Pre-Calc Homework Solutions 221

Pre-Calc Homework Solutions 221 - Section 5.4 3 . 2 221 21....

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Unformatted text preview: Section 5.4 3 . 2 221 21. (a) No, f (x) tan x is discontinuous at x 2 and x 26. First find the area under the graph of y 1 x. x 1/2 dx (b) The integral does not have a value. If 0 b b 2 , then 0 2 3/2 x 3 1 0 2 3 b Next find the area under the graph of y 2 x 2. tan x dx 0 ln cos x 0 ln cos b since the 2 Fundamental Theorem applies for [0, b]. As b cos b 0 so ln cos b b x 2 dx , 1 1 3 x 3 2 1 8 3 1 3 7 3 2 3 7 3 or 0 tan x dx . Area of the shaded region 3 1 cos x. Hence the integral does not exist over a subinterval of [0, 2 ], so it doesn't exist over [0, 2 ]. 22. (a) No, f (x) x x2 1 is discontinuous at x 1 27. First find the area under the graph of y (1 0 cos x) dx x sin x 0 The area of the rectangle is 2 . 1. Area of the shaded region b b 2 . (b) The integral does not have a value. If 0 b 0 1, then 28. First, find the area of the region between y ln b 1, x-axis for 5 /6 x x2 1 dx 1 1 1 b 0 1 x 1 1 1 sin x and the dx ln x 1 0 x since 2 x x and the Fundamental Theorem 1 or /6 6 , 5 . 6 5 /6 applies for [0, b]. As b 1 , ln b b 0 sin x dx cos x /6 3 2 6 2 3 3 2 3 3 x x2 1 dx 1 . Hence the integral does not exist The area of the rectangle is sin Area of the shaded region 29. NINT 1 , x, 0, 10 2 sin x 1 , x, 1 3 over a subinterval of [0, 2], so it does not exist over [0, 2]. 23. (a) No, f (x) (b) NINT sin x is discontinuous at x x 3 3.802 3 2x 4 x4 0. 30. NINT 31. 0.8, 0.8 1, 1) 1.427 0.914 2 and x 8.886 0.6 in a [0, 1] by [0,1] 2 sin x , x, x 1, 2 2.55. The integral exists since sin x is bounded. x 1 NINT( 2 cos x, x, the area is finite because 24. (a) No, f (x) (b) NINT 1 1 32. 8 2x 2 0 between x 2x 2, x, t2 NINT( 8 cos x is discontinuous at x x2 2, 2) 0. 33. Plot y1 x 3 NINT(e , t, 0, x), y2 window, then use the intersect function to find cos x , x, x2 2, 3 2.08. The integral exists 1 cos x is bounded. x2 0.699. 0, x x 1 3 3 since the area is finite because 34. When y y y 1 3 1. 25. First, find the area under the graph of y 1 0 x 2. x3 x 3, x, 0, 1) 0.883 NINT( 1 x 2 dx 1 3 x 3 1 0 1 3 Next find the area under the graph of y 2 2 1 2 5 6 x. (2 1 x) dx 2x 1 2 x 2 2 2 1 3 2 1 2 Area of the shaded region 1 3 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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