# 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 mid-ordinate 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 mid-ordinate rule.

1. 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 = 33.125-32 = 1.125

percentage error = 1.125/32*100% = appr 3.5 %

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