Simple question about antiderivative? So this has been confusing me a lot. Let $f(x)=x^2$ and let $F(x)=\displaystyle \int_{1}^{x} f(x) \, \mathrm{d}x$. Then $F(1)=0$, obviously, but the antiderivative of $f$ (which is the same as $\displaystyle \int_{1}^{x} f(x) \, \mathrm{d}x$) is $F(x)=x^3/3$, so $F(1)=1/3$, not $0$?
What caused the confusion is that I forgot that a function has infinitely many anti derivatives; they are defined in this case as $F(x)=x^3/3 + C$, and in this example $F(x)=x^3/3 -1/3$, $x^3/3$ is just a special antiderivative where $C=0$.

It does not matter what C is equal to.
The antiderivative of f is $F(x) = \int _0 ^x f(t) dt = \int _0 ^x t^2 dt = (\frac{t^3}{3} + C) |_0^x = (\frac {x^3}{3}+ C) - (\frac {0^3}{3} +C) = \frac{x^3}{3}$
$\therefore F(1) = \int _0 ^1 f(t) dt = 1/3$
