# 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.

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Rules of wizardry ? – Belgi Apr 13 '12 at 22:15
Your idea of what should happen is accurate; you just have to initiate it. First off, what is the general form of the upper rectangular Riemann sum of $\int_a^b f(x)dx$ with $n$ pieces? Now set $a=2,b=5$ and $f(x)=4-2x$ (except the $x$ will no longer be just an $x$). – anon Apr 13 '12 at 22:18
The rules of wizardry are what Stewart uses to calculate summation for x, x^2 and x^3 – user138246 Apr 13 '12 at 22:40
Downvoter should explain himself. This is bad MSE behavior in my opinion. A downvote should be instructive ; explain what you dislike about this question. Personally I think everything's okay, so this is a bad downvote. – Patrick Da Silva Apr 13 '12 at 22:56

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 $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)$?

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So the limit would be 1 but then I mutliply the limit be -9 so the answer is -9? I am a little confused by the n(n+1)/2 part. Does that replace K or does it just notate the summation of a single variable? – user138246 Apr 13 '12 at 22:46
@Jordan: The answer is indeed $-9$. $\frac{n(n+1)}2$ is the value of the sum $\sum_{k=1}^nk=1+2+3+\ldots+n$. For instance, when $n=4$ it’s $1+2+3+4=\frac{4(4+1)}2=\frac{20}2=10$. – Brian M. Scott Apr 13 '12 at 22:53
@Brian : You should edit your comment, the $n$ got down the sum because you forgot an ^. – Patrick Da Silva Apr 13 '12 at 22:54
@Patrick: Thanks! Caught it in time. – Brian M. Scott Apr 13 '12 at 22:55
@Brian For my next problem I have a k + k^2 does that mean I use both the summation formulas? – user138246 Apr 13 '12 at 23:29