How do you calculate the left and right Riemann sum for the given function over the interval [0,2], n=4 for # f(x) = (e^x) − 5#?
# L RS = 5.0757#
# R RS = 1.8811 #
We have:
# f(x) = e^x5 #
graph{e^x5 [10, 10, 8, 8]}
We want to calculate over the interval
# Deltax = (20)/4 = 1/2#
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 #
# " " = 0.5 * (4  3.3513  2.2817  0.5183) #
# " " = 0.5 * (10.1513) #
# " " = 5.0757 #
Right Riemann Sum
# R RS = sum_(r=1)^4 f(x_i)Deltax_i #
# " " = 0.5 * (3.3513  2.2817  0.5183 + 2.3891) #
# " " = 0.5 * (3.7623) #
# " " = 1.8811 #
Actual Value
For comparison of accuracy:
# Area = int_0^2 \ e^x5 \ dx #
# " " = [e^x5x]_0^2 #
# " " = (e^210)  (e^00) #
# " " = e^210  1#
# " " = e^211#
# " " = 3.610943 ...#
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To calculate the left and right Riemann sums for the given function ( f(x) = e^x  5 ) over the interval ([0, 2]) with ( n = 4 ), you partition the interval into ( n ) subintervals of equal width and evaluate the function at specific points within each subinterval.

Determine the width of each subinterval: [ \Delta x = \frac{b  a}{n} = \frac{2  0}{4} = \frac{1}{2} ]

Calculate the partition points: [ x_0 = 0, \quad x_1 = \frac{1}{2}, \quad x_2 = 1, \quad x_3 = \frac{3}{2}, \quad x_4 = 2 ]

Evaluate the function at each partition point: [ f(x_0) = f(0), \quad f(x_1), \quad f(x_2), \quad f(x_3), \quad f(x_4) = f(2) ]

For the left Riemann sum, use the left endpoints of each subinterval: [ L_4 = \Delta x \left( f(x_0) + f(x_1) + f(x_2) + f(x_3) \right) ]

For the right Riemann sum, use the right endpoints of each subinterval: [ R_4 = \Delta x \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) \right) ]

Substitute the values of ( f(x) = e^x  5 ) at each endpoint and compute the sums.
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