How do I calculate the answer for this integrals problem? left Riemann sum over [2,3] for f(x)=1/(x-1)

[lasvegas666]

I don't know how to find the approximate answer to this Riemann sum problem. Additionally, I'm not asking for the answer. I'm asking for help with finding out the problem-solving methodology so that I can come to the correct answer on my own.

 

[Dr.Peterson]

View attachment 38026
I don't know how to find the approximate answer to this Riemann sum problem. Additionally, I'm not asking for the answer. I'm asking for help with finding out the problem-solving methodology so that I can come to the correct answer on my own.

Since you've successfully done a left Riemann sum in another thread, please give it a try here, and show your work, so we can see what help you need.

 

[lasvegas666]

Since you've successfully done a left Riemann sum in another thread, please give it a try here, and show your work, so we can see what help you need.

I legit don't know how to solve this problem because it's asking me for an approximate decimal answer.

 

[blamocur]

I legit don't know how to solve this problem because it's asking me for an approximate decimal answer.

Do you know what Riemann sum is? If not, have you searched for the definition?

 

[lasvegas666]

Do you know what Riemann sum is? If not, have you searched for the definition?

A Riemann sum is the area sum of the values on a graph for a given function.

 

[Dr.Peterson]

I legit don't know how to solve this problem because it's asking me for an approximate decimal answer.

But the method is the same as what you did here:

How do I solve this problem? left Riemann sum, 8 subintervals on [0,8] for area under curve

I'm having trouble comprehending how I can solve this problem on my own.

www.freemathhelp.com

The only difference is that you have to calculate the values of f(x), rather than being given them in a graph, and the numbers involved will not be whole numbers. Don't let decimals scare you. (And any Riemann sum is an approximation to the actual area.)

Here is an example of such a calculation:

Left Riemann Sums | Mathmatique

www.mathmatique.com

(See problem 6 at the bottom for maximum similarity.)

 

[lasvegas666]

But the method is the same as what you did here:

How do I solve this problem? left Riemann sum, 8 subintervals on [0,8] for area under curve

I'm having trouble comprehending how I can solve this problem on my own.

www.freemathhelp.com

The only difference is that you have to calculate the values of f(x), rather than being given them in a graph, and the numbers involved will not be whole numbers. Don't let decimals scare you. (And any Riemann sum is an approximation to the actual area.)

Here is an example of such a calculation:

Left Riemann Sums | Mathmatique

www.mathmatique.com

(See problem 6 at the bottom for maximum similarity.)

I tried doing this but I didn't get the right answer.

 

[blamocur]

to
View attachment 38038
I tried doing this but I didn't get the right answer.

The sum in your picture makes no sense to me.
What are the values of 'x' at the left end of each subinterval?
What are the values of [imath]1/(x-1)[/imath] for those values of 'x' ?

BTW, the sum in your picture has 5 terms, but are asked to use 4 subintervals.

 

[lasvegas666]

The sum in your picture makes no sense to me.
What are the values of 'x' at the left end of each subinterval?
What are the values of [imath]1/(x-1)[/imath] for those values of 'x' ?

BTW, the sum in your picture has 5 terms, but are asked to use 4 subintervals.

Alright. I tried to use the [math](b-a) / n[/math] method to get -1/4, and then I just added one to the other four interval points when calculating my answer.

 

[blamocur]

Alright. I tried to use the [math](b-a) / n[/math] method to get -1/4, and then I just added one to the other four interval points when calculating my answer.

Can you write down and post answers to my questions in post #8?
Verbal descriptions of math solutions are rarely useful

 

[lasvegas666]

Can you write down and post answers to my questions in post #8?
Verbal descriptions of math solutions are rarely useful

I answered that.

I would just like help finding the answer to this problem.

 

[blamocur]

I answered that.

I would just like help finding the answer to this problem.

And I've done the best I could to help.

 

[Dr.Peterson]

I tried doing this but I didn't get the right answer.

Let's look through what you wrote bit by bit.



The formula you want to use is [imath]\Delta x\left(f(a)+f(a+\Delta x)+\dots f(a+(n-1)\Delta x)\right)[/imath], where [imath]\Delta x=\frac{b-a}{n}[/imath].

You've made a small error in calculating [imath]\Delta x[/imath]; it should be [imath]\Delta x=\frac{b-a}{n}=\frac{3-2}{4}=\frac{1}{4}=0.25[/imath]. It should be positive (except in very odd cases).

Then the four left end points should be 2, 2+0.25=2.25, 2+2(0.25)=2.5, 2+3(0.25)=2.75; you appear to have used some fractions I can't quite read that are not between 2 and 3, and an incomprehensible extra term at the start.

All you need to do to get this right is to take it slower and make sure you put the right numbers in. Please be patient and give it another try,

 

[mario99]

I would do it this way:

[imath]\displaystyle \sum_{i=1}^{4}\frac{1}{x_i - 1}\Delta x[/imath]

where

[imath]\displaystyle x_i = 2 + (i - 1)\Delta x[/imath]

And

[imath]\displaystyle \Delta x = \frac{3-2}{4} = \frac{1}{4}[/imath]