How do you use Riemann sums to evaluate the area under the curve of #f(x)=x^3# on the closed interval [1,3], with n=4 rectangles using right, left, and midpoints?

1 Answer
Jim H
Oct 3, 2016

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

#int_1^3 x^3 dx#

Note that #f(x) = x^3# and #a=1# and #b=3#

#n=4# So #Deltax = (b-a)/n = (3-1)/4 =1/2#

All endpoints: star with #a# and add #Deltax# successively:

#1# #underbrace(color(white)"XX")_(+1/2)# #3/2# #underbrace(color(white)"XX")_(+1/2)# #2# #underbrace(color(white)"XX")_(+1/2)# #5/2# #underbrace(color(white)"XX")_(+1/2)# #3#

The subintervals are #(1,3/2)#, #(3/2,2)#, #(2,5/2)#, #(5/2,3)#

Now the Riemann sum is the sum of the area of the 4 rectangles. We find the area of each rectangle by

#"height" xx "base" = f("sample point") xx Deltax#

For the left sum, #L_4# we use the left endpoints of the subintervals as our sample points.

Left Endpoints: #1#, #3/2#, #2#, #5/2#

So,
#L_4 = f(1)*1/2+f(3/2)*1/2+f(2)*1/2+f(5/2)*1/2#

#= (f(1)+f(3/2)+f(2)+f(5/2))1/2#

# = (1+27/8+8+125/8)1/2 = 14#

For the right sum, use the right endpoints of the subintervals

Right Endpoints: #3/2#, #2#, #5/2#, #3#

#R_4 = (f(3/2)+f(2)+f(5/2)+f(3))1/2#

I'll leave the arithmetic to the reader.

For midpoints, we take the midpoint of each subinterval.

The subintervals are #(1,3/2)#, #(3/2,2)#, #(2,5/2)#, #(5/2,3)#

The midpoint is the average of the endpoints. Add the endpoints and divide by 2, (Or find the first one and successively add #Deltax = 1/2 = 2/4#.)

Midpoints: #5/4#, #7/4#, #9/4#, #11/4#

#M_4 = (f(5/4)+f(7/4)+f(9/4)+f(11/4))1/2#

The arithmetic is left to the reader.

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