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Math Help - Integrals and Proofs

  1. #1
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    Integrals and Proofs

    Hey there,

    Here's the question:

    a.) Prove that if f is integrable on [a,b] and m <= f(x) <= M for all [a,b], then:
    <br />
\int_a^b f(x) dx = (b-a)k<br />
    for some number "k" with m <= k <= M

    b.) Prove that if f is continuous on [a,b], then
    <br />
\int_a^b f(x) dx = (b-a)f(q)<br />
    for some q in [a,b]


    So for the work I set up that U(f,P) - L(f,P) < E, which implies the usualy equasion of the Sum of (Mi-mi)(Change in ti) < E. But I'm unsure where I should start throwing in this "k" variable? Thanks
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    Consider the function

    F(x)=\int_{a}^{x}f(t)dt

    First note that F(x) is differentiable (why)

    Now use the mean value theorem on [a,b]

    F(b)-F(a)=F'(c)(b-a)

    Can you finish from here
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  3. #3
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    F(x) is differentiable because we have assumed that the function f(t) is integrable?
    From there, using the mean value theorem I would see that in the end we'd get

    <br />
\int_{a}^{b}f(t)dt - \int_{a}^{a}f(t)dt = F'(c)(b-a)

    <br />
\int_{a}^{b}f(t)dt = F'(c)(b-a)<br />

    Does this end up meaning F'(c) = k? and thus we have our result?
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    Remember what the function

    F(x)=\int_{a}^{x}f(t)dt represents and you still need to show that
    m < k < M

    What is F'(x) equal to?

    So the question is why is m< F'(c) < M for some  c \in (a,b)
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    Would F'(x) = f(x), then? So F'c = f(c). In this case does this f(c) become our k? I'm still having some trouble trying to figure out why exactly F'(c) does stay between m and M. Could I say f(c) (or k) would have to be between m and M due to previous theorems regarding continuity (if f(a) < c < f(b), there exists some x with f(x) = c? and that we do obviously have some bounds?
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    Quote Originally Posted by mscbuck View Post
    Would F'(x) = f(x), then? So F'c = f(c). In this case does this f(c) become our k? I'm still having some trouble trying to figure out why exactly F'(c) does stay between m and M. Could I say f(c) (or k) would have to be between m and M due to previous theorems regarding continuity (if f(a) < c < f(b), there exists some x with f(x) = c? and that we do obviously have some bounds?
    hmm. I think that this proof will only work for part b when f is continuous.

    The function F(x)=\int_{a}^{x}f(t)dt can only be differentiated by the FTC when f(t) is continuous.

    So here is a different proof:
    Since
    \displaystyle m < f(x) < M \implies \int_{a}^{b}mdx < \int_{a}^{b}f(x)dx< \int_{a}^{b}Mdx

    \displaystyle m(b-a) < \int_{a}^{b}f(x)dx < M(b-a) \implies m < \frac{\int_{a}^{b}f(x)dx}{b-a} < M

    Now choose \displaystyle k=\frac{\int_{a}^{b}f(x)dx}{b-a}
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  7. #7
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    Thanks EmptySet! I actually figured out part a. right before you posted, but I did the exact same steps so I'm glad. An add-on to this is a question that asks "Show by an example that continuity is essential". I'm unsure of what he is asking, because I know that we can have discontinuous functions that still can be integrated. Continuity is sufficient but not necessary for integrability right?
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    Quote Originally Posted by mscbuck View Post
    Thanks EmptySet! I actually figured out part a. right before you posted, but I did the exact same steps so I'm glad. An add-on to this is a question that asks "Show by an example that continuity is essential". I'm unsure of what he is asking, because I know that we can have discontinuous functions that still can be integrated. Continuity is sufficient but not necessary for integrability right?
    Yes that is correct consider the function

    \displaystyle f(x)=\begin{cases} 0 \text{ if } 0 \le x \le \frac{1}{2} \\ 1 \text{ if } \frac{1}{2} < x \le 1 \end{cases}

    and the integral \displaystyle \int_{0}^{1}f(x)dx
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