Without adding up 117 numbers can you compute the other sum (left or right)? You

Question

Without adding up 117 numbers can you compute the other sum (left or right)?

You want to compute 9 1 x3 dx by approximations using Rieman sums. Let N be the number of rectangles that you will use for your approximation. Estimate the area by computing left sums and right sums for N = 8 and then computing left sums and right sums for N = 16. Why does one type of sum overestimate and the other underestimate? Suppose that a friend computes a Riemann sum with N = 117 and obtains 1615.2. Did your friend use a left or right Riemann sum?

Explanation / Answer

x^3 is monotone increasing. Thus, left Riemann sums underestimate, as the rectangles are underneath the function, and right Riemann sums overestimate, as the rectangles are above the function.

Using equal width intervals, as (9-1)/8 = 1, we calculate 1/1*(f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)) =

1*(1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3) =1*(1+8+27+64+125+216+343+512) = 1296

The right Riemann sum is

1*(f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)) = 1*(2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3) =

1*(8+27+64+125+216+343+512+729) = 2024

For 16 intervals, (9-1)/16 = 8/16 = 1/2, so using equal width intervals, we calculate for the left Riemann sum

1/2*(f(1)+f(3/2)+f(2)+f(5/2+f(3)+f(7/2+f(4)+f(9/2+f(5)+f(11/2+f(6)+f(13/2+f(7)+f(15/2+f(8)+f(17/2)) =

1/2*(1^3+(3/2)^3+2^3+(5/2)^3+3^3+(7/2)^3+4^3+(9/2)^3+5^3+(11/2)^3+6^3+(13/2)^3+7^3+(15/2)^3+8^3+(17/2)^3) =

1/2*(1+27/8+8+125/8+27+343/8+64+729/8+125+1331/8+216+2197/8+343+3375/8+512+4913/8) = 1463

The right Riemann sum is

1/2*(f(3/2)+f(2)+f(5/2+f(3)+f(7/2+f(4)+f(9/2+f(5)+f(11/2+f(6)+f(13/2+f(7)+f(15/2+f(8)+f(17/2)+f(9)) =

1/2*(27/8+8+125/8+27+343/8+64+729/8+125+1331/8+216+2197/8+343+3375/8+512+4913/8+729) = 1827

The integral from 1 to 9 of x^3 = x^4/4 evaluated from 1 to 9 = 9^4/4 - 1^4/4 = (6561-1)/4 = 6560/4 = 1640

As 1615.2 is less than 1640, this is a left Riemann sum.

note that for all but the first and last rectangle, the right value of one interval is the left value of the next, hence the areas cancel; we see that in the examples above with n=8 and 16, where all but f(1) and f(9)* the interval width cancel out when we look at differences. Thus, with n = 8, the difference of areas is 2024 - 1296 = 728. Also(9^3 - 1^3)*1 = (729-1)*1 = 728*1=728. Thus, it checks. With n = 16, 1827 - 1463 = 364 1/2*(9^3-1^3) = 1/2*728 = 364; thus, it again checks.

If the 117 rectangles are all of equal width, then the difference in the right and left Riemann sum would be (9^3-1^3)*8/117 =

448/9

Thus, the right Riemann sum would be 1615.2+448/9 = 1664.978 is approximately 1665.0

Let's check if rectangles of the width 8/117 produced 1615.2.

I did the calculation, and got 1615.20461684564 for the left Riemann sum, so this was indeed what was done.

(You start with 1, then add increments of 8/117 until you get to 9 - 8/117. Then, you cube these values, and sum. Then, multiply by 8/117)

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