Consider the function f(x)=x2 on the interval [1, 9]. Let P be a uniform partition of [1,9] with 16 sub-intervals. Compute the left and right Riemann sum of f on the partition. Use exact values.

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To find left and rigth Riemann sum of f(x)=x2 on the interval [1, 9] with 16 subintervals. The formula for left Riemann sum is, ∫abf(x)dx≈Δx(f(x0)+f(x1)+....f(xn−2)+f(xn−1) where Δx=b−an Here f(x)=x2 and [a,b]=[1,9],n=16 So Δx=9−116=12. Now divide the interval [1, 9] into n=16 subintervals of the length Δx=12. The subintervals are, [1,32],[32,2],[2,52],[52,3],[3,72],[72,4]....[8,172],[17/2,9]. To find left Riemann sum use the left end points of the subintervals. Here the left endpoints are, 1, 32,2,....,8,172. Now evaluate the function at left endpoints, f(x0)=f(1)=12=1 f(x1)=f(32)=(32)2=92 f(x2)=f(2)=22=4 f(x3)=f(52)=(52)2=255 ⋮ f(x14)=f(8)=82=64 f(x15)=f(172)=(172)2=2894 Substitute all values in the formula, ∫19x2dx≈12(1+94+4+254+9+494+....+64+2894) ≈12(1+2.25+4+6.25+9+....+64+72.25) ≈12(446) ≈223 Therefore left Riemann sum is 223. To find right Riemann sum. The formula for right Riemann sum is, ∫abf(x)dx≈Δx(f(x1)+f(x2)+....f(xn−1)+f(xn)) where Δx=b−an Here f(x)=x2and[a,b]=[1,9],n=16. So Δx=9−116=12. Now divide the interval [1, 9] into n=16 subintervals of the length Δx=.12 The subintervals are,

Consider the function f(x) = x^{2} on the interval [1, 9]. Let P be a uniform partition of [1,9] with 16 sub-intervals. Compute the left and right Riemann sum of f on the partition. Use exact values.

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