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Math Help - Fundamental Theorem of Calculus, Part I

  1. #1
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    Fundamental Theorem of Calculus, Part I

    For g(x) = ∫ a to x f(t)dt, I would like to know why there is an x at the upper limit and why it not equal to b, what is the difference?
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  2. #2
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    Quote Originally Posted by SyNtHeSiS View Post
    For g(x) = ∫ a to x f(t)dt, I would like to know why there is an x at the upper limit and why it not equal to b, what is the difference?
    There's a very important difference- \int_a^b f(t)dt is a number, a constant. \int_a^x f(t) dt is a function of x.

    For example if f(t)= 2t+ 1, then \int_1^2 f(t)dt= \int_a^b 2t+ 1 dt= \left[ t^2+ t\right]_1^2= 2^2+ 2- 1^2- 1= 4 while \int_1^x f(t)dt= \left[ t^2+ t\right]_a^x= x^2+ x- 1^2- 1= x^2+ x- 2.

    More subtle is the difference between \int_a^x f(t)dt and the \int f(x)dx, the "indefinite integral" or "anti-derivative". \int 2x dx= x^2+ C where we need the constant C since (x^2+ C)'= 2x for any C. The "a" in \int_a^x f(t)dt determines the specific C: \int_0^x 2t dt= x^2- 0^2= x^2 while \int_1^x 2t dt= x^2- 1^2= x^2- 1.
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    Makes sense. So basically it differs from an indefinite integral by not containing the constant c?
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    Quote Originally Posted by SyNtHeSiS View Post
    Makes sense. So basically it differs from an indefinite integral by not containing the constant c?
    Yes. The specific c is determined by the choice of lower limit.
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