How do you use Riemann sums to evaluate the area under the curve of #f(x)= 3x # on the closed interval [1,5], with n=4 rectangles using right, left, and midpoints?
1 Answer
# L RS = 30 #
# R RS = 42 #
# M RS = 36 #
Explanation:
We have:
# f(x) = 3x #
We want to calculate over the interval
# Deltax = (5-1)/4 = 1#
Note that we have a fixed interval (strictly speaking a Riemann sum can have a varying sized partition width). The values of the function are tabulated as follows;
Left Riemann Sum
# L RS = sum_(r=0)^3 f(x_i)Deltax_i #
# " " = 1 * (3 + 6 + 9 + 12) #
# " " = 1 * (30) #
# " " = 30 #
Right Riemann Sum
# R RS = sum_(r=1)^4 f(x_i)Deltax_i #
# " " = 1 * (6 + 9 + 12 + 15) #
# " " = 1 * (42) #
# " " = 42 #
Mid Riemann Sum
# M RS = sum_(r=0)^3 ((f(x_i)+f(x_(i+1)))/2)Deltax_i #
# " " = 1 * (4.5 + 7.5 + 10.5 + 13.5) #
# " " = 1 * (36) #
# " " = 36 #
Actual Value
For comparison of accuracy:
# Area = int_1^5 \ 3x \ dx #
# " " = [(3x^2)/2]_1^5 #
# " " = 3/2 \ [x^2]_1^5 #
# " " = 3/2( 25-1)#
# " " = 36#