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Math Help - Fundamental Theorem of Calculus Help Needed

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    Fundamental Theorem of Calculus Help Needed

    Hello all, I am new here, and would like some help please. I did look around at LaTex first in an effort to make my questions more readable.

    h(x)=\int_{2}^{x^2}\sqrt{u-1}du
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    Quote Originally Posted by Zeppelin
    Hello all, I amnew here, and would like some help please. I did look around at LaTex first in an effort to make my questions more readable.

    h(x)=\int_{2}^{x^2}\sqrt{u-1}du
    Change of variable. Let y = u - 1. Then
    h(x) = \int_1^{x^2-1} dy\, \sqrt{y}

    = \frac{2}{3}y^{3/2}|_1^{x^2-1}

    = \frac{2}{3}(x^2-1)^{3/2} - \frac{2}{3}(1)^{3/2}

    = \frac{2}{3}(x^2-1)^{3/2} - \frac{2}{3}

    -Dan
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    Quote Originally Posted by Zeppelin
    Hello all, I am new here, and would like some help please. I did look around at LaTex first in an effort to make my questions more readable.

    h(x)=\int_{2}^{x^2}\sqrt{u-1}du
    I think he means he want its derivative (it makes more sense like that because many questions want you to find the derivative of the integral invloving the fundamental theorem).

    If you had,
    f(x)=\int_2^x \sqrt{u-1}du
    Then,
    f'(x)=\sqrt{x-1}
    But they have a square.

    You can image h(x) as a composition of functions.
    f(g(x)) where g(x)=x^2.
    Then its derivative involves the chain rule.

    g'(x)f'(g(x))
    But,
    g'(x)=2x
    And,
    f'(x)=\sqrt{x-1} thus f'(g(x))=\sqrt{x^2-1}
    Thus, the derivative is,
    2x\sqrt{x^2-1}
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  4. #4
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    Yes thank you Perfect Hacker, I apologize, I didn't add the question, just the equation. I was looking for the derivitive, thanks for the tutoring. I'm sure I'll be back!
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