Answer to Question #6242 in Algebra for Brandy
2012-02-02T10:23:15-05:00
Use inscribed rectangles to approximate the area under h(x) = -0.5x^2 + 4 for 0 <= x <= 2 and rectangle width 0.25.
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2012-02-07T14:18:13-0500
Use inscribed rectangles to approximate the area under h(x) = -0.5x² + 4 for 0 <= x <= 2 and rectangle width 0.25. First rectangle: (x1 is the left base point, x2 is the right base point, S1 is rectangle's area) x1 = 0; x2 = 0.25; h(x1) = 4; h(x2) = 3.96875; S1 = min(h(x1),h(x2))*0.25 = 3.96875*0.25 = 0.9921875; Second rectangle: x1 = 0.25; x2 = 0.5; h(x1) = 3.96875; h(x2) = 3.875; S2 = 0.25*3.875 = 0.96875; Third rectangle: x1 = 0.5; x2 = 0.75; h(x1) = 3.875; h(x2) = 3.71875; S3 = 0.25*3.71875 = 0.9296875; Forth rectangle: x1 = 0.75; x2 = 1; h(x1) = 3.71875; h(x2) = 3.5; S4 = 0.25*3.71875 = 0.9296875; Fifth rectangle: x1 = 1; x2 = 1.25; h(x1) = 3.5; h(x2) = 3.21875; S5 = 0.25*3.21875 = 0.8046875; Sixth rectangle: x1 = 1.25; x2 = 1.5; h(x1) = 3.21875; h(x2) = 2.875; S6 = 0.25*2.875 = 0.71875; Seventh rectangle: x1 = 1.5; x2 = 1.75; h(x1) = 2.875; h(x2) = 2.46875; S7 = 0.25*2.46875 = 0.6171875; x1 = 1.75; x2 = 2; h(x1) = 2.46875; h(x2) = 2; S8 = 0.25*2 = 0.5; So,vapproximated area is S = S1+S2+S3+S4+S5+S6+S7+S8 = 6.4609375.
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