Roman C. answered • 04/21/16
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∫08 x2 dx
The cuts are at x = 2,4,6. The width of each interval is 2.
Lower rectangle rule:
2f(0) + 2f(2) + 2f(4) + 2f(6)
= 2·02 + 2·22 + 2·42 + 2·62
= 0 + 8 + 32 + 72
= 112
Upper rectangle rule:
2f(2) + 2f(4) + 2f(6) + 2f(8)
= 2·22 + 2·42 + 2·62 + 2·82
2f(2) + 2f(4) + 2f(6) + 2f(8)
= 2·22 + 2·42 + 2·62 + 2·82
= 8 + 32 + 72 + 128
= 240
Average : (112 + 240) / 2 = 176
Trapezoid rule:
2[(f(0)+f(2))/2] + 2[(f(2)+f(4))/2] + 2[(f(4)+f(6))/2] + 2[(f(6)+f(8))/2]
= 2[(02+22)/2] + 2[(22+42)/2] + 2[(42+62)/2] + 2[(62+82)/2]
= 2[4/2] + 2[20/2] + 2[52/2] + 2[100/2]
= 4 + 20 + 52 + 100
= 176
Which agrees.
By the way, here is the general proof that T = (L + U) / 2
Consider a Riemman integrable function f(x) on [a,b] and evaluating ∫ab f(x) dx
Take intervals [a=x0,x1], [x1,x2], ..., [xn-1,xn=b]
T = (x1 - x0)[(f(x0) + f(x1))/2] + ... + (xn - xn-1)[(f(xn) + f(xn-1))/2]
= (x1 - x0)f(x0)/2 + (x1 - x0)f(x1)/2 + ... + (xn - xn-1)f(xn)/2 + (xn - xn-1)f(xn-1)/2
= [(x1 - x0)f(x0) + ... + (xn - xn-1)f(xn-1)] / 2 + [(x1 - x0)f(x1) + ... + (xn - xn-1)f(xn)] / 2
= L/2 + U/2
= (L + U) / 2
