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Math Help - 2nd Fundamental Theorem of Calculus

  1. #1
    wwilpi
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    2nd Fundamental Theorem of Calculus

    These problems are found in a calculus text at the end of a section on the 2nd FTC. I have examined these for a couple of hours, and cannot come to a solution.

    Given: The indefinite integral[v(t-k)dt] = S(t) + C, where 'k' and 'C' are constants.
    Find: S'(t)+c

    Given: The indefinite integral[a*f(x)dx] = g(ax) + C, where 'C' is a constant and 'a' is a constant <>0.
    Find: g'(x)
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  2. #2
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    Quote Originally Posted by wwilpi View Post
    Given: The indefinite integral[v(t-k)dt] = S(t) + C, where 'k' and 'C' are constants.
    Find: S'(t)+c
    The problem is these are not so well-defined and lack math rigor. So I am going to change it slightly.

    What you should have written is,
    \int _{x_0}^x v(t-k) dt=S(x)+C
    Now, the function on the left is a composition function,
    f(t)=t-k and, g(t)=\int_{x_0}^x v(t) dt
    Then,
    \int_{x_0}^x v(t-k) dt= g\circ f
    Then the derivative of this is evaluated throw the can rule.
    Since f,g are differenciable we have,
    f' \cdot g'\circ f
    Now,
    f'=1
    g'=v(t)----> 2nd Fundamental Theorem.
    Thus,
    v(t-k)=S'(x)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by wwilpi View Post
    These problems are found in a calculus text at the end of a section on the 2nd FTC. I have examined these for a couple of hours, and cannot come to a solution.

    Given: The indefinite integral[v(t-k)dt] = S(t) + C, where 'k' and 'C' are constants.
    Find: S'(t)+c
    <br />
\int v(t-k) dt = S(t)+C<br />

    is a sloppy way of writing:

    <br />
\int_0^t v(\tau-k) d\tau + S(0)+C = S(t)+C<br />

    Now differentiate this with respect to t to get:

    <br />
v(t-k) = S'(t)<br />

    so:

    <br />
S'(t)+c = v(t-k)+c<br />

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by wwilpi View Post

    Given: The indefinite integral[a*f(x)dx] = g(ax) + C, where 'C' is a constant and 'a' is a constant <>0.
    Find: g'(x)
    Similar to last problem, applying the same argumet we get:

    <br />
a f(x) = \frac{d}{dx} g(ax) =a\ g'(ax)<br />

    so:

    <br />
g'(x)=f(x/a)<br />

    RonL
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