G y sn i n 1 n 1 2n 6 n 9 2n2 n2 9 2n2 1 9 27 2n 9

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Unformatted text preview: 4 Area x n 16 32 3 ,c 32 3 tan ci 16 tan 5 32 5 ,c 32 4 7 32 Area 1 2 69 8 67. f x n x i 1 f ci i 1 1 1 16 3 9 16 3 49 16 3 16 tan 32 tan 3 32 tan 7 32 0.345 x on 0, 4 . 4 5.3838 8 5.3523 12 5.3439 16 5.3403 20 5.3384 Approximate area Exact value is 16 3 S ection 4.2 Area 187 69. f x n tan x on 1, 3 . 8 4 2.2223 8 2.2387 12 2.2418 16 2.2430 20 2.2435 Approximate area 71. We can use the line y x bounded by x a and x b. The sum of the areas of these inscribed rectangles is the lower sum. y The sum of the areas of these circumscribed rectangles is the upper sum. y x a x a b b We can see that the rectangles do not contain all of the area in the first graph and the rectangles in the second graph cover more than the area of the region. The exact value of the area lies between these two sums. 73. (a) 8 6 4 2 x 1 2 3 4 1 2 3 4 y (b) 8 6 4 2 y x Lower sum: s4 0 (c) 8 6 4 2 4 51 3 6 151 3 46 3 15.333 Upper sum: 4 S4 51 3 6 62 5 2111 15 326 15 21.733 y (d) In each case, x 4 n. The lower sum uses left endpoints, i 1 4 n . The upper sum uses right endpoints, i 4 n . The Midpoint Rule uses midpoints, i 1 4 n . 2 x 1 2 3 4 Midpoint Rule: M4 22 3 (e) n sn Sn Mn 4 15.333 21.733 19.403 44 5 55 7 8 17.368 20.568 19.201 62 9 6112 315 19.403 100 18.995 19.251 19.125 200 19.06 19.188 19.125 20 18.459 19.739 19.137 (f) s n increases because the lower sum approaches the exact value as n increases. S n decreases because the upper sum approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the exact value, whereas the upper sum is always larger. 188 Chapter 4 Differentiation 75. y 77. True. (Theorem 4.2 (2)) 4 3 2 1 1 2 3 4 x b. A 79. f x 6 square units sin x, 0, y 2 1.0 0.75 0.5 0.25 sin x, the x-axis, x 0 and x 2. Let A2 area of the rectLet A1 area bounded by f x angle bounded by y 1, y 0, x 0, and x 2. Thus, A2 21 1.570796. In this program, the computer is generating N2 pairs of random points in the rectangle whose area is represented by A2. It is keeping track of how many of these points, N1, lie in the region whose area is represented by A1. Since the points are randomly generated, we assume that A1 A2 N1 ⇒ A1 N2 N1 A. N2 2 f (x) = sin(x) ( π , 1) 2 π 4 π 2 x The larger N2 is the better the approximation to A1. 81. Suppose there are n rows in the figure. The stars on the left total 1 n n 1 stars in total, hence 21 1 2 2 ... ... n n nn 1 2 2 ... n, as do the stars on the right. There are 1 1. nn 83. (a) y 4.09 10 5 x3 0.016x2 2.67x 452.9 (b) 500 (c) Using the integration capability of a graphing utility, you obtain A 76,897.5 ft2. 0 0 350 Section 4.3 1. f x xi lim i 1 Riemann Sums and Definite Integrals 0, x 1 n2 n→ 2 x, y 3i2 n2 n 0, x 3 2i n2 n 3, ci 1 3i2 n2 3 2 y 3i y= x n→ f ci xi lim i 1 3i2 3 2i n2 n2 i 1 1 ... 3 n2 3(2)2 . . . 3(n − 1)2 3 n2 n2 x 3 3n lim 2i2 n→ n3 i 1 n→ lim 3 3 nn 2 n3 n 1 2n 6 1 2n 3n2 23 1 1 n nn 2 1 2n2 1 n→ lim 3 3 2 3 0 33 3.464 S ection 4.3 Riemann Sums and Definite Integrals 189 3. y 6 on 4, 10 . n n Note: x f4 i 1 10 n 6 n 4 n 6 , n 6 6 n → 0 as n → n i 1 f ci i 1 10 xi 6 dx 6i n 36 i 1 36 n 36 4 n→ lim 36 5. y x3 on n 1, 1 . n Note: x f i 1 1 n 2 n 24 n3 i 1 n n 1 n 2 , n 1 → 0 as n → 2i n n 3 f ci i 1 xi 2 2 1 12 n2 i n 2i n i 1 i 1 2 n n 1 i 1 6i n 12i 2 n2 8i 3 n3 2 n i2 1 16 n4 i 3 n i3 1 61 2 n 0 42 1 n2 41 2 n 1 n2 2 n 1 x3 dx 1 n→ lim 7. y x2 n 1 on 1, 2 . n Note: x i n 1 n 2 n 1 1 , n 1 i n → 0 as n → 2 n i 1 f ci i 1 xi i 1 f1 1 1 n 1 n 1 2 6 n 1 i 1 2i n 1 n2 i2 n2 10 3 1 3 2n 1 n 1 6n2 2 2 2n i n2 i 1 10 3 5 1n2 i n3 i 1 3 2...
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