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MA114 Exam 2 Solution Fall 2007

This is a solution of the second exam in MA114 Sections 001 to 006. The statements of the problems are not verbatim. Also, problem 8 parts c and d did not actually appear on the exam (Instead the points for them were added to parts a and b of the same problem.)

1. (10 points) Respond as indicated to each of the following. You do not need to show work for this problem.
1. Among the choices below, circle the substitution that would be the most appropriate to compute int sqrt(1-x^2) dx.
1. u=sin(x)
2. x=tan(u)
3. x=sin(u)
4. u = e^x
5. x = u^4
The correct answer is (iii) x=sin(u). This substitution allows sqrt(1-x^2) to be replaced with cos(u) if one chooses the domain to be u in [-pi/2, pi/2].
2. Circle the correct choice. If T(x) = 3 - 7(x - a) + t(x - a)^2 is the degree 3 Taylor polynomial forthe function f(x) at x = a, then the equation of the tangent line to the graph of f(x) at x=a is:
1. y = -7 + 5(x-a)
2. y=3-7(x-a)
3. y=3 - 7x
The correct answer is (ii) y = 3 - 7(x-a) because the right hand side is equal to f(a) + f'(a)(x-a).
3. Answer "true" or "false". If p(x) is any polynomial, then the partial fraction decomposition of f(x) = (p(x))/(x^3+x) can be written in the form f(x) = q(x) + (Ax+B)/(x^2+1) + C/x where A, B, and C are constants and q(x) is a polynomial.

This is true.

4. Answer "true" or "false". If {a_n} is a sequence and lim_(n->oo) a_n = L, then for every epsilon > 0, there is an N such that if |L-a_n| < epsilon, then n > N.

This is false.

5. Answer "true" or "false". The partial fraction decomposition of f(x) = (x^2)/((x-a)^2(x-2)) is of the form f(x) = A/(x-1) + (Bx)/(x-1) + C/(x-2) where A, B, C are constants.

This is false.

2. (10 points) Use partial fractions to evaluate int 1/((x-1)(x-2)) dx. SHOW YOUR WORK.

The partial fraction decomposition is of the form

1/((x-1)(x-2)) = A/(x-1) + B/(x-2)
After clearing denominators, one has 1= A(x-2) + B(x-1). Substituting x=2 gives B=1 and substituting x=1 gives A = -1. So, one has
int (dx)/((x-1)(x-2)) = int -1/(x-1) + 1/(x-2)\ dx = -ln(|x-1|) + ln(|x-2|) + C = ln(|(x-2)/(x-1)|) + C

3. (10 points) Calculate the partial fraction expansion of f(x) = (x^3+x+1)/(x(x^2+1)). DO NOT integrate the result. SHOW YOUR WORK.

Because the numerator has degree greater than that of the denominator, one starts by doing the division to get a quotient of 1 with a remainder of 1. The decomposition would be of the form:

f(x) = 1 + 1/(x(x^2+1)) = 1 + A/x + (Bx+C)/(x^2+1)
where A, B, and C are constants. Clearing the denominators in the remainder part gives:
1 = A(x^2+ 1) + (Bx + C)x
Substituting x= 0 gives A = 1. Substituting x = 1 and x = -1 gives 1 = 2 + B + C and 1 = 2 + B - C. Adding the two equations gives 2 = 4 + 2B or B = -1. Subtracting the two equations gives C = 0. So the partial fraction decomposition is
f(x) = 1 + 1/x - x/(x^2 + 1)

4. Evaluate each of the following integrals. SHOW YOUR WORK.
1. (5 points) int sin^3(x+2) cos(x + 2) dx

Substitute u = sin(x+2). One has du = cos(x+2) dx and so

int sin^3(x+2) cos(x+2) dx = int u^3 du = u^4/4 = (sin^4(x+2))/4 + C

2. (5 points) int sin(3x)cos(5x) dx

The addition formulas for sine give:

sin(3x + 5x) = sin(3x)cos(5x) + cos(3x)sin(5x)
and
sin(3x - 5x) = sin(3x)cos(5x) - cos(3x)sin(5x)
sin(8x) + sin(-2x) = 2sin(3x)cos(5x)
So,
int sin(3x)cos(5x)dx = int 1/2(sin(8x)-sin(2x))dx = -1/16 cos(8x) + 1/4 cos(2x) + C

