@ XB, Y1 = Y2 5+2X=8+1X XB=3
     
    A =  0 ∫3(Y2 – Y1)dx = 0 ∫3[(8+1X) – (5+2X)]dx = 0 ∫3[3X0 – 1 X1]dx
    A = [(3/1) X1 – (1/2) X2]|03 = [(3)(3) – (1/2)(9)] –[(3)(0) – (1/2)(0)]= 4.5
    CHECK: b=(8-5) = 3 h=(3-0) = 3 A = (3)(3)/2 = 4.5
     
    B =  3 ∫5(Y1 – Y2)dx = 3 ∫5[(5+2X) – (8+1X)]dx = 3 ∫5[-3X0 + 1 X1]dx
    B =[(-3/1) X1 + (1/2) X2]|35 = [(-3)(5)+(1/2)(25)] – [(-3)(3)+(1/2)(9)]
    B = [-15 + 12.5] – [-9 + 4.5] = -2.5 + 4.5 = +2
    CHECK: b=(15 - 13) = 2 h=(5 - 3) = 2 A = (2)(2)/2 = +2.0
     
    C =  0 ∫3(Y1)dx = 0 ∫3(5+2X)dx = [(5/1)X1 + (2/2)X2]|03 =
    C = [(5)(3)+(1)(9)] – [(-5)(0)+(1)(0)] = [15 + 9 ] – [0 + 0] = 24
    CHECK: [C is composed of a triangle (T) and a rectangle (R). b=(11 - 5) = 6 h=(3 - 0) = 3 T = (6)(3)/2 = +9R = (5)(3) = 15 C = S + T = 9 + 15 = 24
     
    D =  3 ∫5(Y2)dx = 3 ∫5(8+1X)dx = [(8/1)X1 + (1/2)X2]|35 =
    D = [(8)(5)+(1/2)(25)] – [(8)(3)+(1/2)(9)] = [40 + 12.5 ] – [24 + 4.5] = 24
    CHECK: [D is composed of a triangle (T) and a rectangle (R). b=(13 - 11)=2 h=(5 - 3) = 2 T = (2)(2)/2 = +2 R = (11)(2) = 22 D = S + T = 2 + 22 = 24
     

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