TITLE: Is the following statement true? $\int_{-1}^{x}\ {2x+x^2}\,dx = \int_{-1}^{x}\ F'(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}\,dt$
QUESTION [7 upvotes]: I have trouble understanding easy integral concept.
Here, one of the integral theorem states:

And I was wondering what would happen if $F'(x)$ is back in to the integral.
For example, $F(x) = \int_{-1}^{x}\ {2t+t^2}, x \in [-1,5]$
$$F'(x) = 2x + x^2, x \in (-1,5)$$
Then if $F'(x)$ is back in to $\int_{-1}^{x}$ as in $\int_{-1}^{x}\ {2x+x^2}$, should it return $F(x)$?
I mean is the following statement true? 
$$\int_{-1}^{x}\ {2x+x^2} = \int_{-1}^{x}\ F'(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}$$
What I am confused about is the change of the variable above.

REPLY [5 votes]: I mean is the following statement true? $$\int_{-1}^{x}\ {2x+x^2}\,dx = \int_{-1}^{x}\ F'(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}\,dt$$

Not quite: $\text{If}\; F(x) = \int_{-1}^x \,2t + t^2 \,dt$, while it is true that $F'(x) = 2x + x^2$, we cannot write, meaningfully, $$F(x) = \int_{-1}^\color{blue}{\bf x} 2\color{blue}{\bf x} + \color{blue}{\bf x}^2\,d\color{blue}{\bf x}$$
The reason $t$ is used in the original integral $F(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}\,dt$ is merely to stand in as a "dummy variable" which is needed only until we compute the indefinite integral, evaluate at the bounds, which thus gives us a function of $x$: $F(x)$. $F'(x)$ is also a function of $x$. 
What you can say is that $F(x) + C = \int F'(x) \,dx$, since here we are not using bounds of integration that are nonsensical.
We can integrate the derivative of a function, and we can find the derivative of the integral of a function, yes indeed. One can think of them, informally, as inverse processes given a function like the one you've posted.