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Math Help - FTC2 - Not sure how to deal with this one

  1. #1
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    FTC2 - Not sure how to deal with this one

    Find g'(x) by using Part 2 of the Fundamental Theorem of Calculus.

    g(x) = \int^x_0 (1+\sqrt{t})dt

    Here is what I have done so far:

    g(x) = t + \frac{1}{2\sqrt{t}}]^x_0
    g(x) = x + \frac{1}{2\sqrt{x}} - (0 + \frac{1}{2\sqrt{0}})

    Assuming that these two steps are correct (are they?), how do handle the fact that \frac{1}{2\sqrt{0}} is undefined?

    Thanks!
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  2. #2
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    pickslides's Avatar
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    Those steps aren't quite right, consider

    \displaystyle \int t^{\frac{1}{2}} ~dt= \frac{2}{3}\times t^{\frac{3}{2}}+C
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  3. #3
    Super Member TheChaz's Avatar
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    Obviously, you are able to compute the integral and then take the derivative. This will not always be the case, and eventually (!) you will become familiar with the following:
    g(x) = \int^x_a f(t)dt = f(x)

    Oh... it's on wikipedia!
    Fundamental theorem of calculus - Wikipedia, the free encyclopedia

    Just replace the "t" inside the integral with x.
    If your upper limit has an "interesting" derivative, you'll need to use the chain rule.
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  4. #4
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    I took the derivative instead of the antiderivative. Duh. Thanks, pickslides, for pointing that out. Chaz, the question specifically requires that I use part 2 of the FTC. Your post uses part 1 (at least that's what our book calls them). The question actually makes us do it both ways. I got the answer easily using part 1, but when I did it with part 2, I got hung up. Now I know why.

    Thanks to both of you.
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