math
posted by kg
y = ½ x^2 – x +3 for 0≤x≤6
(b) Calculate the mid ordinates for 5 strips between x = 1 and x= 6, and hence use the midordinate rule to approximate the area under the curve between x = 1, x = 6 and the x = axis.
(c) assuming that the area determined by integration to be the actual area, calculate the percentage error in using the midordinate rule.

Reiny
I will assume that you are calculating the heights at
x = 1.5, 2.5, 3.5, 4.5, and 5.5
which would be :
2.625, 3.625, 5.625, 8.625, and 12.625
approximate area of the 5 parts
= (2.625)(1) + (3.625)(1) + (5.625)(1) + (8.625)(1) + (12.625)(1)
= 33.125
actual area
= ∫((1/2)x^2  x + 3) dx from x = 1 to 6
= [ (1/6)x^3  x^2/2 + 3x ] from x = 1 to 6
= (1/6)(216)  36/2 + 18 (3/2  1/2 + 3)
= 36  18 + 18  3/2 + 1/2  3
= 32
error = 33.12532 = 1.125
percentage error = 1.125/32*100% = appr 3.5 %
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