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Math Help - [SOLVED] Definite Intgral Involving lnx

  1. #1
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    [SOLVED] Definite Intgral Involving lnx

    This is the equation:

     \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt

    and this is what I have:

     (4xsinx)^\frac{1}{3} \Bigg|_{-1}^{lnx^2}

    Now where do I go from here?
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  2. #2
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    Quote Originally Posted by myngo9191 View Post
    This is the equation:

     \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt

    and this is what I have:

     (4xsinx)^\frac{1}{3} \Bigg|_{-1}^{lnx^2}

    Now where do I go from here?
    This follows from the fundemental theorem of calculus and what have there is not correct.

     \frac{d}{dx} \int_{a}^{g(x)} f(t)dt = f[g(x)]g`(x)

    Can you compute knowing the above?
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  3. #3
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    Quote Originally Posted by AllanCuz View Post
    This follows from the fundemental theorem of calculus and what have there is not correct.

     \frac{d}{dx} \int_{a}^{g(x)} f(t)dt = f[g(x)]g`(x)

    Can you compute knowing the above?
    Is this what I should have?

     (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2}
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by myngo9191 View Post
    Is this what I should have?

     (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2}
    From above,

     \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt = [4(lnx^2)sin(lnx^2)]^\frac{1}{3} * \frac{2}{x}

    We do not evaluate here which is what you have in your post. We are finished, this is the answer!
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  5. #5
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    Quote Originally Posted by AllanCuz View Post
    From above,

     \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt = [4(lnx^2)sin(lnx^2)]^\frac{1}{3} * \frac{2}{x}

    We do not evaluate here which is what you have in your post. We are finished, this is the answer!

    0o0h!!! Thank you s0o0 much!!!
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