Answer Key
831
33 . L 100 =3.555, R 100 =3.670. The plot shows that the left Riemann sum is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles. 35 . The sum represents the cumulative rainfall in January 2009. 37 . The total mileage is 7× ∑ i =1 25 ⎛ ⎝ 1+ ( i −1) 10 ⎞ ⎠ =7×25+ 7 10×12×25=385mi. 39 . Add the numbers to get 8.1-in. net increase. 41 . 309,389,957 43 . L 8 =3+2+1+2+3+4+5+4=24 45 . L 8 =3+5+7+6+8+6+5+4=44 47 . L 10 ≈1.7604, L 30 ≈1.7625, L 50 ≈1.76265 49 . R 1 =−1, L 1 =1, R 10 =−0.1, L 10 =0.1, L 100 =0.01, and R 100 =−0.1. By symmetry of the graph, the exact area is zero. 51 . R 1 =0, L 1 =0, R 10 =2.4499, L 10 =2.4499, R 100 =2.1365, L 100 =2.1365 53 . If ⎡ ⎣ c , d ⎤ ⎦ is a subinterval of ⎡ ⎣ a , b ⎤ ⎦ under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is f ( c )( d − c ). But, f ( c ) ≤ f ( x ) for c ≤ x ≤ d , so the area under the graph of f between c and d is f ( c )( d − c ) plus the area below the graph of f but above the horizontal line segment at height f ( c ), which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of f on ⎡ ⎣ a , b ⎤ ⎦ . 55 . L N = b − a N ∑ i =1 N f ⎛ ⎝ a +( b − a ) i −1 N ⎞ ⎠ = b − a N ∑ i =0 N −1 f ⎛ ⎝ a +( b − a ) i N ⎞ ⎠ and R N = b − a N ∑ i =1 N f ⎛ ⎝ a +( b − a ) i N ⎞ ⎠ . The left sum has a term corresponding to i =0 and the right sum has a term corresponding to i = N . In R N − L N , any term corresponding to i =1, 2,…, N −1 occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with R N − L N = b − a N ⎛ ⎝ f ⎛ ⎝ a +( b − a ) ⎞ ⎠ N N ⎞ ⎠ − ⎛ ⎝ f ( a )+( b − a ) 0 N ⎞ ⎠ = b − a N ⎛ ⎝ f ( b )− f ( a ) ⎞ ⎠ . 57 . Graph 1: a. L ( A ) =0, B ( A ) =20; b. U ( A ) =20. Graph 2: a. L ( A ) =9; b. B ( A ) =11, U ( A ) =20. Graph 3: a. L ( A ) =11.0; b. B ( A ) =4.5, U ( A ) =15.5. 59 . Let A be the area of the unit circle. The circle encloses n congruent triangles each of area sin ⎛ ⎝ 2 π n so n
⎞ ⎠ 2 ,
⎛ ⎝ 2 π n
⎞ ⎠ ≤ A .
2 sin
⎛ ⎝ 2 π n
⎞ ⎠ , so
⎛ ⎝ cos
⎛ ⎝ π n
⎞ ⎠ +sin
⎛ ⎝ π n
⎞ ⎠ tan
⎛ ⎝ π n
⎞ ⎠
⎞ ⎠ sin
Similarly, the circle is contained inside n congruent triangles each of area BH 2 =
1 2
π sin ⎛
⎞ ⎠
⎝ 2 π n
⎛ ⎝ 2 π n
⎞ ⎠ =
⎛ ⎝ 2 π n
⎞ ⎠ ⎛
⎛ ⎝ π n
⎞ ⎠ ⎞ ⎠ +sin
⎛ ⎝ π n
⎞ ⎠ tan
⎛ ⎝ π n
⎞ ⎠ . As n →∞, n
A ≤ n
⎞ ⎠ → π , so we conclude π ≤ A . Also, as
⎝ cos
⎛ ⎝ 2 π n
2 sin
2 sin
n →∞, cos ⎛
⎞ ⎠ +sin
⎛ ⎝ π n
⎞ ⎠ tan
⎛ ⎝ π n
⎞ ⎠ →1, so we also have A ≤ π . By the squeeze theorem for limits, we conclude that A = π .
⎝ π n
2 ⎛ ⎝ 5 x 2 −3 x 3 ⎞ 1 cos 2 (2 πx ) dx
61 . ∫ 63 . ∫
⎠ dx
0
0
Made with FlippingBook - professional solution for displaying marketing and sales documents online