Business Calc Homework w answers_Part_45

Business Calc Homework w answers_Part_45 - Section 5.4 21....

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21. (a) No, f ( x ) 5 tan x is discontinuous at x 5 } p 2 } and x 5 } 3 2 p } . (b) The integral does not have a value. If 0 , b , } p 2 } , then E b 0 tan x dx 5 3 2 ln ) cos x ) 4 52 ln ) cos b ) since the Fundamental Theorem applies for [0, b ]. As b } p 2 } 2 , cos b 0 1 so 2 ln ) cos b ) or E b 0 tan x dx . Hence the integral does not exist over a subinterval of [0, 2 p ], so it doesn’t exist over [0, 2 p ]. 22. (a) No, f ( x ) 5 } x x 2 1 2 1 1 } is discontinuous at x 5 1. (b) The integral does not have a value. If 0 , b , 1, then E b 0 } x x 2 1 2 1 1 } dx 5 E b 0 } x 2 1 1 } dx 5 3 ln ) x 2 1 ) 4 5 ln ) b 2 1 ) , since } x x 2 1 2 1 1 } 5 } x 2 1 1 } and the Fundamental Theorem applies for [0, b ]. As b 1 2 , ln ) b 2 1 ) 2‘ or E b 0 } x x 2 1 2 1 1 } dx 2‘ . Hence the integral does not exist over a subinterval of [0, 2], so it does not exist over [0, 2]. 23. (a) No, f ( x ) 5 } sin x x } is discontinuous at x 5 0. (b) NINT 1 } sin x x } , x , 2 1, 2 2 < 2.55. The integral exists since the area is finite because } sin x x } is bounded. 24. (a) No, f ( x ) 5 } 1 2 x c 2 os x } is discontinuous at x 5 0. (b) NINT 1 } 1 2 x c 2 os x } , x , 2 2, 3 2 < 2.08. The integral exists since the area is finite because } 1 2 x c 2 os x } is bounded. 25. First, find the area under the graph of y 5 x 2 . E 1 0 x 2 dx 5 3 } 1 3 } x 3 4 5 } 1 3 } Next find the area under the graph of y 5 2 2 x . E 2 1 (2 2 x ) dx 5 3 2 x 2 } 1 2 } x 2 4 5 2 2 } 3 2 } 5 } 1 2 } Area of the shaded region 5 } 1 3 } 1 } 1 2 } 5 } 5 6 } 26. First find the area under the graph of y 5 ˇ x w . E 1 0 x 1/2 dx 5 3 } 2 3 } x 3/2 4 5 } 2 3 } Next find the area under the graph of y 5 x 2 . E 2 1 x 2 dx 5 3 } 1 3 } x 3 4 5 } 8 3 } 2 } 1 3 } 5 } 7 3 } Area of the shaded region 5 } 2 3 } 1 } 7 3 } 5 3 27. First find the area under the graph of y 5 1 1 cos x . E p 0 (1 1 cos x ) dx 5 3 x 1 sin x 4 5 p The area of the rectangle is 2 p . Area of the shaded region 5 2 p 2 p 5 p . 28. First, find the area of the region between y 5 sin x and the x -axis for 3 } p 6 } , } 5 6 p } 4 . E 5 p /6 p /6 sin x dx 5 3 2 cos x 4 5 } ˇ 2 3 w } 2 1 2} ˇ 2 3 w } 2 5 ˇ 3 w The area of the rectangle is 1 sin } p 6 } 21 } 2 3 p } 2 5 } p 3 } Area of the shaded region 5 ˇ 3 w 2 } p 3 } 29. NINT 1 } 3 1 2 1 sin x } , x ,0 ,10 2 < 3.802 30. NINT 1 } 2 x x 4 4 2 2 1 1 } , x , 2 0.8, 0.8 2 < 1.427 31. } 1 2 } NINT( ˇ co w s w x w , x , 2 1, 1) < 0.914 32. ˇ 8 w 2 w 2 w x 2 w $ 0 between x 2 and x 5 2 NINT( ˇ 8 w 2 w 2 w x 2 w , x , 2 2, 2) < 8.886 33. Plot y 1 5 NINT( e 2 t 2 , t , x ), y 2 5 0.6 in a [0, 1] by [0,1] window, then use the intersect function to find x < 0.699. 34. When y 5 0, x 5 1. y 3 5 1 2 x 3 y 5 ˇ 3 1 w 2 w x w 3 w NINT( ˇ 3 1 w 2 w x w 3 w , x , 0, 1) < 0.883 5 p /6 p /6 p 0 2 1 1 0 2 1 1 0 b 0 b 0 Section 5.4 221
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35. E x a f ( t ) dt 1 K 5 E x b f ( t ) dt K 52 E x a f ( t ) dt 1 E x b f ( t ) dt 5 E a x f ( t ) dt 1 E x b f ( t ) dt 5 E a b f ( t ) dt K 5 E 2 1 2 ( t 2 2 3 t 1 1) dt 5 3 } 1 3 } t 3 2 } 3 2 } t 2 1 t 4 5 3 2} 1 3 } 2 } 3 2 } 1 ( 2 1) 4 2 3 } 8 3 } 2 6 1 2 4 52} 3 2 } 36. To find an antiderivative of sin 2 x , recall from trigonometry that cos 2 x 5 1 2 2 sin 2 x , so sin 2 x 5 } 1 2 } 2 } 1 2 } cos 2 x .
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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Business Calc Homework w answers_Part_45 - Section 5.4 21....

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