M: Solve Riemann Sum Problem Homework

In summary, a Riemann Sum is a method used in calculus to approximate the area under a curve by dividing it into smaller rectangles and finding the sum of their areas. To solve a Riemann Sum problem, you need to find the width and height of each rectangle and then add up their areas. The purpose of using Riemann Sums is to approximate the area when traditional methods are not possible and to introduce the concept of integration. Any number of rectangles can be used, but as the number increases, the approximation becomes more accurate. However, Riemann Sums have limitations in that they can only approximate the area and may become less accurate for complex or sharp functions, requiring more advanced integration techniques.
  • #1
SYoungblood
64
1

Homework Statement


[/B]
Hello, thank you in advance for your help. I am calculating a Riemann sum with right hand endpoints. I hit a small snag, and I appreciate your help in getting me straight.

Homework Equations



f(x) = x2+ 1, over the interval [0,1]. This is problem number such-and-such from a well-known calculus textbook, not anything that is on an exam that I know of. The limit of these sums approaches infinity.

I'll call the Riemann Sum Sp, the length (x value) of the rectangles ∆x, and the height of the rectangles (y value) ck.

The Attempt at a Solution


[/B]
∆x = 1/n, good to go.

ck = a + k[(b - a)/n] = 0 + k (1/n) = k/n

Sp = Σ(k = 1, n) k/n * 1/n + 1

Sp = 1/n2 Σ(k = 1, n) k + 1

And here is where I ran out of juice.

Again, and help is appreciated.

SY
 
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  • #2
SYoungblood said:

Homework Statement


[/B]
Hello, thank you in advance for your help. I am calculating a Riemann sum with right hand endpoints. I hit a small snag, and I appreciate your help in getting me straight.

Homework Equations



f(x) = x2+ 1, over the interval [0,1]. This is problem number such-and-such from a well-known calculus textbook, not anything that is on an exam that I know of. The limit of these sums approaches infinity.

I'll call the Riemann Sum Sp, the length (x value) of the rectangles ∆x, and the height of the rectangles (y value) ck.

The Attempt at a Solution


[/B]
∆x = 1/n, good to go.

ck = a + k[(b - a)/n] = 0 + k (1/n) = k/n

Sp = Σ(k = 1, n) k/n * 1/n + 1

Since ##f(x) = x^2 +1## and ##c_k = \frac k n##, then ##f(c_k) = \frac {k^2} {n^2}+1## so you should be calculating$$
\sum_{k=1}^n(\frac {k^2} {n^2}+1)\frac 1 n$$

You will need the formula for the sum of squares of the first n integers, which you can look up.
 
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  • #3
I believe I just pulled a first-year calculus moron maneuver. Instead of k, it should be k2? And each term would then proceed to be determined using sigma notation for the degree of the individual term?

Here, that would be (as I see it),

Sp = 1/n2 *
∑k=1n k2 + ∑k=1n 1
= 1/n2 * [n(n+1)(2n+1)/6] + (n)-- my apologies for this being superscripted, I am having an issue w/ the formatting

Here, am I correct in understanding with the first sigma term, that reduces to n * 1/n2?

If I am, them we have 1/n * [(2n3+ n2+ 2n + n)/6]

= (2n3+ n2+3n)/6n

= (2n2+n + 3)/6

As I understand it, if i substitute the limit for n, infinity, then I am left with an area of 2/6 = 1/3 sp units. That doesn't pass the sniff test. What did I do wrong?

Thank you for your time.
 
  • #4
SYoungblood said:
I believe I just pulled a first-year calculus moron maneuver. Instead of k, it should be k2? And each term would then proceed to be determined using sigma notation for the degree of the individual term?

Here, that would be (as I see it),

Sp = 1/n2 *
∑k=1n k2 + ∑k=1n 1
= 1/n2 * [n(n+1)(2n+1)/6] + (n)-- my apologies for this being superscripted, I am having an issue w/ the formatting

Here, am I correct in understanding with the first sigma term, that reduces to n * 1/n2?

If I am, them we have 1/n * [(2n3+ n2+ 2n + n)/6]

= (2n3+ n2+3n)/6n

= (2n2+n + 3)/6

As I understand it, if i substitute the limit for n, infinity, then I am left with an area of 2/6 = 1/3 sp units. That doesn't pass the sniff test. What did I do wrong?

Thank you for your time.
Read post #2.
 

1. What is a Riemann Sum?

A Riemann Sum is a method used in calculus to approximate the area under a curve. It involves dividing the area into smaller rectangles and finding the sum of their areas.

2. How do I solve a Riemann Sum problem?

To solve a Riemann Sum problem, you need to first find the width of each rectangle by dividing the total width of the interval by the number of rectangles. Then, you need to find the height of each rectangle by evaluating the function at specific points within each interval. Finally, you add up the areas of all the rectangles to get an approximation of the area under the curve.

3. What is the purpose of using Riemann Sums?

Riemann Sums are used to approximate the area under a curve when it is not possible to find the exact value using traditional methods. They are also used to introduce the concept of integration in calculus.

4. Can I use any number of rectangles in a Riemann Sum?

Yes, you can use any number of rectangles in a Riemann Sum. However, as the number of rectangles increases, the approximation becomes more accurate.

5. Are there any limitations to using Riemann Sums?

One limitation of Riemann Sums is that they can only approximate the area under a curve, not find the exact value. They also become less accurate when the function is more complex or has sharp turns. In these cases, more advanced integration techniques may be necessary.

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