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5.6-5practice

5.6-5practice - 370 Chapter 5 Integration 94 A oe A b A...

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370 Chapter 5 Integration 94. A A A œ " # Limits of integration: y x and y x x x œ œ Ê œ \$ & \$ & x x 0 x (x 1)(x 1) 0 a 1, b 0 Ê œ Ê œ Ê œ œ & \$ \$ " " and a 0, b 1; f (x) g (x) x x and # # " " \$ & œ œ œ f (x) g (x) x x by symmetry about the origin, # # & \$ œ Ê A A 2A A 2 x x dx 2 " # # \$ & " ! œ Ê œ œ ' 0 1 a b x x 4 6 2 œ œ ˆ " " " 4 6 6 95. A A A œ " # Limits of integration: y x and y x , x 0 œ œ Ê œ Á " " x x x 1 x 1 , f (x) g (x) x 0 x Ê œ Ê œ œ œ \$ " " A x dx ; f (x) g (x) 0 Ê œ œ œ œ " # # " ! " " # ' 0 1 x 2 x x A x dx 1 ; œ Ê œ œ œ œ # # # " " " # " # # ' 1 2 x A A A 1 œ œ œ " # " " # # 96. Limits of integration: sin x cos x x a 0 œ Ê œ Ê œ 1 4 and b ; f(x) g(x) cos x sin x œ œ 1 4 A (cos x sin x) dx [sin x cos x] Ê œ œ ' 0 4 1 Î% ! (0 1) 2 1 œ œ Š È È È 2 2 # # 97. (ln 2x ln x) dx ( ln x ln 2 ln x) dx (ln 2) dx (ln 2)(5 1) ln 2 ln 16 ' ' ' 1 1 1 5 5 5 œ œ œ œ œ % 98. A tan x dx tan x dx dx dx ln cos x ln cos x œ œ œ ' ' ' ' 4 0 4 0 0 3 0 3 ! Î% Î\$ ! sin x sin x cos x cos x c d c d k k k k 1 1 ln 1 ln ln ln 1 ln 2 ln 2 ln 2 œ œ œ Š ˆ È " " # # È 2 3 99. e e dx e e e 3 1 2 2 ' 0 ln 3 a b Š Š ˆ ˆ 2x x x ln 3 e e e 9 8 ln 3 0 œ œ œ œ œ 2x 2 ln 3 # # # # # # ! " 100. e e dx 2e 2e 2e 2e 2e 2e (4 1) (2 2) 5 4 1 ' 0 2 ln 2 ˆ ˆ a b x 2 x 2 x 2 x 2 ln 2 ln 2 2 ln 2 0 Î Î Î Î ! ! œ œ œ œ œ 101. A dx 2 dx; u 1 x du 2x dx; x 0 u 1, x 2 u 5 œ œ œ Ê œ œ Ê œ œ Ê œ ' ' 2 0 2 2 2x 2x 1 x 1 x # c d A 2 du 2 ln u 2(ln 5 ln 1) 2 ln 5 Ä œ œ œ œ ' 1 5 " & " u c d k k 102. A 2 dx 2 dx 2 2 œ œ œ œ œ œ ' ' " " # # # # # # " " 1 1 1 1 1 x x ˆ ‰ ˆ ˆ ‰ ˆ Š Š x ln 2 2 3 3 ln ln ln

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Section 5.6 Substitution and Area Between Curves 371 103. (a) The coordinates of the points of intersection of the line and parabola are c x x c and y c œ Ê œ „ œ # È (b) f(y) g(y) y y 2 y the area of œ œ Ê È È È ˆ the lower section is, A [f(y) g(y)] dy L 0 c œ ' 2 y dy 2 y c . The area of œ œ œ ' 0 c c È 2 4 3 3 \$Î# \$Î# ! the entire shaded region can be found by setting c 4: A 4 . Since we want c to divide the œ œ œ œ ˆ ‰ 4 4 8 32 3 3 3 \$Î# region into subsections of equal area we have A 2A 2 c c 4 œ Ê œ Ê œ L 32 4 3 3 ˆ \$Î# #Î\$ (c) f(x) g(x) c x A [f(x) g(x)] dx c x dx cx 2 c œ Ê œ œ œ œ # # \$Î# L c c c c c c ' ' a b x c 3 3 È c . Again, the area of the whole shaded region can be found by setting c 4 A . From the œ œ Ê œ 4 32 3 3 \$Î# condition A 2A , we get c c 4 as in part (b). œ œ Ê œ L 4 32 3 3 \$Î# #Î\$ 104. (a) Limits of integration: y 3 x and y 1 œ œ # 3 x 1 x 4 a 2 and b 2; Ê œ Ê œ Ê œ œ # # f(x) g(x) 3 x ( 1) 4 x œ œ a b # # A 4 x dx 4x Ê œ œ ' 1 2 a b # # # x 3 8 8 16 œ œ œ ˆ ˆ 8 8 16 32 3 3 3 3 (b) Limits of integration: let x 0 in y 3 x œ œ # y 3; f(y) g(y) 3 y 3 y Ê œ œ È È ˆ 2(3 y) œ "Î# A 2 (3 y) dy 2 (3 y) ( 1) dy ( 2) 0 (3 1) Ê œ œ œ œ ' ' 1 1 3 3 "Î# "Î# \$Î# \$ " ˆ 2(3 y) 3 3 4 (8) œ œ ˆ ‰ 4 32 3 3 105. Limits of integration: y 1 x and y œ œ È 2 x È 1 x , x 0 x x 2 x (2 x) Ê œ Á Ê œ Ê œ È È 2 x È # x 4 4x x x 5x 4 0 Ê œ Ê œ # # (x 4)(x 1) 0 x 1, 4 (but x 4 does not Ê œ Ê œ œ satisfy the equation); y and y œ œ Ê œ 2 x 2 x x x 4 4 È È 8 x x 64 x x 4. Ê œ Ê œ Ê œ È \$
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