REVIEW, FINAL (Key): MATH 1B ADDITIONAL PROBLEMS

Warning: I have not seen a copy of the final examination. Concepts and problems reviewed here should not be taken as an exclusive list! It should give you a good sense, though, of what will be asked.

Problem 15. The integral

∫_{a}^{b} f(x) dx

was approximated using the Trapezoidal rule and n = 10. Using the error bound it was found that |E_{T} | ≤ 1. Which of the following is the smallest value of n in the list for which |E_{T} | ≤ 10^{−6}?

(a) 999

(b) 10074

(c) 1053

(d) 60

(e) The answer cannot be determined from the information given.

Solution. We know that

|E_{T} | ≤ \frac{K(b − a)^{3}}{12n^{2}}

where K = max_{x∈[a,b]} |f′′(x)|. Therefore the error bound for n = 10 gives

\frac{K(b − a)^{3}}{1200} = 1

so K(b − a)^{3} = 1200. Therefore

|E_{T} | ≤ \frac{K(b − a)^{3}}{12n^{2}} = \frac{1200}{12n^{2}} = \frac{100}{n^{2}} ≤ 10^{−6}

so n ≥ 10^{4} = 10000. The answer is (b).

Problem 16. Find the area of the region bounded by the curve

y = sin^{−1} x

and y = 0, x = 1/2.

Solution. Drawing a picture we see that we are to compute the integral

∫_{0}^{1/2} sin^{−1} x dx,

since y = 0 = sin^{−1} x gives x = 0.

To do this integral, we (sneakily) use integration by parts, with u = sin^{−1} x, dv = dx, so that du = 1/\sqrt{1 − x^{2}} dx and v = x, and then we have

∫ sin^{−1} x dx = x sin^{−1} x − ∫ \frac{x}{\sqrt{1 − x^{2}}} dx.

For the second integral, we substitute u = 1 − x^{2}, so then du = −2x dx, and hence we have

∫ \frac{x}{\sqrt{1 − x^{2}}} dx = \frac{−1}{2} ∫ \frac{du}{\sqrt{u}} = −\sqrt{u} + C = −\sqrt{1 − x^{2}} + C

and in sum

∫ sin^{−1} x dx = x sin^{−1} x + \sqrt{1 − x^{2}} + C.

Therefore

∫_{0}^{1/2} sin^{−1} x dx = \frac{1}{2} sin^{−1}(1/2) + \sqrt{\frac{3}{4}} − 1 = \frac{π}{12} − 1 + \frac{1}{2}\sqrt{3}.

Problem 17. Let z_{1}, z_{2} be the solutions to the equation z^{2} − z + 7 = 0. What is the value of z_{1} + z_{2} + 7/(z_{1}z_{2})?

(a) 0

(b) 2

(c) 1 + i

(d) −1

(e) None of the above.

Solution. We have z = (1 ± \sqrt{1 − 28})/2 = 1/2 ± (3/2)\sqrt{3}i. Therefore

z_{1} + z_{2} = (1/2 + (3/2)\sqrt{3}i) + (1/2 − (3/2)\sqrt{3}i) = 1

and

z_{1}z_{2} = (1/2 + 3/2\sqrt{3}i)(1/2 − 3/2\sqrt{3}i) = 1/4 + 9/4(3) = 7,

hence

z_{1} + z_{2} + 7/(z_{1}z_{2}) = 1 + 1 = 2.

The answer is (b).

(A previous version had z^{2} − z − 7 = 0, which had two real roots, something you should already know how to compute with.)

Problem 18. Solve the differential equation

y′′ − 2y′ + y = x.

Solution. The homogeneous problem y′′−2y′+y = 0 has characteristic equation r^{2} − 2r + 1 = (r − 1)^{2} = 0, so r = 1 is a double root, hence y_{h}(x) = c_{1}e^{x} + c_{2}xe^{x}.

We use the method of undetermined coefficients to find the particular solution, and guess y_{p}(x) = Ax + B, so y′_{p}(x) = A and y′′_{p}(x) = 0, so

y′′_{p} − 2y′_{p} + y_{p} = 0 − 2A + Ax + B = Ax + (−2A + B) = x

So A = 1 and −2A + B = −2 + B = 0, so B = 2. Therefore

y(x) = y_{h}(x) + y_{p}(x) = c_{1}e^{x} + c_{2}xe^{x} + x + 2.

Problem 19. Determine if the series

∑_{n=3}^{∞} \frac{1}{n(ln n)^{3}}

is convergent or divergent.

Solution. We use the integral test. The function f(x) = 1/x(ln x)^{3} is continuous, positive, and decreasing since x(ln x)^{3} is obviously increasing (or just differentiate). Hence we look at

∫_{3}^{∞} \frac{1}{x(ln x)^{3}} dx.

