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Math Help - 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]
    \frac{d}{dx} \int_x^{e^{x^2}} dt \, ln(t)

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

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

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

    = 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,
    e^{x^2}>0 this function is on the entire real domain.

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

    You can consider this function as a composition of functions.
    g(x)=\int_2^x \ln tdt,\, x>0
    h(x)=e^{x^2}
    Then,
    f =g\circ h
    Since, h is differenciable everywhere and g is differenciable for x>0 for that is the fundamental theorem. Thus,
    f'=h' g'\circ h
    Thus,
    f'=2xe^{x^2} \ln (e^{x^2})=2x^3e^{x^2} for 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]
    \frac{d}{dx} \int_2^{e^{x^2}} dt \, ln(t)

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

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

    = 2xe^{x^2} + 2x^3e^{x^2} - 2xe^{x^2}

    = 2x^3e^{x^2}

    -Dan
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