How do you use Riemann sums to evaluate the area under the curve of #f(x)=cosx+0.5# on the closed interval [0,2pi], with n=pi rectangles using midpoints?

1 Answer
Feb 26, 2016

#pi * f((0+pi)/2)+pi * f((pi+2pi)/2)=pi#, so the area under the curve is about #pi#. (In fact, it is exactly #pi#).

Explanation:

Since #n=pi#, we will be measuring rectangles with length #pi#. Since we're operating on the interval #[0,2pi]#, we will only have two rectangles.

The first rectangle is on the interval #[0,pi]#. The midpoint of #x=0# and #x=pi# is #x=(0+pi)/2=pi/2#.

At #x=pi/2# we have a rectangle with a base of #pi# and height of #f(pi/2)=cos(pi/2)+1/2=1/2#.

This rectangle has an area of #A=pi(1/2)=pi/2#.

The next rectangle will be on the interval #[pi,2pi]#, of which the midpoint is located at #x=(3pi)/2#.

The height of this rectangle is #f((3pi)/2)=cos((3pi)/2)+1/2=1/2#.

The area of the rectangle with base #pi# and height #1/2# is again #A=pi(1/2)=pi/2#.

The two rectangles we've found both have area #pi/2#, so the total area of both rectangles is #pi#, the approximate area under the curve.

wolframalpha.com