New  York  City  College  of  Technology,  CUNY                          Mathematics  Department     Fall  2012  

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MAT 1575 Final Exam Review Problems

Revised by Prof. Kostadinov, Fall 2012, Fall 2011, Spring 2011, Spring 2010

1. Evaluate the following indefinite integrals:

a. x

x2 + 9dx∫                                              b.        

3x2

x3 − 53∫ dx                        

     

2. Evaluate the following indefinite integrals:

               a.     x2 ln(x)dx∫                    b.     x2e− x dx∫                                            c.     xcos(3x)dx∫    

3. Evaluate the following indefinite integrals:

a. 1

x2 36 − x2dx∫ b.

x2 − 9x4∫ dx c.

9x 2 x 2 + 9

dx∫                  

d. 6

x 2 x 2 − 36dx∫  

4. Evaluate the following indefinite integrals:

a.  3− 5x 2

x 3 + 6xdx∫                                                  b.  

5x + 6x 2 − 36

dx∫                                    c.  3x + 2

x2 + 2x −8dx∫                  

5. Find the area of the region enclosed by the graphs of:

a. y = 3− x2 and y = −2x b. y = x2 − 2x and y = x + 4

6. Find the volume of the solid, by using disks, obtained by revolving about the x-axis. The region is bounded by the graphs of: a. y = x2 − 9 and y = 0 . b. y = −x2 + 9 and y = 0 .

7. Determine whether the integral is convergent or divergent. Evaluate the integral if convergent:

a. 2

(x + 2)3dx

0

∞

∫ b. 5x + 55

0

∞

∫ dx c.

3

(x − 3)43dx

0

5

∫

8. Decide if the following series converges or not. Justify your answer using an appropriate test:

a. 9n2

3n5 + 5n=1

∞

∑ b. 510nn=1

∞

∑ c. 5n10nn=1

∞

∑ d. 5n 0.1( )nn=1

∞

∑

9. Determine whether the series is absolutely or conditionally convergent or divergent:

a.       (−1)n 107n + 2n=1

∞

∑                  b.  

(−1)n

n5n=1

∞

∑

c.         (-1)n5−nn=0

∞

∑                        d.    

(−1)n n2 − n −12n2 + n +1n=1

∞

∑  

10. Find the radius and the interval of convergence of the following power series:

a.      

xn

n+ 2n=0

∞

∑                                        b.      

(n+ 2)(x + 3)n

n=0

∞

∑                      c.      

xn

n5nn=1

∞

∑                                          d.      (−1)n+1xn

5nn=0

∞

∑            

11. Find the Taylor polynomial of degree 2 for the given function, centered at the given number a:

a.       f (x) = e−2x  at   a = −1 .                  b.     f (x) = cos(5x)  at   a = 2π .    

12. Find the Taylor polynomial of degree 3 for the given function, centered at the given number a. Graph the

function and the Taylor polynomial on the same screen:

a. f (x) = 1+ e− x at a = −1. b. f (x) = sin(x) at a =π2.

Answers to questions:

(1a). x2 + 9 +C                                (1b). 32(x3 − 5)23 +C

                                             

(2a). x3 ln(x)3

− x3

9+C              (2b). −(x2 + 2x + 2)e− x +C

           (2c).

13xsin(3x)+ 1

9cos(3x)+C      

(3a).  − 36 − x2

36x+C                    (3b).

(x2 − 9)3/2

27x3+C

             (3c).  − x2 + 9

x+C              (3d).  

x2 − 366x

+C  

(4a).  ln x2

− 114ln(x2 + 6)+C          (4b).    3ln x − 6 + 2 ln x + 6 +C          (4c).  

