Calculus - Chapter 5 Applications of the Definite Integral...

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Chapter 5 Applications of the Definite Integral 153 Chapter 5 Section 5.1 5. 32 ,1 , 1 3 yxyx x == () 3 1 3 43 1 1 81 27 1 1 31 4 3 160 12 40 3 xx d x x  −−   =− +   + + = = 7. , 2 0 x yeyx x 0 2 0 2 2 2 2 (1 ) 2 4 0 0) 2 2 5 x x ex d x x e e + =−+ − 9. 2 1, 1 , 0 2 yx y x x = 12 22 01 23 11 1 ( 1 ) 84 2 ( 000 ) 4 2 3 x x dx x x dx xxd x x x d x + −− − + + ++ −−+ = ∫∫ 11. 3 1 , 2 2 y x x =− = −−≤≤ 33 21 24 4 2 1 ( 1 ) 4 2 1 1 2 (424 )( 442 ) 2 4 2 29 2 d x x x d x x x + + − + + + + =
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Chapter 5 Applications of the Definite Integral 154 13. 22 1, 7 yx y x =− = () 2 2 2 3 2 71 2 8 3 16 16 16 16 33 64 3 xx d x x x ªº −− − ¬¼ §· ¨¸ ©¹ § · =−− + ¨ ¸ © ¹ = ³ 15. 2 3 1 y x =+ 2 2 1 2 23 1 31 1 3 2 12 8 3 1 42 2 32 3 1 6 xxd x x −− + § · ¨ ¸ © ¹ = ³ 17. 3 ,3 2 yxy x == + 2 3 1 2 24 1 3 2 (6 4 4) 2 27 4 d x x +− =+−− −− = ³ 19. , yxyx 1 34 1 0 0 11 (0 0) 1 12 d x −= = ³ 21. 2 ,1 x yey x 0 2 .7145 0 3 .7145 [(1 ) ] 3 ( 1 0 0) ( 1.08235) .08235 x x xe d x x ex =− + − =−+ − = ³ 23. 2 sin , y x .8767 3 .8767 2 0 0 sin cos 3 .135697 x x x = ³ 25. 4 ,2 yxy x + 1.3532 25 1.3532 4 1 1 4.01449 x x = + = ³ 27. ,0 2 y x y xy 00 1 2 0 [(2 ) ] (2 2 ) 2 (2 1) (0 0) 1 yy d y y d y −=− =−−− = ³³ 29. 2 3, 2 x y + 2 2 1 2 3 2 1 3 2 8 64 2 3 1 6 d y y −+ § · ¨ ¸ © ¹ = ³ 31. 1 1 2 0 0 ,, 1 2 101 x y x xdx x = = ³ 33. ,6 y y x y === 2 2 0 (6 ) (6 2 ) 6 (12 4) (0 0) 8 d y y d y = =
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Chapter 5 Applications of the Definite Integral 155 () 3 3 3 2 0 0 3 3 3 2 3 3 33 3 3 ( 0 0 ) 23 3 (9 9) 3 3 3 x xd x x x x x §· −= ¨¸ ©¹ =− = −=− =−− = ³ ³ 37. .07(4) .04(4 4) (4) 16.1 21.3 (4) 21.3 21.3 fe ge == 21.3 million barrels per year represents the consumption for 1974. 10 .07 .04( 4) 4 10 .07 .04( 4) 4 16.1 21.3 230 532.5 14.4 14.4 million barrels saved tt ee d t = ³ 39. For 0, t .04 .02 () 2 2 bt e e dt =≥= 10 .04 .02 0 10 .04 .02 0 22 50 100 2.45 million people per year d t = ³ 41. .4 0 .4 ( ) (0) 4 (.1) 2 (.2) 4 (.3) (.4) 291.67 3(4) cc c c c c fxd x f f f f f =+ + + + ³ .4 0 .4 .4 00 4 0 . ( ) (0) 4 (.1) 2 (.2) 4 (.3) (.4) 102.33 3(4) 291.67 102.33 .6491 291.67 1 .6491 .3508, so the proportion of energy retained is about 35.08%. e e e e ce c x f f f f f x x x 4 ≈+ + + + ≈= ³ ³³ ³ 43. [] 3 0 3 ( ) (0) 4 (.75) 2 (1.5) 4 (2.25) (3) 860 3(4) ss s s s s x f f f f f + + + = ³ 3 0 3 ( ) ( 0 )4( . 7 5 )2( 1 . 5 2 . 2 5 ) ( 3 ) 8 0 0 3(4) rr r r r r x f f f f f + + + = ³ 860 800 1 .9302 860 . Energy returned by the tendon is 93.02%.
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Chapter 5 Applications of the Definite Integral 156 45. 10 10 00 () 40 32; () 30 32 [ ( ) ( )] 10 100 ft t gt t g t f t dt dt =− + −= = ³³ The area between the curves is 100, representing the 100 feet that separate the objects after 10 seconds. This is just the difference in their initial velocities multiplied by 10 seconds. 47. () 401 ; () 20 t e gt t = ( ) ( ) at 0 and 1.593624. t t == Therefore the largest lead is 1.594 0 [ () ] 6 .476m i les ft gt d t −≈ ³ By trial and error we find that 0 [( ) ( ) ] 0w h e n 2 . 5 5 7 , T t T ³ so the second car catches up after 2.557 hours.
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This homework help was uploaded on 04/19/2008 for the course MATH 105 taught by Professor Schroder during the Spring '08 term at Millersville.

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Calculus - Chapter 5 Applications of the Definite Integral...

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