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 $C=0$.
 A: The integral $\int_1^x x^2\,dx$ is equal to 
$$\left.\frac{x^3}{3}\right|_1^x.$$
This is $\frac{1}{3}x^3-\frac{1}{3}$. Note that  $\frac{1}{3}x^3-\frac{1}{3}$ is $0$ at $x=1$, as you pointed out it should be.
Remark: When we are finding definite integrals with limits of integration that involve variables, it is best to use a different "dummy" variable of integration. Things may be clearer if you write
$$F(x)=\int_1^x t^2\,dt.$$
While in principle the notation you used is not wrong, it is, for good reason, frowned on.
Note also that talking about "the" antiderivative of a function $f(t)$ is not quite right. The function has infinitely many (very closely related) antiderivatives.
A: What your example shows is that the anti-derivative of $\operatorname{f}$ is not, in general, given by
$$\int_1^x \operatorname{f}(t) \, \operatorname{d}\!t $$
Take your example, let $\operatorname{f}(x) = x^2$. We know that the anti-derivative if $\tfrac{1}{3}x^3+c$, yet
$$\int_1^x t^2 \, \operatorname{d}\!t = \left[ \tfrac{1}{3}t^3\right]_0^x = \tfrac{1}{3}x^3-\tfrac{1}{3} \equiv \tfrac{1}{3}(x^3-1).$$
A: It does not matter what C is equal to.
The antiderivative of f is $F(x) = \int _0 ^x f(t) dt = \int _0 ^x t^2 dt = (\frac{t^3}{3} + C) |_0^x = (\frac {x^3}{3}+ C) - (\frac {0^3}{3} +C) = \frac{x^3}{3}$
$\therefore F(1) = \int _0 ^1 f(t) dt = 1/3$