5. Evaluate each of the following. SHOW YOUR WORK!
1. (5 points) int sin^2(x) dx

Subtract cos(2x) = cos^2(x) - sin^2(x) from 1 = cos^2(x) + sin^2(x) to get

1 - cos(2x) = 2 sin^2(x)
Using this, one can integrate:
int sin^2(x) dx = int (1 - cos(2x))/2 dx = x/2 - (sin(2x))/4 + C

2. (5 points) int_0^(1/3) 1/sqrt(9x^2+1) dx

Substitute 3x = tan(theta) for theta in (-pi/2, pi/2). One has 3dx = sec^2(theta) d theta and the limits x = 0, 1/3 correspond to theta = 0, pi/4 respectively. The substitution yields:

int_0^(1/3) 1/sqrt(9x^2+1) dx = int_0^(pi/4) (sec^2(theta) d theta)/(3 sqrt(tan^2(theta) + 1)) = int_0^(pi/4) (sec^2(theta) d theta)/(3 |sec(theta)|)
Since theta in [-pi/2,pi/2], one has |sec(theta)| = sec(theta) and so the integral simplifies to
1/3 int_0^(pi/4) sec(theta) d theta = 1/3 int_0^(pi/4) (sec^2(theta) + sec(theta)tan(theta))/(sec(theta) + tan(theta)) d theta = 1/3{:ln(|sec(theta) + tan(theta)|)|_0^(pi/4) = 1/3 ln(sqrt(2) + 1) - ln(1) = 1/3 ln(sqrt(2)+1)
So, the value of the integral is 1/3 ln(sqrt(2)+1) approx 0.2937911956.

6. SHOW YOUR WORK on all parts of this problem.
1. (4 points) Calculate the Taylor polynomial of degree 3 at x=0 for f(x) = ln(1-x).

Start by calculating the derivatives and their values at x=0:

FunctionValue at 0
f(x) = ln(1-x) 0
f'(x) = -1/(1-x) -1
f''(x) = -1/((1-x)^2) -1
f'''(x) = -2/((1-x)^3) -2
The Taylor polynomial of order 3 at x=0 is
T_3(x) = f(0) + f'(0)x + f''(0)(x^2)/2 + f'''(0) (x^3)/6 = 0 - x - (x^2)/2 - (x^3)/3

2. (4 points) Calculate the Taylor polynomial of degree 2 for f(x) = 1 + 2x + 3x^2 + 5x^3 at x = 1.

The table of derivatives and their values at x = 1 is:

FunctionValue at 1
f(x) = 1 + 2x + 3x^2 + 5x^3 11
f'(x) = 2 + 6x + 15 x^2 23
f''(x) = 6 + 30 x 36
The Taylor polynomial of degree 2 at x=1 is:
T_2(x) = f(1) + f'(1)(x-1) + f''(0)((x-1)^2)/2 = 11 + 23(x-1) + 18(x-1)^2

3. (2 points) Calculate the remainder R_2(x) at x=1 when f(x) = 1 + 2x + 3x^2 + 5x^3. You can either give the exact value or you can provide an upper bound for its absolute value.

We already have T_2(x) and we can extend it to T_3(x): One has f'''(x) = 30 and so

f(x) = T_3(x) = T_2(x) + f'''(1)((x-1)^3)/6 = T_2(x) + 5(x-1)^3
Now, we have
f(x) = T_2(x) + R_2(x) = T_3(x) = T_2(x) + 5(x-1)^3
and so R_2(x) = 5(x-1)^3.

Alternatively, one can use |f'''(x)| = 30 = M to get

|R_2(x)| <= (M |x-1|^(3))/(3!) = 5|x-1|^3

7. SHOW YOUR WORK on all parts of this problem.
1. (5 points)
1. Calculate the trapezoid rule estimate for int_(-1)^3 x^2 + x dx with two subintervals.

Let f(x) = x^2 + x. One has (b-a)/n = 2 and so one needs to calculate: f(-1) = 0, f(1) = 2, and f(3) = 12. The trapezoid rule gives the approximation:

(f(-1) + 2 f(1) + f(3)) (b-a)/(2n) = (0 + 4 + 12)4/4 = 16

2. Use the appropriate error formula to calculate the maximum error in this estimate.

One has f''(x) = 2 and so one can let K=2 in the estimate:

K/12 ((b-a)^3)/(n^2) = 2/12 (4^3)/4 = 8/3
Note that the exact value of the integral is 40/3 and so this is an example where the error is the largest possible.

