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Thread: fundamental theorem of calculus

  1. #1
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    fundamental theorem of calculus

    compute the following
    (a) d/dx[ integrate(x to e^(x^2)) (ln(t))dt]
    (b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by gracy View Post
    compute the following
    (a) d/dx[ integrate(x to e^(x^2)) (ln(t))dt]
    $\displaystyle \frac{d}{dx} \int_x^{e^{x^2}} dt \, ln(t)$

    = $\displaystyle \frac{d}{dx} \left ( t \cdot ln(t) - t \right )|_x^{e^{x^2}}$

    = $\displaystyle \frac{d}{dx} \left ( e^{x^2} \cdot x^2 - e^{x^2} - x \cdot ln(x) + x \right )$

    = $\displaystyle 2xe^{x^2} + 2x^3e^{x^2} - 2xe^{x^2} - ln(x) - x \cdot \frac{1}{x} + 1$

    = $\displaystyle 2x^3e^{x^2} - ln(x)$

    -Dan
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  3. #3
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    Quote Originally Posted by gracy View Post
    (b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]
    Again I am so angry that you do not mention a function on what domain.
    But since,
    $\displaystyle e^{x^2}>0$ this function is on the entire real domain.

    You have,
    $\displaystyle f(x)=\int_2^{e^{x^2}}\ln t dt$

    You can consider this function as a composition of functions.
    $\displaystyle g(x)=\int_2^x \ln tdt,\, x>0$
    $\displaystyle h(x)=e^{x^2}$
    Then,
    $\displaystyle f =g\circ h$
    Since, $\displaystyle h$ is differenciable everywhere and $\displaystyle g$ is differenciable for $\displaystyle x>0$ for that is the fundamental theorem. Thus,
    $\displaystyle f'=h' g'\circ h$
    Thus,
    $\displaystyle f'=2xe^{x^2} \ln (e^{x^2})=2x^3e^{x^2}$ for $\displaystyle x>0$.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by gracy View Post
    (b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]
    $\displaystyle \frac{d}{dx} \int_2^{e^{x^2}} dt \, ln(t)$

    = $\displaystyle \frac{d}{dx}(t \cdot ln(t) - t)|_2^{e^{x^2}}$

    = $\displaystyle \frac{d}{dx} \left ( e^{x^2} \cdot x^2 - e^{x^2} - 2 \cdot ln(2) + 2 \right )$

    = $\displaystyle 2xe^{x^2} + 2x^3e^{x^2} - 2xe^{x^2}$

    = $\displaystyle 2x^3e^{x^2}$

    -Dan
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