How do you use Riemann sums to evaluate the area under the curve of #f(x)= In(x)# on the closed interval [3,18], with n=3 rectangles using right, left, and midpoints?

1 Answer
Jan 29, 2016

Please see the explanation section below.

Explanation:

I will use what I think is the usual notation throughout this solution.

Note that #f(x) = lnx# and #a=3# and #b=18#

#n=3# So #Deltax = (b-a)/n = (18-3)/3 =5#

The endpoints: start with #a# and add #Deltax# successively:

#3# #underbrace(color(white)"XX")_(+5)# #8# #underbrace(color(white)"XX")_(+5)# #13# #underbrace(color(white)"XX")_(+5)# #18#

The subintervals then are:

#[3,8]#, #[8,13]#, and #[13,18]#

The left endpoints are: #3#, #8# and #13#.

The right endpoints are #8#, #13#, and #18#

The midpoints may be found by averaging the endpoints.
They are: #5.5#, #10.5#, #15.5#
(As an alternative method, find the first midpoint and add #Deltax# successively.)

Now the Riemann sum is the sum of the area of 3 rectangles. We find the area of each rectangle by
#"height" xx "base" = f("sample point") xx Deltax#

So, using left endpoints, we have

#L = f(3)*5+f(8)*5+f(13)*5#

#= (f(3)+f(8)+f(13))*5#

The arithmetic is left to the student.

Using right endpoints we have

#R = f(8)*5+f(13)*5+f(18)*5#

#= (f(8)+f(13)+f(18))*5#

The arithmetic is left to the student.

Using midpoints we have

#M = f(5.5)*5+f(10.5)*5+f(15.5)*5#

#= (f(5.5)+f(10.5)+f(15.5))*5#

The arithmetic is left to the student.