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Math Help - [SOLVED] Derivative of an integral

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Derivative of an integral

    From galactus,
    \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)
    . It helped me verifying my result and confirmed it.
    I got that the derivative of \int_a^{x^3} sin^3(t) dt=3x^2{sin(x^3)}^3.
    While from Mathematica, it gives the result \frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big) which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!
    Last edited by arbolis; June 29th 2008 at 06:23 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    From galactus, . It helped me verifying my result and confirmed it.
    I got that the derivative of \int_a^{x^3} sin^3(t)^3 dt=3x^2{sin(x^3)}^3.
    While from Mathematica, it gives the result \frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big) which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!
    Your answer is correct.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
    Yeah, most likely.
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  5. #5
    Senior Member nikhil's Avatar
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    Quote Originally Posted by arbolis View Post
    Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
    I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
    Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nikhil View Post
    I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
    Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
    What are you talking about? The constant terms will drop out due to the derivative.
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  7. #7
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    Quote Originally Posted by arbolis View Post
    Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
    One way to check this would be to let a = 0, say, and then put Mathematica on the job. Note that there are various equivalent forms of the correct answer.
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  8. #8
    MHF Contributor arbolis's Avatar
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    I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
    Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
    I don't understand all what you mean, but the formula galactus gave works here. (Now I realize I proved it about 9 months or so ago.) And if both of the limit of the integral were constants, say 0 and 10, then the derivative would be 0, which is true. Think about it, you have F(x)=\int_0^{10} f(t)dt. It's obvious that the integral is equal to a constant, whatever x is, since it doesn't depends of x. So the derivative of F is 0 (since the derivative of a constant is always 0).
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  9. #9
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    Quote Originally Posted by nikhil View Post
    I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
    Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
    The formula given by galactus works fine, including the case where one or both of h(x) and g(x) are equal to a constant.
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  10. #10
    Senior Member nikhil's Avatar
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    Talking Ooops

    Quote Originally Posted by Mathstud28 View Post
    What are you talking about? The constant terms will drop out due to the derivative.
    oops sorry dude forgot we were talking about its derivative. Next time i will be carefull for sure
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  11. #11
    Senior Member nikhil's Avatar
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    Talking Ooops

    Quote Originally Posted by Mathstud28 View Post
    What are you talking about? The constant terms will drop out due to the derivative.
    oops sorry dude forgot we were talking about its derivative at the end. Next time i will be carefull for sure
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  12. #12
    MHF Contributor arbolis's Avatar
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    One way to check this would be to let a = 0, say, and then put Mathematica on the job. Note that there are various equivalent forms of the correct answer.
    If I typed it well, I got that the derivative equals 8 x^2 Cos(\frac{x^3}{2})(2+Cos(x^3))Sin^3(\frac{x^3}{2})-4x^2Sin^4(\frac{x^3}{2})Sin(x^3). It surprises me more and more. If somebody has Mathematica and has some time to spend, I'll be glad if you could check out the derivative of the integral!
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  13. #13
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    Quote Originally Posted by arbolis View Post
    From galactus, . It helped me verifying my result and confirmed it.
    I got that the derivative of \int_a^{x^3} sin^3(t)^3 dt=3x^2{sin(x^3)}^3.
    While from Mathematica, it gives the result \frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big) which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!
    You have as the integrand sin^3(t)^3 .....? Is the integrand \sin^3 (t) or \sin^3(t^3)?
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  14. #14
    MHF Contributor arbolis's Avatar
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    You have as the integrand .....? Is the integrand or ?
    Sorry about that, it is \sin^3 (t), that is (sin(t))^3.

    EDIT: Finally I don't mind what Mathematica thinks. Since I didn't made this program, it doesn't interest me that much. It may has some conventions I don't know yet, that's all. Thread solved.
    Last edited by arbolis; June 30th 2008 at 10:13 AM.
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