How would I evaluate this integral using the 2nd Fundamental Theorem of Calculus? Thanks in advance.
Ahh I love these questions.
Let $\displaystyle \frac{2i}{n}=x$. Thus $\displaystyle dx=\frac{2}{n}$.
Now rewrite to an integral integral with this substitution.
$\displaystyle \int_{L} 1 +x +x^2dx$
Now for the bounds.
For the first bound let $\displaystyle i=0$ and for the second let $\displaystyle i=n$. So the bounds are from 0 to 2.
Got it?
The standard form the the Riemann Ingegral is,
$\displaystyle \lim_{n\to \infty} \sum_{k=1}^n f(a+k\Delta x)\Delta x$
In this case,
$\displaystyle \Delta x=\frac{2}{n}$
Thus,
$\displaystyle b-a=2$
But there is not "a" term,
$\displaystyle a=0$ thus, $\displaystyle b=2$.
And the function is,
$\displaystyle f(x)=1+x+x^2$
Thus we need to find,
$\displaystyle \int_0^2 1+x+x^2 dx$
Which we can easily evaluate by Fundamental theorem because this function is continous on the closed interval.