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 + 4 4. 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 + 9 6 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 + 5 0 (x ) 4 evised by Prof. Kostadinov (spring 04), Prof. Africk (fall 06, spring 07), Prof. ElHitti (spring 07, summer 07)
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 5 + 5 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
(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
(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 + 4 + ln x + C x + (6b) ln x 6 + ln x + 6 + 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
(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
(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