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227Hw4solns

# 227Hw4solns - Math 227 Sections 4 and 5 Solultions to...

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Math 227 Sections 4 and 5 Solultions to Written HW 4 Section 5.5 # 11,12,38 Section 5.6 # 24,25,26 5.5.11 Apply Formula 26 with a = p 7 : Z 1 x p 7 + x 2 dx = 1 p 7 ln p 7 + p 7 + x 2 x + C 5.5.12 Apply Formula 28 with a = p 7 : Z 1 x p 7 x 2 dx = sin 1 ± x p 7 ² + C 5.6.24 The tank must hold 5000 lbs of fuel having a density of 42 pounds/cubic foot. Rounding to the nearest hundreth of a cubic foot, its volume V is given by V = 5000 lb 42 lb/ft 3 = 119 : 05 ft 3 : Let A denote the cross-sectional area of the tank and let L denote the length of the tank. Then A ± L = V so L = V A = 119 : 05 A feet. We can approximate A using Simpson±s Method then solve for L: From the given data, x = 1 ft and n = 6 : A ² S 6 = 1 3 [ y 0 + 4 y 1 + 2 y 1 + 4 y 3 + 2 y 4 + 4 y 5 + y 6 ] = 1 3 [1 : 5 + (4)(1 : 6) + (2)(1 : 8) + (4)(1 : 9) + (2)(2 : 0) + (4)(2 : 1) + 2 : 1] = 11 : 2 ft 2 : Thus L ² 119 : 05 11 : 2 = 10 : 6 feet. 5.6.25

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227Hw4solns - Math 227 Sections 4 and 5 Solultions to...

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