Fundamental Theorem of Calculus, Part I

Apr 2010
156
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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?
 

HallsofIvy

MHF Helper
Apr 2005
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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- \(\displaystyle \int_a^b f(t)dt\) is a number, a constant. \(\displaystyle \int_a^x f(t) dt\) is a function of x.

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

More subtle is the difference between \(\displaystyle \int_a^x f(t)dt\) and the \(\displaystyle \int f(x)dx\), the "indefinite integral" or "anti-derivative". \(\displaystyle \int 2x dx= x^2+ C\) where we need the constant C since \(\displaystyle (x^2+ C)'= 2x\) for any C. The "a" in \(\displaystyle \int_a^x f(t)dt\) determines the specific C: \(\displaystyle \int_0^x 2t dt= x^2- 0^2= x^2\) while \(\displaystyle \int_1^x 2t dt= x^2- 1^2= x^2- 1\).
 
Apr 2010
156
0
Makes sense. So basically it differs from an indefinite integral by not containing the constant c?