Finding a definite integral from the Riemann sum

In summary, the conversation discusses an interval with given values, and using those values, the expression (1 + i/n)(1/n)ln[(n + i)/n] is rearranged using logarithmic properties. The question of what goes in the logarithmic function is raised, and it is suggested to simplify the argument of the logarithm. Eventually, it is realized that the argument can be simplified to (1 + i/n), resulting in the function f(x) = xlnx + (1/n)ln(1+i/n).
  • #1
crememars
15
2
Homework Statement
Consider the following limit of a Riemann sum for a function f on [a, b]. Identify f, a, and b,
and express the limit as a definite integral.

*see actual expression in the description below. it was too complicated to type out so I included a picture instead.
Relevant Equations
∆x = (b-a)/n
xiR = a + i∆x
1679251462458.png


Hi! I am having trouble finalizing this problem.

The interval is given so we know that a = 1 and b = 2. From there you can figure out that ∆x = 1/n, xiR = 1 + i/n.
Using logarithmic properties, I rearranged the expression and wrote (1 + i/n)(1/n)ln[(n + i)/n].
I can guess that the function is going to look something like this: f(x) = xln(...) but I don't know what goes in the logarithmic function...

I have been trying to rewrite it in terms of xiR but no luck :(
any help would be really appreciated. thank you !
 
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  • #2
Can you simplify the argument of the logarithm any further?
 
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  • #3
pasmith said:
Can you simplify the argument of the logarithm any further?
Yes, I realized I was overcomplicating things. If I simplify (n+i)/n to (n/n +i/n) to (1+ i/n), I get f(x) = xlnx
 

1. What is a Riemann sum?

A Riemann sum is a method of approximating the area under a curve by dividing it into smaller rectangles and summing their areas. It is used to find the definite integral of a function.

2. How is a Riemann sum used to find a definite integral?

The Riemann sum is used to approximate the area under a curve, and as the number of rectangles used increases, the approximation becomes more accurate. In the limit, as the width of the rectangles approaches zero, the Riemann sum becomes equal to the definite integral of the function.

3. What is the difference between a left, right, and midpoint Riemann sum?

A left Riemann sum uses the left endpoint of each rectangle to calculate its area, a right Riemann sum uses the right endpoint, and a midpoint Riemann sum uses the midpoint of each rectangle. The choice of which type of Riemann sum to use can affect the accuracy of the approximation.

4. Can a Riemann sum be used for any type of function?

Yes, a Riemann sum can be used for any continuous function. However, for some functions, the number of rectangles needed to achieve a desired level of accuracy may be very large.

5. What is the relationship between the Riemann sum and the definite integral?

The Riemann sum is an approximation of the definite integral, with the accuracy increasing as the number of rectangles used increases. In the limit, as the number of rectangles approaches infinity, the Riemann sum becomes equal to the definite integral.

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