Riemann Sum Integration Problem

[joe19]

I'm fairly new to this site, so I don't have an idea of how things work around here so just bear with me. I'm stuck on this Riemann sum problem.


I'm a bit lost and I'm looking for some guidance on where to go and what my answer should look like, thanks!

 

[MarkFL]

Hello, and welcome to FMH!

I would look at the definition for the left-hand sum:

[MATH]\int_a^b f(x)\,dx=\lim_{n\to\infty}\left(\frac{b-a}{n}\sum_{k=0}^{n-1}f\left(a+k\frac{b-a}{n}\right)\right)[/MATH]
What must \(a\) be?

 

[joe19]

Hello, and welcome to FMH!

I would look at the definition for the left-hand sum:

[MATH]\int_a^b f(x)\,dx=\lim_{n\to\infty}\left(\frac{b-a}{n}\sum_{k=0}^{n-1}f\left(a+k\frac{b-a}{n}\right)\right)[/MATH]
What must \(a\) be?

After doing some digging, I found a to be 2 and b to be 6, I don't know if it's right, but after finding a more simplified version of the definition that's what I found them to be. Two questions, do you know if my answer happens to be right? And I'm a bit lost on how I'd go about extracting a function from here.

 

[MarkFL]

After doing some digging, I found a to be 2 and b to be 6, I don't know if it's right, but after finding a more simplified version of the definition that's what I found them to be. Two questions, do you know if my answer happens to be right? And I'm a bit lost on how I'd go about extracting a function from here.

Yes, I also concluded that \(a=2\) and \(b=6\). Let's put in what we have now:

[MATH]\int_a^b f(x)\,dx\approx\frac{4}{3}\sum_{k=0}^{n-1}f\left(2+k\frac{4}{3}\right)=\frac{4}{3}\sum_{k=0}^{n-1}\left(2+k\frac{4}{3}\right)^2[/MATH]
Can you now see what \(f(x)\) is?

 

[joe19]

Yes, I also concluded that \(a=2\) and \(b=6\). Let's put in what we have now:

[MATH]\int_a^b f(x)\,dx\approx\frac{4}{3}\sum_{k=0}^{n-1}f\left(2+k\frac{4}{3}\right)=\frac{4}{3}\sum_{k=0}^{n-1}\left(2+k\frac{4}{3}\right)^2[/MATH]
Can you now see what \(f(x)\) is?

I think so, I'm not on top of my game right now as we haven't worked on this subject in quite some time in my class, but I'm assuming after seeing the information that you provided that f(x) = 2 + (4/3)x?

 

[MarkFL]

If you look at this part:

[MATH]\frac{4}{3}\sum_{k=0}^{n-1}f\left(2+k\frac{4}{3}\right)=\frac{4}{3}\sum_{k=0}^{n-1}\left(2+k\frac{4}{3}\right)^2[/MATH]
We see by equating the corresponding summands that we have

[MATH]f\left(2+k\frac{4}{3}\right)=\left(2+k\frac{4}{3}\right)^2[/MATH]
That implies that \(f\) is a function that outputs the square of its input, or:

[MATH]f(x)=x^2[/MATH]
Does that make sense?

 

[joe19]

If you look at this part:

[MATH]\frac{4}{3}\sum_{k=0}^{n-1}f\left(2+k\frac{4}{3}\right)=\frac{4}{3}\sum_{k=0}^{n-1}\left(2+k\frac{4}{3}\right)^2[/MATH]
We see by equating the corresponding summands that we have

[MATH]f\left(2+k\frac{4}{3}\right)=\left(2+k\frac{4}{3}\right)^2[/MATH]
That implies that \(f\) is a function that outputs the square of its input, or:

[MATH]f(x)=x^2[/MATH]
Does that make sense?

Yes! Thanks for the clear explanation, I'm better able to understand how you were able to utilize 'a' and 'b' and put them into the definition of a left-hand sum to find the answer, so just to clarify, the equation is x^2, correct? If so, then it looks like I have everything I was looking for.