53ln x + 4 + 4

3ln x − 2 +C    

(5a). The area of the region between the two curves is:

 

Area = 3− x2 − −2x( )( )dx = 323−1

3

∫  

(5b). The area of the region between the two curves is:

Area = x + 4 − x2 − 2x( )( )dx = 1256−1

4

∫  

New  York  City  College  of  Technology,  CUNY                          Mathematics  Department     Fall  2012  

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(6a). Approximate the volume of the solid by vertical disks with radius y = x2 − 9  between x = −3 and x = 3  ; in the

limit, the volume is V = π (x2 − 9)2 dx = 12965

π−3

3

∫  

(6b). The solid of revolution is the same by symmetry, therefore the volume must be the same as well:

V = π (−x2 + 9)2 dx = 12965

π−3

3

∫

(7a). 14

(7b). The integral does not converge (7c). The integral does not converge

(8a). The sum converges by the Comparison (or Limit Comparison) Test and the p-Test.

(8b). This is a geometric series, which converges to 5/9:

510nn=1

∞

∑ = 510

+ 5102

+ 5103

+= a1− r

= 5 /10

1− 110

= 5 /109 /10

= 59

(8c). The series converges by the Ratio Test. (8d). This is the same series as in (8c).

(9a). Conditionally convergent: convergent by the Alternating Series Test but not absolutely convergent since 107n + 2n=1

∞

∑    diverges (like the Harmonic series, by the Limit Comparison Test).

(9b). Absolutely convergent: convergent by the Alternating Series Test and absolutely convergent since 1

n5= 1

n5/2n=1

∞

∑n=1

∞

∑

converges by the p-Test with p = 2.5.

(9c). Absolutely convergent: convergent by the Alternating Series Test and absolutely convergent since

5−n = 11−1/ 5

= 54n=0

∞

∑ converges as a 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.

(10a). The power series converges when x <1 by the Ratio Test, which gives a radius of convergence 1 and

interval of convergence centered at 0 with radius 1: −1< x <1 .

(10b). The power series converges when x + 3 <1 by the Ratio Test, which gives a radius of convergence 1 and

interval of convergence centered at -3 with radius 1: −4 < x < −2 . Bonus: show that the given power series

represents the Taylor’s expansion of y = −x −1(x + 2)2

about x = −3 .

(10c). The power series converges when x < 5 by the Ratio Test, which gives a radius of convergence 5 and

interval of convergence centered at 0 with radius 5: −5 < x < 5 . Bonus: show that the given power series

represents the Taylor’s expansion of y = − log 5 − x5

⎛⎝⎜

⎞⎠⎟ about x = 0 .

(10d). The radius of convergence is 5 and the interval of convergence is x < 5 or −5 < x < 5 . This follows from

either using the Ratio Test or from recognizing that the power series is a geometric series − −x5

⎛⎝⎜

⎞⎠⎟

n

n=0

∞

∑ , which

converges when −x5

<1 thus giving the interval of convergence x < 5 . The series does not converge at the end

points x = ±5 (why?). Inside the interval of convergence, one can sum up the geometric series to get the function:

− −x

5⎛⎝⎜

⎞⎠⎟

n

n=0

∞

∑ = − 11− −x / 5( ) = − 1

1+ x / 5= − 5

x +5

(11a). The Taylor polynomial of degree 2, about a = −1 is given by:

T2(x) = f (−1)+ f '(−1)(x +1)+ f (2) (−1)

2(x +1)2 = e2 − 2e2(x +1)+ 2e2(x +1)2 = e2 + 2e2x + 2e2x2

(11b). The Taylor polynomial of degree 2, about a = 2π is given by:

T2(x) = f (2π )+ f '(2π )(x − 2π )+ f (2) (2π )

2(x − 2π )2 = 1− 25

2(x − 2π )2

(12a). Graphs of the function f (x) and its Taylor

polynomial of degree 3, T3(x) :

(12b). T3(x) = 1−12

x − π2

⎛⎝⎜

⎞⎠⎟2

T3(x) = f (−1)+ f '(−1)(x +1)+ f (2) (−1)

2(x +1)2 + f (3) (−1)

6(x +1)3 = 1+ e− e(x +1)+ e

2(x +1)2 − e

6(x +1)3 = 1+ e

3− e

2x − e

6x3

f(x) = 1 + e

�x

f(x) = sin(x)