Definite integral with Riemann sums

[TheFallen018]

Heya,

So, I know this is a pretty simple problem, but I seem stuck on it nevertheless.

Here's the question
Calculate the upper and lower sums , on a regular partition of the intervals, for the following integrals
\begin{align*}
\int_{1}^{3}(1-7x)dx
\end{align*}

Please correct me if I'm doing this wrong. Here's what I did
\begin{align*}\Delta x = \frac{3-1}{n} = \frac{2}{n}\end{align*}
\begin{align*}x_{i}=1+\Delta x*i = 1 + \frac{2}{n}i\end{align*}
\begin{align*}f(x_{i}) = (1-7x_{i})) = (1-7(1+\frac{2}{n}i))\end{align*}
\begin{align*}\frac{2}{n}\sum_{i=1}^{n}f(x_{i}) = \frac{2}{n}\sum_{i=1}^{n}(1-7(1+\frac{2}{n}i))\end{align*}

Considering the question asked for it on a regular partition of intervals, I assumed that meant 3, so I set n=3. Doing this, I get the answer -92/3, however apparently this isn't the correct answer for the lower sum. If I run n up to approach infinity, it goes to -26, which seems right, but that isn't accepted either.

Have I done something terribly wrong? Thanks

 

[HOI]

No, a "regular partition" means just that each subinterval has the same length so your $\Delta x= \frac{2}{n}$ is correct.

You have, correctly, [tex]\frac{2}{n}\sum_{i=1}^n (1- 7(1+ \frac{2}{n}i)[/tex]. That can be written as $\frac{2}{n}\sum_{i=1}^n (-6 - \frac{14}{n}i)= \frac{-12}{n}\sum_{i=1}^n 1- \frac{28}{n^2}\sum_{i= 1}^n i$.

You know that the sum of "1" from 1 to n, just adding 1 n times, is n and that the sum of "i" from 1 to n is $\frac{1}{2}n(n+ 1)$. So $\frac{-12}{n}\sum_{i=1}^n 1- \frac{28}{n^2}\sum_{i=1}^n i= -12- 14- \frac{14}{n}= -26- \frac{14}{n}$. As n goes to infinity, that limit is -26 which is the correct answer for the integral:
$\int_1^3 1- 7x dx= \left[x- \frac{7}{2}x^2\right]_1^3= 3- \frac{63}{2}- 1+ \frac{7}{2}= 2- \frac{56}{2}= 2- 28= -26$.

However, after re-reading your post you appear to have misunderstood what is asked. The problem asks you to "calculate the upper and lower sums". You must give two answers. The "upper sum" is from i= 1 to n while the "lower sum" is the same but taken from i= 0 to n- 1. The upper sum is what I calculated above, $-26- \frac{14}{n}$.

 

[TheFallen018]

No, a "regular partition" means just that each subinterval has the same length so your $\Delta x= \frac{2}{n}$ is correct.

You have, correctly, [tex]\frac{2}{n}\sum_{i=1}^n (1- 7(1+ \frac{2}{n}i)[/tex]. That can be written as $\frac{2}{n}\sum_{i=1}^n (-6 - \frac{14}{n}i)= \frac{-12}{n}\sum_{i=1}^n 1- \frac{28}{n^2}\sum_{i= 1}^n i$.

You know that the sum of "1" from 1 to n, just adding 1 n times, is n and that the sum of "i" from 1 to n is $\frac{1}{2}n(n+ 1)$. So $\frac{-12}{n}\sum_{i=1}^n 1- \frac{28}{n^2}\sum_{i=1}^n i= -12- 14- \frac{14}{n}= -26- \frac{14}{n}$. As n goes to infinity, that limit is -26 which is the correct answer for the integral:
$\int_1^3 1- 7x dx= \left[x- \frac{7}{2}x^2\right]_1^3= 3- \frac{63}{2}- 1+ \frac{7}{2}= 2- \frac{56}{2}= 2- 28= -26$.

However, after re-reading your post you appear to have misunderstood what is asked. The problem asks you to "calculate the upper and lower sums". You must give two answers. The "upper sum" is from i= 1 to n while the "lower sum" is the same but taken from i= 0 to n- 1. The upper sum is what I calculated above, $-26- \frac{14}{n}$.

Thanks man, I really appreciate the help. Yes, there were two separate answers to give. I guess I didn't mention that part. It turns out what was giving me the trouble was that they wanted an algebraic solution, where n was not assigned any number at all. Why that wasn't stated, I don't know. It ended up being $\frac{14}{n}-26$ for the upper sum and $-\frac{14}{n}-26$ for the lower sum, which is what you came up with.

Thanks again :)