Evaluate definite integral using the definition: $\int_{-3}^{1}(x^2-2x+4) dx$? Can someone walk me through finding the value of a definite integral by using the definition itself? In this case:
`The definition of the definite integral: let $f$ be a function that is defined on the closed interval $[a,b]$. The definite integral of $f$ from $a$ to $b$, denoted $$\int_a^b f(x),dx  $$ is $$\int_a^b f(x),dx = \lim_{|P|->0} \Sigma^n_{k=1} f(c_k)\ \triangle x_k$$ provided the limit exists 
(NOTE: I messed up notation on the sigma. n should be over top and k=1 on bottom.
Evaluate $$ \int_{-3}^{1}(x^2-2x+4) dx$$ the definite integral using the definition.
I've got a feeling something similar will be on my Final on Thursday and while I think I can work most integral problems, I'm not sure I fully understand how to do so via the definition explicitly.
EDIT:
Forgot to include. I know the below, just not if it's exactly what is being asked for:
$$ \int_{-3}^{1}(x^2-2x+4) dx$$
$$ (1/3)(x)^3 - (x)^2 + 4x + c$$
$$ (1/3)(1)^3 - (1)^2 + 4(1) = (10/3)$$
$$ (1/3)(-3)^3 - (-3)^2 + 4(-3) = -30$$
$$(10/3) - (-30) = 100/3$$

Consider $\int_{-3}^{1}x dx$. Divide $[-3, 1]$ in a partition $x_0= -3, x_1=-3 + 4/n, x_2=-3+2(4/n)$ ... $x_n=-3+n(4/n)=1$. The lower sum L and the upper sum S  are given by $$L_n=\Sigma_{t=1}^{n}\Big(-3+ (t-1)\frac{4}{n}\Big)\frac{4}{n}$$ $$U_n=\Sigma_{t=1}^{n}\Big(-3+ t\frac{4}{n}\Big)\frac{4}{n}$$ In above the quantities inside the bigger brackets are the minimum and maximum value of the function( $x$ in this case) in the respective cases of $L_n$ and $U_n$. $\frac{4}{n}$ is the partition size. It easily follows that $$L_n=-12 + \frac{8(n-1)}{n}$$ and $$U_n=-12 + \frac{8(n+1)}{n}$$Both $L_n$ and $U_n$ tend to $-4$ as $n\rightarrow \infty$. $-4$ is also the value of the $\int_{-3}^{1}x dx$. In a similar fashion, one could consider the other terms in the integral $\int_{-3}^{1}(x^2 - 2x +4) dx$
