\displaystyle a>b$
    If you find $\displaystyle a<b$ then you'll find the other two, you may consider the proof for $\displaystyle a=b$ 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$\displaystyle \leq$b$\displaystyle \leq$x. 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 $\displaystyle f(x)$ be integrable function on $\displaystyle [a,b]$, then we can easily prove the following:

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

    Proof:

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

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

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

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

    The proof is identical to proof in i).


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

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

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


    Now, to your question:

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


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

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

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