or - CHAPTER 7 Applications of Integration Section 7.1 Area of a Region Between Two Curves 1. A= ~2[0- (x :2-6x)] dr=-~2(x 2-6x) dr

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1 CHAPTER 7 Applications of Integration Section 7.1 Area of a Region Between Two Curves 1. A= ~[0- (x : 6x)] dr=-~(x 6x) dr 6~ 1356 or - 6. A: ~[(x- 1) 3 -(x-1)]dx 11. [~/3 ( - see x) dx I tr I ~" 3 3 I 3 I 3! I 1. ~.~/ (sec x - cos x) dr I I ~x ~r

2 Section 7.1 Area of a Region Between Two Curves 3 3 z(x) = - ½x g(x) = - ~ A~,I Matches (a) " f(x) = x + 1 g(x) = (x -1)= Az Matches (d) 17. (a) x = x=- = =- +-6 = 0 (*3)(-) = 0 Intersection points: (0, )and (-5,-3) A fo =a-sl = 6._[1 + 3.~ = 1 ~ ) + v/-~-xdx + lo "/-~ -xdx 6 I 3 (b) A = ~[( )_ ( - )] d -,:s -7- (c) The second method is simpler. Explanations will var. 18. (a) = x and = 6 - x x = 6-x => x ~ *x-6=o~(x+3)(x-)=o Intersection points: (, )and (-3, 9) -~_;~ ~ ~ 6~~-x -9. :,j_l.~ 113/I,6 -x - xldx = 1~5 6 (b) A = x/td* 6-),~d = ~, 6~ = 1~ 6 6 (c) The first method is simpler. Explanations will va~.

3 Chapter 7 Applications of Integration :ioo_ xi_(+,x_,, I:/3 ~x - 7--x+ lo/dx I. X +lox.~ (611+80)-(1-7 +0) = x+ 3x o / -0=~ 6 ], 0) 3. The points of intersection are given b: x x = 0 x(x-) = 0 whenx = 0, A = Io[g(x) f(x)~ dx 3 =-1(x x)dx =-i ~ - x~ L ~,.~o 3 A = 11I(-x3 + 3) - x]dx =I:f- ax-;1~ ~ _L~ =6 -, 1. A= x -(x+ = o-~x-x+ ~ +X o + ) 0. The points of intersection are given b: -x ~ +x+l = x+l -x + 3x = 0 x = 3x whenx = 0,3 A= 1[(-x+x+ 1)-(x+ 1)]dx = 1(_; + 3x) x 6" (, 6) = - + Jo -9 + (o, ) 3 6~ 5-

4 Section 7.1 Area of a Region Between Two Curves 5 5. The points of intersection are given b: x + x = x + x +x- = 0 (x+z)(x-1) = 0 whenx =-,1 1, (1, 1) ~ x The points of intersection are given b: #~+3 = i~ +3 X x =-- whenx = 0, The points of intersection are given b: -x = + 1 x ~ -x = 0 whenx = 0, 3 = X3/ 16 Jo (0, li A = -x x+l dx = I~(-x + xldx + x -6 3 = = ~+3 = The points of intersection are given b: 3x/~ - 1 =x-1 X- 1 = (X-- 1) 3 = X 3-3X + 3X- 1 X X + x = 0 x(x -3x+) = 0 x(x-)(x-1) = 0 whenx = 0,1, The points of intersection are given b: x = x and x = 0 and x = 0 x=l x=0 x= 1 A = I~[( - )- ()] d = E - ]; =1 Note that if ou integrate with respect to x, ou need two integrals. Also, note that the region is a triangle. 3 /3 = x x-l) o 3. ), -1)

5 6 Chapter 7 Applications of Integration 31. The points of intersection are given b: =+ (-)( +1) = 0 when =-1, f()] + = A = I[f()- g()]d _ ~) (-)d _- [-,/16-o =.3s The points of intersection are given b: - = - ( - 3) = 0 when = 0, 3 10 lo 35. =~x=~ x A= ~ [~o10 --fd = [10 In ]~ =- lo(in 10 - In ) = 10 In 5 ~ (o, / - 1 (1, lo) - -J Jo - ~ 8-6 (o, ) - -: A = l_1[-f()- g()] d =1n 1.7 =!_~,[(~ + 1)-O]d = + =6-1 i0, ) (5, )~ 1 ~ (o, - 1) -~

6 Section 7.1 Area of a Region Between Two Curves (a) , 1) i 1-1 (b) The points of intersection are given b: X x + 3x = x x(x-1)(x-3) = 0 whenx = 0,1,3 = ~:(x3_xz+3x)~ + ;(_x3+x~_3x)~=[ %_~x3 + 3x ~] [~ ~x 3 3x z]3 3 o+ +3, (c) Numerical approximation: = , (a) a 0. (a) lo (b) The point of intersection is given b: x 3 -x+l =-x x 3 +1 = 0 whenx =-1 A = I~,[f(x)- g(x)]dx =!~,[(x x + 1)-(-x)] dx - (o, 0) (b) The points of intersection are given b: X _ x = X x (x) = 0 whenx =0,_+ A = Io~ [x (X x)] dr = 1(X -x)dx 1-1 (c) Numerical approximation: (a) 9 = x_._sj (c) Numerical approximation: (b) The points of intersection are given b: x -x+3 = 3+x-x x(x-) = 0 whenx = 0, A= I][(3 + x - x) - (x x+ 3)] dr =!(-x + 8x) dx = _~ + x ~ = 6 _~0 3 (c) Numerical approximation: 1.333

