# Fundamental Theorem of Calculus, Part I

#### 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?

#### HallsofIvy

MHF Helper
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 $$\displaystyle \int_1^x 2t dt= x^2- 1^2= x^2- 1$$.

• Also sprach Zarathustra

#### SyNtHeSiS

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

#### HallsofIvy

MHF Helper
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.