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Math Help - Integral proofs

  1. #1
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    Integral proofs

    Actually I have 2 questions?

    1) Prove that the integral of (df/dx) from a to be is equal to f(b) - f(a) using the definition of Riemann Integration and the definition of derivative.

    I tried using the antiderivative proof but I dont think that is the right direction on for this one.

    2) Prove that the integral of f(x)= x^2 sin(1/x) on the interval (-pi, pi) is beteen -2pi^3/3 and 2pi^3/3.

    Bit clueless on this one.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Caity View Post
    Actually I have 2 questions?

    1) Prove that the integral of (df/dx) from a to be is equal to f(b) - f(a) using the definition of Riemann Integration and the definition of derivative.

    I tried using the antiderivative proof but I dont think that is the right direction on for this one.
    where are you stuck on this one? you've learned about partitions and all that, right?

    2) Prove that the integral of f(x)= x^2 sin(1/x) on the interval (-pi, pi) is beteen -2pi^3/3 and 2pi^3/3.

    Bit clueless on this one.
    note that -1 \le \sin \left( \frac 1x \right) \le 1 for x \ne 0

    thus, -x^2 \le x^2 \sin \left( \frac 1x \right) \le x^2

    and so  \int_{- \pi}^{\pi}-x^2~dx \le  \int_{- \pi}^{\pi} x^2 \sin \left( \frac 1x \right) ~dx \le  \int_{- \pi}^{\pi} x^2~dx \Longleftrightarrow 2 \int_0^{\pi}-x^2~dx \le  \int_{- \pi}^{\pi} x^2 \sin \left( \frac 1x \right) ~dx \le  2 \int_0^{\pi} x^2~dx since x^2 is even
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  3. #3
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    Quote Originally Posted by Caity View Post
    1) Prove that the integral of (df/dx) from a to be is equal to f(b) - f(a) using the definition of Riemann Integration and the definition of derivative.
    \int_a^b f'(x) dx = f(b) - f(a)
    This is because f(x) is antiderivative of f'(x).
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  4. #4
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    Ok so its not the antiderivative thing. I'll go reread those definitions about partitions and all that.
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    Quote Originally Posted by ThePerfectHacker View Post
    \int_a^b f'(x) dx = f(b) - f(a)
    This is because f(x) is antiderivative of f'(x).
    i might be misunderstanding something here, but aren't you using what you're supposed to be proving? namely the fundamental theorem of calculus. i would think we had to go through all that stuff about finding a partition and estimating upper and lower Riemann sums etc etc
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  6. #6
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    I think I got it now... I used the definition of the derivative in the integral then used (b-a) for delta x. I replaced x for b. The (b-a)'s cancelled out and I ended up with f(b)-f(a)
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  7. #7
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    If f:[a,b]\to\mathbb R is an integrable function in the interval [a,b] and if \varphi:[a,b]\to\mathbb R is a primitive function of f in [a,b], then it verifies that \int_a^b f(x)\,dx=\varphi(b)-\varphi(a).

    Proof

    For any P=\{x_0,x_1,\ldots,x_n\} partition of [a,b], by applyin' the MVT to \varphi in each one of intervals [x_{i-1},x_i], we can ensure that it does exist a point \mu_i\in[x_{i-1},x_i] (for each i=1,2,\ldots,n) such that \varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right)=\varphi '\left( \mu _{i} \right)\left( x_{i}-x_{i-1} \right)=f\left( \mu _{i} \right)\cdot \Delta x_{i}. Summing member to member the previous equalities, we get \varphi (b)-\varphi (a)=\sum\limits_{i=1}^{n}{\big( \varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right) \big)}=\sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}.

    Let m_i and M_i be the infimum and supremum of f in [x_{i-1},x_i] and according to lower and upper sums s(P) and S(P) of f to the P partition, respectively, as one wants it m_i=f\left( \mu _{i} \right)\le M_i, it yields

    s(P)\le \sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}\le S(P)\implies s(P)\le \varphi (b)-\varphi (a)\le S(P).
    Hence, \varphi (b)-\varphi (a) is included between the lower and upper sums of f for any P partition, and this property has the integral \int_a^b f, from here \int_a^b f(x)\,dx=\varphi(b)-\varphi(a).\quad\blacksquare
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    If f:[a,b]\to\mathbb R is an integrable function in the interval [a,b] and if \varphi:[a,b]\to\mathbb R is a primitive function of f in [a,b], then it verifies that \int_a^b f(x)\,dx=\varphi(b)-\varphi(a).

    Proof

    For any P=\{x_0,x_1,\ldots,x_n\} partition of [a,b], by applyin' the MVT to \varphi in each one of intervals [x_{i-1},x_i], we can ensure that it does exist a point \mu_i\in[x_{i-1},x_i] (for each i=1,2,\ldots,n) such that \varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right)=\varphi '\left( \mu _{i} \right)\left( x_{i}-x_{i-1} \right)=f\left( \mu _{i} \right)\cdot \Delta x_{i}. Summing member to member the previous equalities, we get \varphi (b)-\varphi (a)=\sum\limits_{i=1}^{n}{\big( \varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right) \big)}=\sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}.

    Let m_i and M_i be the infimum and supremum of f in [x_{i-1},x_i] and according to lower and upper sums s(P) and S(P) of f to the P partition, respectively, as one wants it m_i=f\left( \mu _{i} \right)\le M_i, it yields

    s(P)\le \sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}\le S(P)\implies s(P)\le \varphi (b)-\varphi (a)\le S(P).
    Hence, \varphi (b)-\varphi (a) is included between the lower and upper sums of f for any P partition, and this property has the integral \int_a^b f, from here \int_a^b f(x)\,dx=\varphi(b)-\varphi(a).\quad\blacksquare
    so you're great on the theory of integration as well i see
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  9. #9
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    What about the definition of the derivative that I'm supposed to use along the the definition of Riemann integration??
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    where are you stuck on this one? you've learned about partitions and all that, right?

    note that -1 \le \sin \left( \frac 1x \right) \le 1 for x \ne 0

    thus, -x^2 \le x^2 \sin \left( \frac 1x \right) \le x^2

    and so  \int_{- \pi}^{\pi}-x^2~dx \le \int_{- \pi}^{\pi} x^2 \sin \left( \frac 1x \right) ~dx \le \int_{- \pi}^{\pi} x^2~dx \Longleftrightarrow 2 \int_0^{\pi}-x^2~dx \le \int_{- \pi}^{\pi} x^2 \sin \left( \frac 1x \right) ~dx \le 2 \int_0^{\pi} x^2~dx since x^2 is even
    I see on the graph that x^2 sin (1/x) is between -x^2 and x^2 but how did you know that? Also I dont quite understand the last part and how it relates to being between -2pi^3/3 and 2pi^3/3.
    Last edited by Caity; October 1st 2008 at 09:05 AM.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Caity View Post
    I see on the graph that x^2 sin (1/x) is between -x^2 and x^2 but how did you know that? Also I dont quite understand the last part and how it relates to being between -2pi^3/3 and 2pi^3/3.
    i just multiplied my first inequality through by x^2. we know \sin \left( \frac 1x \right) is between -1 and 1 (for x \ne 0). if you multiply that through by x^2, you get x^2 \sin \left( \frac 1x \right) is between -x^2 and x^2

    as far getting to the claim, just integrate each term in the last inequality i gave.
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