How do you Find the Riemann sum for f(x)=sinx on the interval [0,pi]?

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The Riemann Sum for this is: ##sum_(i=1)^n(pi/n)(sin((ipi)/n))##

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As ##nrarroo## this sum seems to be approaching 2.

A Riemann Sum can be thought of as a generalization of the “area problem” that begins the subject of Integral Calculus. The area problem asks how one can find the area under a curve that lies above the ##x##-axis over any given interval.

This is solved by dividing the interval into smaller “subintervals” of equal widths and finding the area of a rectangle for each subinterval with the width being the equal ##Delta x## we have chosen and the height being the ##y##-coordinate at some point (the left-endpoint, the right-endpoint, the midpoint, etc.) in the particular subinterval. We then add together all of these areas to approximate the area under the curve. We then decide that the actual area can be found by taking the limit of this area as the width of the rectangles shrinks (i.e., the number of rectangles approaches ##oo##).

In Riemann Sums, we realize that some of the restrictions in the above problem can be relaxed, with no impact on the final limit process. Specifically, the function does not need to always be positive (and you will note that your function is not on your interval), and the subintervals do not need to be “uniform” (of equal width), though we often choose uniform subintervals for ease of computation. Also, the point we choose for the height could vary from one interval to the next.

For your function, begin by dividing the interval ##[0,pi]## into ##n## subintervals which we will make uniform to make the rest of the problem simpler. This makes their width: ##Delta x = (pi-0)/n=pi/n##.

Now select a place to measure the height of each rectangle. We choose to use only right-hand endpoints, again to make our lives easier. This makes the height of the first rectangle: ##y=sin(pi/n)##, the second rectangle is ##y=sin((2pi)/n)##, then ##y=sin((3pi)/n)##, etc.

In other words, we get the sum: ##sum_(i=1)^n(pi/n)(sin((ipi)/n))##.

If we evaluate this for increasing values of ##n##, we get:

##n=10##, Sum=1.9835235. . .
##n=100##, Sum=1.9998355. . .
##n=1000##, Sum=1.999998355. . .

leading to the conclusion that the limit appears to be ##2##.



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