Evaluating integrals with sigma notation I am suppose to use Theorem 4 in my books (just the limit as n approaches infinity of a summation representing the function).
I am trying to evaluate
$$ \int_2^5 (4-2x) dx$$
I really have no idea what to do. I know $dx$ is $3/n$ and that the summation should look something like $\frac{n(n+1)}{2}$ after using the rules of wizardry that I memorized from earlier in the chapter. 
As far as what to do now I tried to put that back into the equation and got nothing logical.
$4-2 n(n+1)$ doesn't really make any sense to me. 
From here I do not know what to do.
 A: Let’s build up the Riemann sum first. The interval is $[2,5]$, so its length is $3$, and when you divide it into $n$ equal subintervals, each will be of length $\frac3n$, so $\Delta x$ (not $dx$) is indeed $\frac3n$. The ends of the subintervals $-$ the $x_k$’s $-$ are $2+\frac3n,2+2\left(\frac3n\right)$, and so on, with $x_k=2+\frac{3k}n$. Thus, your $n$-th Riemann sum, $R_n$, is 
$$\begin{align*}
R_n=\sum_{k=1}^n(4-2x_k)\frac3n&=\sum_{k=1}^n\left(4-2\left(2+\frac{3k}n\right)\right)\frac3n\\
&=\sum_{k=1}^n\left(4-4-\frac{6k}n\right)\frac3n\\
&=\sum_{k=1}^n\left(-\frac{6k}n\right)\frac3n\\
&=\sum_{k=1}^n\left(-\frac{18k}{n^2}\right)\;.
\end{align*}$$
Now that’s just $$-\frac{18}{n^2}-\frac{18\cdot 2}{n^2}-\frac{18\cdot 3}{n^2}-\ldots-\frac{18n}{n^2}\;,$$
with a factor of $-\dfrac{18}{n^2}$ in every term that we can factor out to get $$-\frac{18}{n^2}(1+2+3+\ldots+n)=-\frac{18}{n^2}\sum_{k=1}^nk\;.$$ As you said in the question, you know that $\sum_{k=1}^nk$, the sum of the first $n$ positive integers, is $\frac{n(n+1)}2$, so $$\begin{align*}R_n&=-\frac{18}{n^2}\sum_{k=1}^nk=-\frac{18}{n^2}\cdot\frac{n(n+1)}2\\&=-\frac{9(n+1)}n=-9\cdot\frac{n+1}n\\&=-9\left(1+\frac1n\right)\;.\end{align*}$$
Finally, $$\begin{align*}\int_2^5(4-2x)dx&=\lim_{n\to\infty}R_n\\&=\lim_{n\to\infty}-9\left(1+\frac1n\right)\\&=-9\lim_{n\to\infty}\left(1+\frac1n\right)\;.\end{align*}$$
Now, what’s $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)$?