Substitute u = ln x to get du = 1/x dx, then the integral is just

∫ \frac{1}{x(ln x)^{3}} dx = ∫ \frac{du}{u^{3}} = \frac{−1}{2u^{2}} = \frac{−1}{2(ln x)^{2}}

hence

∫_{3}^{∞} \frac{1}{x(ln x)^{3}} dx = 0 − \frac{−1}{2(ln 3)^{2}} < ∞

so the integral and hence the sum are convergent.

Problem 20. Use Euler’s method with step size 1/2 to estimate y(1) where y(x) is the solution to the intial-value problem y′ = x + 2y^{2}, y(0) = 0.

Solution. We start with x_{0} = 0, y_{0} = 0, with h = 1/2, and

y_{n} = y_{n−1} + hF (x_{n−1}, y_{n−1}) = y_{n−1} + \frac{1}{2}(x_{n−1} + 2y_{n−1}^{2}).

Hence

y_{1} = 0 + \frac{1}{2}(0 + 0) = 0

and

y(1) ≈ y_{2} = 0 + 1/2(1/2 + 0)^{2} = 1/8.

Problem 21. What is the value of

∑_{n=1}^{∞} \frac{(−1)^{n}π^{2n}}{2^{2n+1}(2n + 1)!}?

(a) 0

(b) −1

(c) 1/π − π/2

(d) 1/π + π/2

(e) (2 − π)/2π.

Solution. We recognize the series as almost

sin(x) = ∑_{n=0}^{∞} \frac{(−1)^{n}x^{2n+1}}{(2n + 1)!} .

To do this, we first need to add on the n = 0 term, which is (−1)^{0}π^{0}/2(1)! = 1/2, hence

∑_{n=1}^{∞} \frac{(−1)^{n}π^{2n}}{2^{2n+1}(2n + 1)!} = ∑_{n=0}^{∞} \frac{(−1)^{n}π^{2n}}{2^{2n+1}(2n + 1)!} − \frac{1}{2}.

Now the second series is just by multiplying in a π

\frac{1}{π} ∑_{n=0}^{∞} \frac{(−1)^{n}π^{2n+1}}{2^{2n+1}(2n + 1)!} = \frac{1}{π} sin(π/2) = \frac{1}{π}

so the sum is 1/π − 1/2 = (2 − π)/2π. The answer is (e).

Problem 22. Evaluate the limit

lim_{x→0} \frac{(sin 2x − 2x)^{2}}{x^{2}(e^{x} − 1)^{3}} .

Solution. We have sin 2x = 2x − (8/3)x^{3} + … and e^{x} = 1 + x + …, so we get

\frac{(sin 2x − 2x)^{2}}{x^{2}(e^{x} − 1)^{3}} = \frac{((−8/3)x^{3} + …)^{2}}{x^{2}(x + …)^{3}} = \frac{(64/9)x^{6} + …}{x^{5} + …} = \frac{(16/9)x + …}{1 + …} → 0

as x → 0.

Problem 23. Consider the series

∑_{n=0}^{∞}(−1)^{n} \frac{n − 1}{n} sin^{2} n.

Which of the following statements is true?

(a) The series is absolutely convergent by the integral test.

(b) The series is convergent by the alternating series test.

(c) The series is divergent by the test for divergence.

(d) The series is convergent by the comparison test.

(e) None of the above.

Solution. The answer is (c), since

lim_{n→∞}(−1)^{n} \frac{n − 1}{n} sin^{2} n ≠ 0.

Statement (a) is false: we end up with (n−1)/n(sin^{2} n) which is positive but not decreasing (it is oscillating, note that (n − 1)/n → 1 as n → ∞. Statement (b) is false, again we need the terms (n − 1)/n sin^{2} n to be decreasing. Statement (d) is false because the series is divergent.

Problem 24. Evaluate

∫ \frac{dx}{\sqrt{x^{2} − 2x}}.

Solution. We need to complete the square in the denominator so we can use a trigonometric substitution. We note that x^{2} − 2x = (x − 1)^{2} − 1, so the integral becomes

∫ \frac{dx}{\sqrt{(x − 1)^{2} − 1}} = ∫ \frac{du}{\sqrt{u^{2} − 1}}

where u = x − 1. Now we substitute u = sec θ, so du = sec θ tan θ dθ, and \sqrt{u^{2} − 1} = tan θ, hence we are left with

∫ \frac{sec θ tan θ dθ}{tan θ} = ∫ sec θ dθ = ln | sec θ + tan θ| + C

= ln |u + \sqrt{u^{2} − 1}| + C = ln |(x − 1) + \sqrt{x^{2} − 2x}| + C.

