New York City College of Technology, CUNY Mathematics Department. MAT 1575 Final Exam Review Problems. 3x (a) x 2 (x 3 +1) 3 dx (b) dx.

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1 New York City College of Technology, CUNY Mathematics Department MAT 575 Final Exam eview Problems. Evaluate the following definite integrals: x x (a) x (x +) dx (b) dx 0 0 x + 9 (c) 0 x + dx. Evaluate the following indefinite integrals: (a) x ln x dx (b) x e x dx (c) x cos(x) dx. Find the area of the region enclosed by the graphs of: (a) y = x and y = x (b) y = x x and y = x Find the volume of the solid obtained by rotating the region bounded by the graphs of: (a) y = x 9, y = 0 about the x-axis. (b) y = 6 x, y = x +, x = about the x-axis. (c) y = x +, y = x + 0, x 0 about the y-axis. 5. Evaluate the following indefinite integrals: 9 (a) dx (c) dx x 6 x x x x 9 (b) dx (d) dx x4 x x 6 6. Evaluate the following indefinite integrals: x + 7 5x + 6 x + (a) dx (b) dx (c) dx x + 6x + 9 x 6 x + x 8 7. Evaluate the following improper integrals: 5 5 (a) dx (b) dx (c) p dx 0 (x + ) 0 5 x (x ) 4 evised by Prof. Kostadinov (spring 04), Prof. Africk (fall 06, spring 07), Prof. ElHitti (spring 07, summer 07)

2 8. Decide if the following series converges or not. Justify your answer using an appropriate test: (a) (b) (c) P 9n n! (d) n= n n= n 5 n 5 n n + (e) n= 0 n n= n + P 5n 5 P n= 0 n 9. Decide whether the series is absolutely or conditionally convergent or divergent: 0 (a) ( ) n (c) ( ) n 5 n n= 7n + n=0 ( ) n n n (b) (d) ( ) n n= n 5 n= n + n + 0. Find the radius and interval of convergence of the following power series: (a) (b) (x ) n (x + ) n (c) n=0 n + n= n5 n ( ) n (x ) n ( ) n (x + ) n (d) n + n5 n n=0 n=. Find the Taylor polynomial of degree for the given function, centered at the given number a: (a) f(x) = e x at a = (b) f(x) = cos(5x) at a = π. Find the Taylor polynomial of degree for the given function, centered at the given number a: (a) f(x) = + e x at a = (b) f(x) = sin(x) at a = π

3 (a) 5 4 (a) x ln(x) (b) (x + x + )e x + C Answers (b) 0 (c) ( ) x 9 + C (c) x sin(x) + 9 cos(x) + C (a) The area of the region between the two curves is: Area = ( x ( x)) dx = (b) The area of the region between the two curves is : 4 5 Area = (x + 4 (x x)) dx = 6

4 (4a) Approximate the volume of the solid by vertical disks with radius y = x 9 between x = and x = ; gives the volume: 96 V = π(x 9) dx = π 5 (4b) Using a washer of outer radius outer = 6 x and inner radius inner = x + at x, gives the volume: V = π ((6 x) (x + ) ) dx = 656π, where the upper limit is obtained from solving 6 x = x+ (4c) 6π 6 x x 9 (5a) + C (5c) + C 6x x (x 9) x 6 (5b) + C (5d) + C 7x 6x 5 4 (6a) + ln x + + C (6c) ln x ln x + C x + (6b) ln x 6 + ln x C (7b) The integral does not converge. (7a) 4 (7c) The integral does not converge. (8a) Diverges by the nth Term Test for Divergence. 4

5 (8b) This is a geometric series, which converges to 5/9: 5 a 5/0 5/0 5 = = = = n= 0 n r 9/0 9 0 (8c) Converges absolutely by the atio Test. (8d) Diverges by the atio Test. (8e) Converges absolutely by the nth oot Test. (9a) Conditionally convergent: convergent by the Alternating Series Test P 0 but not absolutely convergent since diverges (like the harn= 7n + monic series, by the Limit Comparison Test). (9b) Absolutely convergent: convergent by the Alternating Series Test and absolutely convergent since = converges by the p Test n= n5 n= n 5/ with p =.5. (9c) Absolutely convergent: convergent by the Alternating Series Test and P 5 absolutely convergent since 5 n = = converges as a n=0 /5 4 geometric series. (9d) Divergent: by the Test for Divergence. The limit of the general term does not exist. Note that the Alternating Series Test does not apply. (0a) The power series converges when x < by the atio Test, which gives a radius of convergence and an interval of convergence centered at. The series diverges at x = (harmonic series) but converges at x = 0 (alternating harmonic series), so the interval of convergence is 0 x <. (0b) The power series converges when x < by the atio Test, which gives a radius of convergence and an interval of convergence centered at. The series diverges at x = 0 (harmonic series) but converges at x = (alternating harmonic series), so the interval of convergence is 0 < x. 5

6 (0c) The power series converges when x + < 5 by the atio Test, which gives a radius of convergence 5 and an interval of convergence centered at. The series diverges at x = 4 (harmonic series) but converges at x = 6 (alternating harmonic series), so the interval of convergence is 6 x < 4. (0d) The power series converges when x + < 5 by the atio Test, which gives a radius of convergence 5 and an interval of convergence centered at. The series diverges at x = 6 (harmonic series) but converges at x = 4 (alternating harmonic series), so the interval of convergence is 6 < x 4. (a) p (x) = e + e x + e x (b) p (x) = 5 (x π) e e e (a) p (x) = + x x 6 π (b) p (x) = x 6