# Finding area by using a summation

1. Jul 6, 2014

### MMM

Hello everyone, I've been working on an area summation problem in my book for quite a bit and I can't solve it.
Find the area under the straight line $y=2x$ between $x = 1$ and $x = 5$
The book shows the answer as 24 and Maple does as well, but I'm not getting 24, I'm getting 8.

Area summation formula given by $\lim_{n\rightarrow +\infty}\sum_{i=1}^\infty\2(1 + 4i/n)4/n$

I got the problem to $\lim_{n\rightarrow +\infty}(4/n) * \sum_{i=1}^\infty\2$ I evaluated that and got 8 then
I tried evaluating the 2nd part $\lim_{n\rightarrow +\infty}(4/n) * \sum_{i=1}^\infty\frac{8i}{n}$ and I got that limit to be 0. That is where I'm messing up at I think, I have no idea what I'm doing wrong. Any help would be greatly appreciated.

Last edited: Jul 6, 2014
2. Jul 6, 2014

### LCKurtz

Please use standard font, not bolded. You need to figure out what is $x_i$ and $f(x_i)$ when you have $n$ equal subintervals on $[1,5]$.

3. Jul 6, 2014

### MMM

f(xi) = 2(1 + 4i/n)
I edited the first post with that information.
I just don't understand why the answer to the problem is 24 and not 8, I have no idea what I did wrong when evaluating the limits.
Isn't this $\lim_{n\rightarrow +\infty}(4/n) * \sum_{i=1}^\infty\frac{8i}{n}$ 0? I just don't understand how the book is getting 24 as the final answer.

Last edited: Jul 6, 2014
4. Jul 6, 2014

### LCKurtz

Your corrected first post now reads$$\lim_{n\rightarrow +\infty}\sum_{i=1}^\infty 2(1 + 4i/n)4/n$$The upper limit of the sum is incorrect. It should be$$\lim_{n\rightarrow +\infty}\sum_{i=1}^n 2(1 + 4i/n)4/n$$Now, simplify that sum as a function of $n$ and then take the limit as $n\to \infty$.

5. Jul 7, 2014

### MMM

I split the sum into two different parts.

$\lim_{n\rightarrow +\infty}\sum_{i=1}^n 8/n$ and I got that limit to be 8.

Now the second sum is $\lim_{n\rightarrow +\infty}\sum_{i=1}^n 32i/n^2$
and I can't figure out what that limit is. How do I rewrite it as a function of n? Since the final answer is 24 the limit must evaluate to 16, but I'm not sure how to get to that conclusion. Any help is greatly appreciated, thanks.4

I figured it out I factored out the $\frac{32}{n^2}$ and used the sum of a finite arithmetic sequence formula that had $n$ terms. Thanks for the help.

Last edited: Jul 7, 2014
6. Jul 7, 2014

### LCKurtz

Good work. Adding up that arithmetic sequence is the key to the problem. You're welcome.