Definite integral with Riemann sums

• MHB
• TheFallen018
In summary, the problem asks to calculate the upper and lower sums for the following integral: $\int_{1}^{3}(1-7x)dx$ and the correct answer is -26.
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

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, $$\frac{2}{n}\sum_{i=1}^n (1- 7(1+ \frac{2}{n}i)$$. 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, $-26- \frac{14}{n}$.

Last edited:
Country Boy said:
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, $$\frac{2}{n}\sum_{i=1}^n (1- 7(1+ \frac{2}{n}i)$$. 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 :)

Q: What is a definite integral with Riemann sums?

A: A definite integral with Riemann sums is a method used in calculus to approximate the area under a curve by dividing it into smaller rectangles and summing their areas.

Q: How is a Riemann sum calculated?

A: A Riemann sum is calculated by dividing the interval of integration into smaller subintervals, determining the width of each subinterval, and using the height of the function at various points within each subinterval to calculate the area of the corresponding rectangle. The sum of these areas gives an approximation of the definite integral.

Q: What is the significance of using Riemann sums in calculating definite integrals?

A: Riemann sums allow for the calculation of definite integrals when the function being integrated is not easily integrable. It also provides a method for approximating the area under a curve when the exact value cannot be determined.

Q: What is the difference between a left, right, and midpoint Riemann sum?

A: Left Riemann sums use the left endpoint of each subinterval to determine the height of the corresponding rectangle, right Riemann sums use the right endpoint, and midpoint Riemann sums use the midpoint of each subinterval. These variations may result in slightly different approximations of the definite integral.

Q: How does increasing the number of subintervals affect the accuracy of a Riemann sum?

A: Increasing the number of subintervals increases the number of rectangles used in the approximation, resulting in a more accurate estimation of the definite integral. However, too many subintervals can lead to a large number of calculations and may not be computationally efficient.

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