How to Evaluate and Solve Riemann Sums for Integration: Step-by-Step Guide

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In summary, the conversation discusses evaluating the integral of cos^2(2*Pi*n*x) from x=0 to x=Pi/2, where n is a positive integer. The conversation also touches on using Riemann sums to solve the integral, but ultimately recommends using the trig identity cos^2(\theta)= (1/2)(1+ cos(2\theta)) to simplify the integration process.
  • #1
Samuelb88
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Homework Statement


Evaluate and give an exact answer.

[tex]$\int_{0}^{\frac{Pi}{2}\right)}cos^2(2*Pi*n*x)dx[/tex] where n is a positive integer.


Homework Equations





The Attempt at a Solution


[tex](change)x=\frac{b-a}{n}\right)=\frac{Pi}{2n}\right)[/tex]
and...
[tex]x_i=a+i(change)x[/tex]

I'm not sure how to write a Riemann Sum in latex code, so I'll be using "S" as the notation for the Riemann Sum (i=1, n): f(x_i)*(b-a)/n

I used the limit as n -> oo definition of an integral:

[tex]$\int_{0}^{\frac{Pi}{2}\right)}cos^2(2*Pi*n*x)dx=lim(n->oo)S[f(x_i)(\frac{b-a}{n}\right))][/tex]

Next I substituted the values of [tex]x_i[/tex]:

[tex]lim(n->oo)S[f(\frac{ib}{n}\right))(\frac{Pi}{2n}\right))][/tex]

Then evaluated [tex]f(x_i)[/tex]:

[tex]lim(n->oo)S[cos^2(2*Pi*n*\frac{Pi}{2n}\right)][/tex]

Here is where i get a bit confused. In the example with n-subintervals that I worked through in my book, it substituted the value of the Riemann sum of i using the power of sums formula, however, I end up with "i" inside of the cosine function, and am not sure how to finish evaluating the integral.

Other examples in the book had the value of n defined, thus making the integration process much more simple.

Am I even doing the right thing? Bare with me, my integration is self taught.
 
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  • #2
Why are you trying to change to Riemann sums? That is, at best, a very difficult way of evaluating integrals! (Though excellent for thinking how to set up integrals for applications.)

Do you know the anti-derivatives of sine and cosine themselves? If so you can use the trig identity [itex]cos^2(\theta)= (1/2)(1+ cos(2\theta)[/itex].
 

1. What is a Riemann sum?

A Riemann sum is a method used in calculus to approximate the area under a curve by dividing the region into smaller rectangles and summing the areas of these rectangles.

2. How do you evaluate a Riemann sum?

To evaluate a Riemann sum, you must first determine the width of each rectangle by dividing the interval into equal subintervals. Then, you calculate the height of each rectangle by plugging in the x-value of a given point into the function. Next, you multiply the width and height of each rectangle and sum them together to find the approximate area under the curve.

3. What is the purpose of a Riemann sum?

The purpose of a Riemann sum is to approximate the area under a curve when it is not possible to find the exact value using traditional integration methods. It is also used as a building block for more advanced techniques, such as the definite integral.

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

A left Riemann sum uses the left endpoint of each subinterval to determine the height of the rectangles, a right Riemann sum uses the right endpoint, and a midpoint Riemann sum uses the midpoint of each subinterval. These different methods can give slightly different approximations, but as the number of subintervals increases, the difference becomes negligible.

5. How do you improve the accuracy of a Riemann sum?

The accuracy of a Riemann sum can be improved by increasing the number of subintervals used, thus creating smaller rectangles and providing a better approximation of the area under the curve. This can be done manually or by using technology, such as a graphing calculator or computer program.

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