1. ## riemann sum

Write down the Riemann sum for int_2^3 2x dx using N equal subintervals, simplify, and find the limit as N approaches infinity.

I am not sure if I set the sum up properly and I am having trouble with the simplification. I don't really know what it is I am trying to isolate... N? i?

I am using the formula:

(b-a)/N Sigma_i=1^N [ f(a + i(b-a / N))]

so I have:

1/N Sigma_i=1 ^ N [ 2(2 + i(1/N))]

So where do I go from here? And also, when I take the limit as N approaches infinity, what do I do with terms including i?

Thanks!

2. Originally Posted by littlejodo
Write down the Riemann sum for int_2^3 2x dx using N equal subintervals, simplify, and find the limit as N approaches infinity.

I am not sure if I set the sum up properly and I am having trouble with the simplification. I don't really know what it is I am trying to isolate... N? i?

I am using the formula:

(b-a)/N Sigma_i=1^N [ f(a + i(b-a / N))]

so I have:

1/N Sigma_i=1 ^ N [ 2(2 + i(1/N))]

So where do I go from here? And also, when I take the limit as N approaches infinity, what do I do with terms including i?

Thanks!
Assume that $f$ is integrable over $[a,b]$ then if we define $\Delta x=\frac{b-a}{n}$ and $M_i=a+\Delta x\cdot i$ then

$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{i=1}^n f\left(M_i\right) \Delta x$. So in your case

$\Delta x=\frac{3-2}{n}=\frac{1}{n}$
$f\left(M_i\right)=f\left(2+\frac{i}{n}\right)=2\le ft(2+\frac{i}{n}\right)$

So then your integral is equal to

\begin{aligned}\int_2^3 2x~dx&=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n}\ left\{2+\frac{i}{n}\right\}\\
&=\frac{2}{n}\left\{2n+\frac{n+1}{2}\right\}{\colo r{red}\star}\\
&=4+1\\
&=5\end{aligned}

$\color{red}\star$ was gotten by the formulas $\sum_{i=1}^{n}c=cn$ and $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

So, is it safe to say that in taking the limit as N --> inf I can treat each N as a 1? I'm trying to see how taking the limit leads to 4 + 1...

I had thought that when the variable was in the denominator that term would equal 0...

The only way I see it leading to 4 + 1 is to have N = 1 but I'm just not sure if that's the correct thing to do. Clearly, I can see that the limit is 5 by looking at graph, but I was hoping to understand the algebra behind the concept. Thanks!

4. Originally Posted by littlejodo

So, is it safe to say that in taking the limit as N --> inf I can treat each N as a 1? I'm trying to see how taking the limit leads to 4 + 1...

I had thought that when the variable was in the denominator that term would equal 0...

The only way I see it leading to 4 + 1 is to have N = 1 but I'm just not sure if that's the correct thing to do. Clearly, I can see that the limit is 5 by looking at graph, but I was hoping to understand the algebra behind the concept. Thanks!
Word to the wise, I think most MHFers look at a response once usually. So if you need to ask a question don't edit your response make a new one if you had asked your previous one a while ago.

$\frac{2}{n}\left\{2n+\frac{n+1}{2}\right\}=4+\frac {n+1}{n}$

Now taking the limit of this we get $\lim\left\{4+\frac{n+1}{n}\right\}=4+1=5$

I assume you know that $\lim\frac{n+1}{n}=1$ right?