2. (5 points)
1. Calculate Simpson's rule estimate for int_(-1)^3 x^3 + x\ dx for 4 subintervals.

Let f(x) = x^3 + x. One has Delta = (b-a)/n = 1 and so, one needs to calculate: f(-1) = -2, f(0) = 0, f(1) = 2, f(2) = 10, and f(3) = 30. The Simpson's rule estimate is:

(Delta)/3 (f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)) = 1/3(-2+0+4+40+30) = 72/3 = 24

2. Use the appropriate error formula to calculate the maximum error in this estimate.

One has f^(4)(x) = 0 since f(x) is only of degree 3. So, the error upper bound of K/180 ((b-a)^5)/n^4 is zero because we can take K = 0.

8. The table below gives the values of a function f(x) and its first four derivatives at selected values of x. Use this information to answer the four parts of the question that follow. Provide a brief explanation and/or calculations that make it clear how you arrived at your answer.
x-2-1012
f(x)7/63/25/61/6-15/2
f'(x)-5/60-5/6-4/3-39/2
f^(2)(x)22/3-22/3-14/3-38
f^(3)(x)-2202-16-54
f^(4)(x)3212-8-28-48
1. (3 points) Calculate the Taylor polynomial of degree 3 for f(x) at x=0. Do not simplify your answer.

One has

T_3(x) = f(0) + f'(0)x + f''(0)(x^2)/2 + f'''(0)(x^3)/6 = 5/6 -5/6 x + 1/3 x^2 + 1/3 x^3

2. (3 points) Calculate the Taylor polynomial of degree 3 for f(x) at x=1. Do not simplify your answer.

One has

T_3(x) = f(1) + f'(1)(x-1) + f''(1)((x-1)^2)/2 + f'''(1)((x-1)^3)/6 = 1/6 -4/3 (x-1) - 7/3 (x-1)^2 - 8/3 (x-1)^3

3. (2 points) Let h(x) = 3(x-1) f(x). Calculate the Taylor polynomial of degree 3 for h(x) at x=1. Do not simplify your answer.

The Taylor polynomial is 3(x-1) times the answer to the last part:

T_3(x) = 3(x-1)T_2(x) = 3(x-1)(1/6 -4/3 (x-1) - 7/3 (x-1)^2) = (x-1)/2 - 4(x-1)^2 -7(x-1)^3

4. (2 points) Calculate the Taylor polynomial of degree 3 for f'(x) at x=0. Do not simplify your answer.

One has:

T_3(x) = f'(0) + f''(0)x + f'''(0)(x^2)/2 + f^(4)(0)(x^3)/6 = -5/6 + 2/3 x + x^2 - 4/3 x^3

9. (10 points) On the line after each of the following, enter one of the following: (a) the value to which the sequence converges, (b) oo, (c) -oo, or (in case none of the others apply) (d) "does not exist" or simply "divergent". In each case provide a brief expllanation and/or supporting calculations for your answer.
1. lim_(n->oo) ((n-3)n)/(2n^2+1)

The value is 1/2 because

lim_(n->oo)((n-3)n)/(2n^2+1) = lim_(n->oo) (1 - 3/n)/(2+1/(n^2)) = 1/2

2. lim_(n->oo) 2^(cos(n pi))

Since cos(n pi) alternates between -1 and 1, the terms alternate between 1/2 and 2. So the limit does not exist.

3. lim_(n->oo) 3^(sin((n pi)/(2n+5)))

One has lim_(n->oo) (n pi)/(2n+5) = lim_(n->oo) pi/(2+5/n) = pi/2 and sin(pi/2) = 1. So, the limit is 3.

4. lim_(n->oo) (2 sin(n))/sqrt(n+1)

The numerator is at most 2 in absolute value and the denominator approaches infinity as n->oo. So, the limit is 0.

5. lim_(n->oo) (-1)^n e^((n+1)/n)

One has lim_(n->oo) (n+1)/n = lim_(n->oo) (1+1/n) = 1` and so the subsequence of even index terms approach e whereas the subsequence of odd index terms approach -e. These not being equal, the limit does not exist.

 modified: Thursday, October 18, 2007

Modified 1/18/2008 12:55