# Thread: Fundamental Theorem of Calculus, Part I

1. ## 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?

2. Originally Posted by SyNtHeSiS
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 $\int_1^x 2t dt= x^2- 1^2= x^2- 1$.

3. Makes sense. So basically it differs from an indefinite integral by not containing the constant c?

4. Originally Posted by SyNtHeSiS
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.