7 8 Chapter 7 Applications of Integration 1. (a) f(x) = x - x ~, g(x) = x ~ - (b) The points of intersection are given b: x - x ~ = x ~ - X -- 5X -I- = 0 (x )(x1) =0 whenx = +,_+1 B smmetr: A I[(X X) - (X )] dx+ Ilrz[(X )-(X X)] dx = 1(X 5Xb ) dx q-i~(--x q-5x -- ) dx (c) Numerical approximation: = 8.0 +x +-m+---x 3 o 5 3 ~ =8 ~o (a) f(x) = x - x, g(x) = x 3 - x (-, o) (, o) -5 (1, -3) (b) The points of intersection are given b: x - x ~ = x 3 - x X -- X 3 -- X + x = 0 x(x-1)(x+)(x-) = 0 whenx =-,0,1, A = I~I(X x) -(X - x~)l dx + 1[(x x~) - (x x)] dx+ I;I(x x)-(x X)] d-x = _ 93 3"-d- ~ ~ (c) Numerical approximation: ~ 9.767

8 Section 7.1 Area of a Region Between Two Curves 9 3. (a) 3 6. (a) -1 (o, o) (, O) 5. (a) (b) The points of intersection are given b: 1 x l+x x + x - = 0 x=+l A = l~ E f(x) - g(x)~ dx =!~ 1 + x [1 dx = arctan x, I 6 o X ~ 1.37 (c) Numerical approximation: (b) and (c) ou must use numerical integration: A = J0 x~ - ~x dx "~ 3.3 ~7.,~ : I~[(:~-,:os x)- ~os x~ = f~ ~(1 - cos x) dx do = Ix- sin X]0 r = x ~ [~/6 (cos x - sin x) dx COS = 1 sin x + xj_~/ -1 (b) A = x = 31n10,~ (c) Numerical approximation: (a) 5 = j.~/3 ( sin x - tan x) dx - #-1 (b) and (c) x/i + x 3 < ½x + on [0, ] ou must use numerical integration because = x/i + x 3 does not have an elementar antiderivative. A= I1½x xfi-~x3; dx ~ = E- cos x + In[ cos x I]~,3 = (1 -ln) ~ " " 1". o,o) _~

9 10 Chapter 7 Applications of Integration 50. A = -x+sec--tan-- dx - = x + x - -- see ~rx rc -~o ~++~(1--~1~.1797 o o (b) A = ~" ( sin x + sin x) dx ~ E-~ oo~x- ~oo~ (c) Numerical approximation: (a) l) = r_l~-.~1 = ~-(1 --le) ~ L Jo (b) A = ~( sin x + cos x) dx ~ [-~ oo~ + ~.in ~]~ - (c) Numerical approximation: (0, it (1,~) 5. From the graph we see thatfand g intersect twice at x = Oandx = 1. (b) A=j1~ [3 1 (c) Numerical approximation: (a) ~ 1 3~ = x + x In o (1, 3) ~ (b) ~ = i~s In x ~ & x /(o, {) (c) Numerical approximation: 5.181

10 Section 7.1 Area of a Region Between Two Curves 57. (a) t 63l. F(x) = t +1 dt = + t = --+ x o (a) F(O) = 0 5a. (a) (b) The integral A--;~ dx does not have an elementar antiderivative. (c) A ~.771 (b) The integral //~1 e) ~ (b) F() =--+=3 6~ (a) does not have an elementar antiderivative. (c) (c) F(6)= --+6= (a) (b) The intersection points are difficult to determine b hand. (c) Area = ;_~ E cos x - x~ dx ~ where c c ~ F(x) = t 1 + dt= t +t =--+x 1 3 x3 o 6 (a) F(O) = (b) The intersection points are difficult to determine. (c) Intersection points: ( , ) and (1.5669,.11018) /~ : ~I 5669 L~/-~-1,16035 q- X- x1dx,~ 3,0578 " (b) F()= () = I l~-x 0t0 Brooks/Cole, Cengage Learning

11 1 Chapter 7 Applications of Integration (c) F(6) = = 8 0- _ _ 8- " 6. F()= (a) F(-1) = " = Be/ _ 8e-l/ (a) F(-1) = 0 T]_l. ~ro~ = -- sin ~ + -- ~ ~r (b) F(0) = 8-8e -1/ ~ (b) F(0) =- ~ , (c) F() = 8e -8e -1/ ~ 5.60 ~o! =7+7=1 =---9 x-1 I (, 6) =-@x_+ 11

12 Section 7.1 Area of a Region Between Two Curves ~,) (, ) Left boundar line: = x +. :> x = - Right boundar line: = x - ~, x = + l~ d []~= = 8 (-8) = 16 (o, o) (a, O) = (1, ) ! :~_ 3 (1, -3) (3, -) 69. Answers will var. Sample answer: If ou let boc = 6 and n = 10, b - a = 10(6) = 60. (a) Area ~ ~E0+ 60 (1) + (1) + (1) + (1) + (15) + (0) + (3) + (5) + (6) + 0~=313]=966ft ~ 60 (1) + (1) (3) + (5) + (6) + 0~ 150] 100 ft (b) Area -~E0+ (1) (1) (15) (0)+ = = 70. Answers will var. Sample answer: Ax =, n = 8, b - a = (8)() = 3 ~ 3 (11) + (13.5)+ (1) + (1.)+ + + (a) Area ~EO + (1.) + 05 ) (13.5) O~ = [190.8] = mi: (b) Area ~, S~S)EO + (11) + (13.5) + (1.)+ (1) + (1.) + (15) + (13.5)+ O~ = -~[96.6] = mi: