APPM 1350 Review Solutions 3

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online