Evaluating a definite integral by definition: $A(x) = \int_{-1}^x (t^2 + 1)\space dt$ I have an area function $A(x)$ defined as
$$A(x) = \int_{-1}^{x} (t^2 + 1)\space dt$$
... and I would like to use the definition of definite integral to evaluate it. 
I started this way
$$A(x) = \lim_{n \to \infty} \sum_{k=1}^n { (t^2 + 1) {x - (-1) \over n}}\tag 1$$
I thought that $$\lim_{n \to \infty} \sum_{k=1}^n {x +1 \over n }\tag 2$$ turns into $x + 1$, so that
$$A(x) = \lim_{n \to \infty} \space\sum_{k=1}^n { t^2}{x + 1 \over n}  + x + 1\tag 3$$
and saying that $t = k{x + 1 \over n}$ it would become
 $$A(x) = \lim_{n \to \infty} \space\sum_{k=1}^n { k^2 (x+1)^3 \over n^3} + x + 1\tag4$$
I'm not sure if I didn't make any mistake until here, and even if I didn't, I don't know what to do now. Maybe I could use 
$$\sum_{k = 1}^{n} k^2 = {n(n + 1)(2n + 1) \over 6}\tag 5$$
but in order to do that, I would have to first get rid of $n^2$ in denominator.
Could you please give me a friendly kick in the head to make it correct? Any help will be highly appreciated.
 A: Remember that
$$\int_a^b dt \, f(t) = \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=0}^n f \left [ a + (b-a) \frac{k}{n} \right ] $$
Here, $a=-1$ and $b=x$, so that the sum we are interested in is
$$ \begin{align} \int_{-1}^x dt \, (1+t^2) &= \lim_{n \to \infty} \frac{x+1}{n} \sum_{k=0}^n \left [ 1+ \left ( -1 + \frac{x+1}{n} k \right )^2 \right ]  \\ &= x+1 + \lim_{n \to \infty} \frac{x+1}{n} \sum_{k=0}^n \left [1 - 2 \frac{x+1}{n} k + \left ( \frac{x+1}{n} \right )^2 k^2 \right ]\\ &= 2 (x+1) + \lim_{n \to \infty} \left [\frac{-2 (x+1)^2}{n^2} \sum_{k=0}^n k + \frac{(x+1)^3}{n^3} \sum_{k=0}^n k^2 \right ]\\ &= 2 (x+1) + \lim_{n \to \infty} \left [\frac{-2 (x+1)^2}{n^2} \frac{n (n+1)}{2} + \frac{(x+1)^3}{n^3} \frac{n (n+1) (2 n+1)}{6} \right] \\ &= 2 (x+1) - (x+1)^2 + \frac13 (x+1)^3 \\ &= \frac13 (x+1) (x^2-x+4)\end{align}$$
You can verify that this is what you get by integrating using the Fundamental theorem.
A: There is my calculation to make compare it with what you did, we will  use the definition, let $f(t)=t^2+1$ and $x_k=-1+k\frac{x-(-1)}{n}$:
$$\begin{align}A(x)&= \int_{-1}^x (t^2+1)\\ \\
&=\lim_{n\to \infty}\frac{x-(-1)}{n}\sum_{k=1}^nf(x_k)\\ \\ 
&=\lim_{n\to \infty}\frac{x+1}{n}\sum_{k=1}^n \left(\left(k\frac{x+1}{n}-1\right)^2+1\right)\\ \\
&=\lim_{n\to \infty}\frac{x+1}{n}\sum_{k=1}^n \left(k^2\left(\frac{x+1}{n}\right)^2-2k\frac{x+1}{n}+2\right)\\ \\
&=\lim_{n\to \infty}\frac{x+1}{n} \left(\left(\frac{x+1}{n}\right)^2\sum_{k=1}^nk^2-\frac{x+1}{n}\sum_{k=1}^n2k+\sum_{k=1}^n2\right)\\ \\
&=\lim_{n\to \infty}\frac{x+1}{n} \left(\left(\frac{x+1}{n}\right)^2{n(n + 1)(2n + 1) \over 6}-\frac{x+1}{n}n(n+1)+2n\right)\\ \\
&=(x+1)\left(\frac{(x+1)^2}{3}-(x+1)+2\right)\\ \\
&=\frac{(x+1)(x^2-x+4)}{3}\end{align}$$
