Estimate the area under the curve #f(x) = x^2# over the interval #[0,10]# with #5# strips using Left Riemann Sums?

2 Answers
Jul 16, 2017

Since you are using 5 sub intervals, your intervals would be divided like this:
#[0,2], [2,4], [4,6], [6,8], [8,10]#

The left riemann sum requires you to use the lower number in every interval listed above as your #x# term, and then evaluate the functional value at that particular #x# value, like this:

Note that #2# would be the width of every rectangle formed, since the intervals are of equal width.

#1. Area_1 = f(0)*2 = 0^2*2 = 0#

#2. Area_2 = f(2)*2 = 2^2*2 = 8#

#3. Area_3 = f(4)*2 = 4^2*2 = 32#

#4. Area_4 = f(6)*2 = 6^2*2 = 72#

#5. Area_5 = f(8)*2 = 8^2*2 = 128#

Upon finding the individual areas of each of the five rectangles formed by the five sub-intervals evaluated at the "left side" #x# value, all that is left to do is add all the individual areas up:

#Total Area = 0 + 8 + 32 + 72 + 128 = 240#

Alternatively, you could use the fact that all the rectangles are the same width, keeping #Deltax# the same:
#Total Sum = 2*(f(0) + f(2) + f(4) + f(6) + f(8))#
#Total Sum = 2*(0 + 4 + 16 + 36 + 64) = 2*(120) = 240#

Therefore your left riemann sum would be #240# for the graph #y = x^2#on the interval #[0,10]#

Jul 16, 2017

# LRS = 240 #

Explanation:

We have:

# f(x) = x^2 #

We want to calculate over the interval #[0,10]# with #5# strips; thus:

# Deltax = (10-0)/5 = 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;

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Left Riemann Sum

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# LRS = sum_(r=0)^4 f(x_i) \ Deltax_i #
# " " = 2 * (0 + 4 + 16 + 36 + 64) #
# " " = 2 * (120) #
# " " = 240 #

Actual Value

For comparison of accuracy:

# Area = int_0^10 \ x^2 \ dx #
# " " = [x^3/3]_0^10 #
# " " = 1/3[x^3]_0^10 #
# " " = 1/3(1000) #
# " " = 1000/3 #
# " " = 333.3333... #