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Math Help - integral problem...

  1. #1
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    integral problem...

    Q: let f(x)=integral of e^(-t^2) limit from sin(x) to cos(x).Then find f'(pi/4)=
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    Quote Originally Posted by Mathventure View Post
    Q: let f(x)=integral of e^(-t^2) limit from sin(x) to cos(x).Then find f'(pi/4)=
    f(x) = \int_{\sin x}^{\cos x} e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt + \int_{\sin x}^a e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt - \int_a^{\sin x} e^{-t^2}\,dt so

    f'(x) = - e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x

    Then sub. in your value.
    Last edited by Jester; April 24th 2009 at 07:43 AM. Reason: I missed a negative - how'd that happen?
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    f(x) = \int_{\sin x}^{\cos x} e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt + \int_{\sin x}^a e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt - \int_a^{\sin x} e^{-t^2}\,dt so

    f'(x) = e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x

    Then sub. in your value.
    after putting value i got zero but options available are
    a)sqrt(1/e)
    b)-sqrt(2/e)
    c)sqrt(2/e)
    d)-sqrt(1/e)
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  4. #4
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    Quote Originally Posted by Mathventure View Post
    after putting value i got zero but options available are
    a)sqrt(1/e)
    b)-sqrt(2/e)
    c)sqrt(2/e)
    d)-sqrt(1/e)
    that was just a typo and you should have noticed it. what danny meant was this: f'(x) = -e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x.
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    Quote Originally Posted by NonCommAlg View Post
    that was just a typo and you should have noticed it. what danny meant was this: f'(x) = -e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x.
    but how he solved integral??
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  6. #6
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    you don't need to solve the integral ... use the 2nd FTC.

    f(x) = \int_{\sin{x}}^{\cos{x}} e^{-t^2} \, dt

    f'(x) = -e^{-\cos^2{x}} \sin{x} - e^{-\sin^2{x}} \cos{x}

    f'\left(\frac{\pi}{4}\right) = -e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2} - e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2}<br />

    f'\left(\frac{\pi}{4}\right) = -\sqrt{\frac{2}{e}}<br />
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  7. #7
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    Quote Originally Posted by skeeter View Post
    you don't need to solve the integral ... use the 2nd FTC.

    f(x) = \int_{\sin{x}}^{\cos{x}} e^{-t^2} \, dt

    f'(x) = -e^{-\cos^2{x}} \sin{x} - e^{-\sin^2{x}} \cos{x}

    f'\left(\frac{\pi}{4}\right) = -e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2} - e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2}<br />

    f'\left(\frac{\pi}{4}\right) = -\sqrt{\frac{2}{e}}<br />
    i was really silly that i forgot FTC....
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  8. #8
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    Quote Originally Posted by Mathventure View Post

    but how he solved integral??
    let \int_a^x f(t) \ dt = g(x), where a is a constant. by the fundamental theorem of calculus we have g'(x) = f(x). now if u(x),v(x) are differentiable functions, then:

    h(x)=\int_{u(x)}^{v(x)} f(t) \ dt = g(v(x)) - g(u(x)). so by the chain rule: h'(x)=v'(x)g'(v(x)) - u'(x)g'(u(x))=v'(x)f(v(x))-u'(x)f(u(x)).
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  9. #9
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    Edit: It really takes too long typing latex sometimes. =p

    Quote Originally Posted by Mathventure View Post
    but how he solved integral??
    What's to solve?

    f(x) = \int_{sinx}^{cosx} e^{-t^2}dt

    If that's f(x), that little function in the integral looks like f'(x)

    e^{-(cosx)^2}(-sinx) - e^{-(sinx)^2}(cosx)
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  10. #10
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    Quote Originally Posted by NonCommAlg View Post
    let \int_a^x f(t) \ dt = g(x), where a is a constant. by the fundamental theorem of calculus we have g'(x) = f(x). now if u(x),v(x) are differentiable functions, then:

    h(x)=\int_{u(x)}^{v(x)} f(t) \ dt = g(v(x)) - g(u(x)). so by the chain rule: h'(x)=v'(x)g'(v(x)) - u'(x)g'(u(x))=v'(x)f(v(x))-u'(x)f(u(x)).
    is there any way to solve the value of integral directly??
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  11. #11
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    Quote Originally Posted by Mathventure View Post
    is there any way to solve the value of integral directly??
    NO!
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  12. #12
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    Quote Originally Posted by Mathventure View Post
    is there any way to solve the value of integral directly??
    If you allow error functions then

    f(x) = \frac{\sqrt{\pi}}{2} \left( \text{erf}(\cos x) - \text{erf}(\sin x)\right)
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  13. #13
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    integral problem...

    Quote Originally Posted by NonCommAlg View Post
    NO!
    then how can we solve double integral of x*e^(y^2) over region R, where R is region bounded by lines x=0,y=1 and the parabola y=x^2.
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  14. #14
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    Quote Originally Posted by Mathventure View Post
    then how can we solve double integral of x*e^(y^2) over region R, where R is region bounded by lines x=0,y=1 and the parabola y=x^2.
    By making sure you integrate in an order that makes the problem tractable:

    \int_{y = 0}^1 \int_{x = 0}^{x = + \sqrt{y}} x e^{y^2} \, dx \, dy.

    There is a world of difference between double integrals and single integrals.


    (I can't resist adding that you should have asked danny "How on erf do you do it?" .... Memo to baldeagle: I am the funniest )
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  15. #15
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    Quote Originally Posted by mr fantastic View Post

    (I can't resist adding that you should have asked danny "How on erf do you do it?" .... Memo to baldeagle: I am the funniest )
    Now that's funny
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