This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 212 Section 5.2
44. (a) The function has discontinuities at x 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5. (b) 37. Observe from the graph below that the region between the graph of f (x) x 3 1 and the xaxis for 0 x 1 cuts out a region R from the square identical to the region under the graph of g(x) x 3 for 0 x 1.
y 1 x [ 6, 5] by [ 18, 4]
5 2 int(x
6 3) dx ( 10) 88 ( 18) ( 8) ( 16) ( 6) ( 14) ( 4) ( 2) y = f(x) R ( 12) 0
1 4 3 4 1
1 0 2 45. (a) The function has a discontinuity at x (b) 1. (x 3 1) dx 1 38. Observe from the graph below that the region between the graph of f (x)
3 x and the xaxis for 0 x 1 cuts out a
[ 3, 4] by [ 4, 3]
4 region R from the square identical to the region under the graph of g(x)
y 1 R y = f(x) x 3 for 0 x 1. 3 x2 x 1 dx 1 1 (4)(4) 2 1 (3)(3) 2 7 2 46. (a) The function has a discontinuity at x (b) 3. [ 5, 6] by [ 9, 2]
6 1
1 x 9 5x x2 dx 3 1 (2)(2) 2 1 (9)(9) 2 77 2 47. (a) As x approaches 0 from the right, f (x) goes to . (b) Using right endpoints we have x dx
0 1
x 4 1 4 3 4 39. NINT 40. 3 x2 , x, 0, 5 0.9905
3 1 2 NINT tan x, x, 0, x 2, x, 2, 2) 1, 3) 4.3863 0 1 dx x2 lim
n k 1 n n 1
k 2 n 1 n 41. NINT(4 10.6667 1.8719 0. lim
n n 42. NINT(x 2e x, x, (b) k 1 n k2 1 22 43. (a) The function has a discontinuity at x lim n 1 Note that n 1 n1
1 22 1 ,x n k ... 1 . n2 1 1 ... 22 n2 1 ... . n2 k n 1 n 2 2 n 1 n n and n , so [ 2, 3] by [ 2, 2]
3 2 48. (a) 1 x x x dx 2 3 RRAM k 1 n 1 2 n k 2 1 n n 1 n 1 n3 ... n 2 n 1 n (b) (c) k 1 n k 2 n k 1 n k2 n3 1 n3 k 1)(2n 6 k2
1 n 1 n3 k k2
1 n n(n 1) n(n 1)(2n 6n 3 1) ...
View
Full
Document
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.
 Spring '08
 GERMAN

Click to edit the document details