Problem 25. Consider the sequence defined by

a_{n} = \frac{ne^{1/n}}{3n − 1}.

What is

lim_{n→∞} a_{n}?

(a) 0

(b) ∞

(c) 1/3

(d) 1

(e) The sequence does not have a limit.

Solution. We use L’Hopital’s rule:

lim_{x→∞} \frac{xe^{1/x}}{3x − 1} = \frac{∞}{∞} = lim_{x→∞} \frac{e^{1/x} + x(−1/x^{2})e^{1/x}}{3} = \frac{1 + 0}{3} = \frac{1}{3}.

The answer is (c).

Problem 26. Evaluate

∫_{0}^{π} sec x dx.

Solution. The integral is improper (!) since sec π/2 = 1/(cos π/2) and cos π/2 = 0. So we write this as the limit of proper integrals:

∫_{0}^{π} sec x dx = lim_{t→π/2^{−}} ∫_{0}^{t} sec x dx + lim_{t→π/2^{+}} ∫_{t}^{π} sec x dx.

We have

∫ sec x dx = ln | sec x + tan x| + C

so

lim_{t→π/2^{−}} ∫_{0}^{t} sec x dx = lim_{t→π/2^{−}} ln | sec t + tan t| − 0.

Now as t → π/2, sec t → ∞ and tan t → ∞ as well, so ln | sec t+tan t| → ∞. The integral is divergent.

Problem 27. Find the area of the surface obtained by rotating the parabola y = x^{2} from x = 0 to x = 1 around the y-axis.

Solution. We use the formula for surface area

A = ∫_{C} r ds.

Drawing a picture, we see that the radius is x since we are rotating around the y-axis, so we end up with

A = ∫_{0}^{1} x\sqrt{1 + (\frac{dy}{dx})^{2}} dx = ∫_{0}^{1} x\sqrt{1 + 4x^{2}} dx.

Substitute u = 1 + 4x^{2} to get du = 8x dx, so

∫ x\sqrt{1 + 4x^{2}} dx = ∫ \frac{1}{8}\sqrt{u} du = \frac{1}{12}u^{3/2} + C = \frac{1}{12}(1 + 4x^{2})^{3/2} + C.

Hence

∫_{0}^{1} x\sqrt{1 + 4x^{2}} dx = \frac{1}{12}(5\sqrt{5} − 1).

Problem 28. Find the particular solution of

y′ + y = x + e^{x}

satisfying y(0) = 0.

Solution. This is a linear first-order equation. We have P (x) = 1 so I(x) = e^{∫ P (x) dx} = e^{x}, and

d(I(x)y) = d(e^{x}y) = e^{x}(x + e^{x}) dx

so

e^{x}y = ∫ (xe^{x} + e^{2x}) dx = (x − 1)e^{x} + \frac{1}{2}e^{2x} + C,

the first integral done by parts. Therefore

y(x) = x − 1 + \frac{1}{2}e^{x} + Ce^{−x}.

So

y(0) = −1 + 1/2 + C = 0

so C = 1/2, and

y(x) = x − 1 + \frac{1}{2}(e^{x} − e^{−x}) = x − 1 + sinh x.

Problem 29. Find the twelfth derivative of (x + 1)^{3}e^{x} at x = −1.

Solution. We must find a Taylor series for e^{x} at x = −1. You can do this by computing derivatives: if f(x) = e^{x}, then f^{(n)}(x) = e^{x} so f^{(n)}(−1) = 1/e, hence

e^{x} = ∑_{n=0}^{∞} \frac{f^{(n)}(−1)}{n!}(x + 1)^{n} = ∑_{n=0}^{∞} \frac{1}{en!}(x + 1)^{n}.

Hence

(x + 1)^{3}e^{x} = ∑_{n=0}^{∞} \frac{1}{en!}(x + 1)^{n+3}.

Since we want the twelfth derivative, we want the coefficient of (x + 1)^{12} in this series, which is 1/(e9!). At the same time, if g(x) = (x + 1)^{3}e^{x} then

g(x) = ∑_{n=0}^{∞} \frac{g^{(n)}(−1)}{n!}(x + 1)^{n}

tells us that the coefficient of (x + 1)^{12} is g^{(12)}(−1)/12!, so equating these two, we get

g^{(12)}(−1) = \frac{12!}{e9!} = 1320/e.

Problem 30. Determine if the series

∑_{n=1}^{∞} \frac{cos(1/n)}{n}

is convergent or divergent.

Solution. Since cos(1/n) → 1 as n → ∞, we compare the above series to ∑_{n=1}^{∞} 1/n. According to the limit comparison test, we should calculate

lim_{n→∞} \frac{cos(1/n)/n}{1/n} = lim_{n→∞} cos(1/n) = 1.

Therefore the series diverges since the harmonic series diverges.