APPM 1350 Review Solutions 3

Z2 7 a find an approximation to the integral 2 x2 dx

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Unformatted text preview: with n = 5 using: (i) right endpoints, (ii) left endpoints, (iii) midpoints (b) Evaluate the lower sums (under estimate) for approximating the area of the region bounded by f (x) = x2 4, 2 x 2 and the x-axis using 4 approximating rectangles of equal width. Solution: (a) (i) Note that Z 2 (2 0 xi = 2 x ) dx ⇡ 5 X x = (2 0)/5 = 2/5 and xi = 0 + i x = 2i/5 and here f (x) = 2 x⇤ =xi i f ( x⇤ ) i xi i=1 z}|{ = = = = (a) (ii) Here, Z 2 (2 0 x = 2/5 and xi 2 x ) dx ⇡ 5 X i=1 1 f ( x⇤ ) i = 2(i 5 X ✓ 2i ◆ 2 f 55 i=1 ✓◆ ✓◆ ✓◆ ✓◆ ✓◆ 2 2 4 6 8 10 f +f +f +f +f 5 5 5 5 5 5 2 46 34 14 14 50 + + 5 25 25 25 25 25 2 30 12 = 5 25 25 1)/5, thus x⇤ = xi i xi x2 , thus z}|{ = = = = 1 5 X ✓ 2(i f i=1 1) 5 ◆ 2 5 ✓◆ ✓◆ ✓◆ ✓◆ 2 2 4 6 8 f (0) + f +f +f +f 5 5 5 5 5 2 50 46 34 14 14 + + + 5 25 25 25 25 25 2 130 52 = 5 25 25 (a) (iii) We have, Z 2 (2 0 x = 2/5 and xi = [xi 2 x ) dx ⇡ 5 X f ( x⇤ ) i 1 + xi ]/2 = [2(i 5 X ✓ 2i f x⇤ = xi i z}|{ = xi i=1 1)/5 + 2i/5]/2 = (2i ◆ 2 5 5 i=1 ✓◆ ✓◆ ✓◆ ✓◆ ✓◆ 2 1 3 5 7 9 f +f +f +f +f 5 5 5 5 5 5 2 49 41 25 1 31 + + + 5 25 25 25 25 25 2 85 34 = 5 25 25 = = = 1 1)/5, thus (b) We have, xi = x = [2 ( 2)]/4 = 1, and since x2 4 0 for 2 x 2, the height of each approximating rectangle is h = 0 f (x⇤ ) = f (x⇤ ), thus i i 2 3 Z2 4 X 6 7 (x2 4) dx ⇡ f (x⇤ ) xi = 1 · 4 f ( 2) f ( 1) f (1) f (2) 5 = [0 + 3 + 3 + 0] = 6 i | {z }| {z } 2 i=1 8. Given...
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