Consider improper integral 1 2 1 x dx x x Which one of the following argument

Consider improper integral 1 2 1 x dx x x which one

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7.Consider improper integral 1201xdxxx. Which one of the following argument is true? 1xxx< 2xin (0, 1) and 1100212dxdxxxconverges, 1201xdxxxconverges. 1xxx< 2xin (0, 1) and 1100212dxdxxxdiverges, 1201xdxxxdiverges. 1xxx< 2xxin (0, 1) and 111/200212xdxdxxxconverges, 1201xdxxxconverges. 1xxx> 22xxin (0, 1) and 1123/2001122xdxdxxxdiverges, 1201xdxxxdiverges. 1xxx> 12xin (0, 1) and 110011122dxdxxxdiverges, 1201xdxxxdiverges. 1xxx> 12xin (0, 1) and 110011122dxdxxxconverges, 1201xdxxxconverges. Answer. (E). 8. Suppose Euler's method with step size h= 0.05 is used to find an approximation of y(0.1), where y(t) is the solution to the initial-value problem y'= (2t 1)(y+ 1), y(0) = 1. Which one of the following is closest to the answer? y(0.1)
MAT1322D Solution to Final Examination December 2017 5 Solution. (C) i ti yi 0 0 1 1 0.05 1 + 0.05 (2 0 −1) (1 + 1)= 0.900 2 0.10 0.900 + 0.05 (2 0.05 1) (0.9 + 1) = 0.815 9. The sum of the series 1202( 1) 53nnnnn is S= (A) 4314; (B) 4514; (C) 299; (D) 257; (E) 139; (F) 237. Solution. (B) 112220002( 1) 52( 1) 5333nnnnnnnnnnnn . The first series is a geometric series with first term a1= 2 and common ratio r1= 29. The second series is also a geometric series with first term a2= 1 and common ratio r2= 59. Hence, S= 21189452571414119910. The interval of convergence of the series 023nnnnxis (A) 1.5 x1.5; (B) 1.5 < x< 1.5; (C) 1.5 < x1.5; (D) 1.5 x< 1.5; (E) −∞< x< ∞; (F) only at x= 0. Solution. (B) 1112322limlim3233nnnnnnnnxxxx. When | x| < 32, this series is absolutely convergent. When | x| > 32, this series is divergent. When x= −32, this series becomes 0023( 1)32nnnnnn; when x= 32, this series becomes0023132nnnnn. In both cases, the series diverges. Hence, the interval of convergence is 33,22= (−1.5, 1.5).

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