Mathematics 126–Calculus for Social Science 2 Solutions – Final Examination December 10, 2001

I. A)

∫_{1}^{2} x^{6} − x^{1/2} + x^{−1} dx = (\frac{x^{7}}{7} − \frac{2}{3}x^{3/2} + ln |x|) |_{1}^{2} ≐ 17.62

B) (This one had the typo.) The correct integral is

∫_{0}^{1}(x^{3} + 1)e^{x^{4}+4x} dx

Let u = x^{4} + 4x, so du = 4x^{3} + 4 = 4(x^{3} + 1) dx and (x^{3} + 1) dx = \frac{1}{4}du. Substituting the whole integral to the new variable u, we have

∫_{u=0}^{u=5} \frac{1}{4}e^{u} du = \frac{1}{4}e^{u}|_{0}^{5} = \frac{1}{4}(e^{5} − 1) ≐ 36.85

C) Integrate by parts with u = (ln(x))^{2}, dv = xdx. Then du = 2 ln(x)\frac{1}{x} and v = \frac{x^{2}}{2} . So after simplifying ∫ v du,

∫ x(ln(x))^{2} dx = \frac{x^{2}}{2} (ln(x))^{2} − ∫ x ln(x) dx

To finish, we must integrate by parts again, with u = ln(x), dv = xdx. This time du = \frac{1}{x}, and v = \frac{x^{2}}{2} , so

∫ x(ln(x))^{2} dx = \frac{x^{2}}{2} (ln(x))^{2} − ∫ x ln(x) dx

= \frac{x^{2}}{2} (ln(x))^{2} − \frac{x^{2}}{2} ln(x) + ∫ \frac{1}{2}x dx

= \frac{x^{2}}{2} (ln(x))^{2} − \frac{x^{2}}{2} ln(x) + \frac{x^{2}}{4} + C

D) Let u = e^{x}. Then du = e^{x}dx, so dx = \frac{du}{e^{x}} = \frac{du}{u}. Substituting, the integral has the form of # 6 in the table with a = 4, b = 7:

∫ \frac{dx}{\sqrt{4 + 7e^{x}}} = ∫ \frac{du}{u\sqrt{4 + 7u}} = \frac{1}{2} ln |\frac{\sqrt{4 + 7u} − 2}{\sqrt{4 + 7u} + 2}| + C = \frac{1}{2} ln |\frac{\sqrt{4 + 7e^{x}} − 2}{\sqrt{4 + 7e^{x}} + 2}| + C

A) The value of the midpoint Riemann sum with n = 4 is

\frac{(.25)}{2\sqrt{2π}} (e^{−(5.125−6)^{2}/8} + e^{−(5.375−6)^{2}/8} + e^{−(5.625−6)^{2}/8} + e^{−(5.875−6)^{2}/8})
≐ \frac{(.25)}{2\sqrt{2π}} (e^{−.0957} + e^{−.0488} + e^{−.0176} + e^{−.0020})
≐ .1916

B) The Simpson’s Rule approximation is

\begin{aligned}\frac{(.25)}{3 · 2\sqrt{2π}} & (e^{−(5−6)^{2}/8} + 4e^{−(5.25−6)^{2}/8} + 2e^{−(5.5−6)^{2}/8} + 4e^{−(5.75−6)^{2}/8} + e^{−(6−6)^{2}/8}) \\ & ≐ .1915 \end{aligned}

C) Note: only one person in the class saw how to do this part, so I adjusted the exam scores by adding 5 points to everyone’s score (including the person who did this correctly). The idea is that the given function

f(x) = \frac{1}{2\sqrt{2π}} e^{(x−6)^{2}/8}

is the probability density function for a normal random variable x with mean μ = 6, and σ = 2. The given integral computes the probability P (5 ≤ x ≤ 6) = P (\frac{5−6}{2} ≤ z ≤ \frac{6−6}{2}), where z is the standard normal variable with μ = 0, and σ = 1. From the table

P (−.5 ≤ z ≤ 0) = P (z ≤ 0) − P (z ≤ −.5) = .5000 − .3085 = .1915

III. The phrase “indefinitely into the future” indicates that we want the value of the improper integral

∫_{0}^{∞}(30 + 2t)e^{(−.08)t} dt

Using parts with u = 30+2t and dv = e^{(−.08)t} dt, we have du = 2dt, and v = \frac{1}{(−.08)}e^{(−.08)t}, so

\begin{aligned}∫ (30 + 2t)e^{(−.08)t} dt & = \frac{(30 + 2t)}{−.08}e^{(−.08)t} + ∫ \frac{2}{.08}e^{(−.08)t} dt \\ & = \frac{(30 + 2t)}{−.08}e^{(−.08)t} − \frac{2}{(.08)^{2}} e^{(−.08)t} \end{aligned}

Then to evaluate the improper integral, we want:

lim_{b→∞} \frac{(30 + 2t)}{−.08}e^{(−.08)t} − \frac{2}{(.08)^{2}} e^{(−.08)t}|_{0}^{b}

= lim_{b→∞} (\frac{(30 + 2b)}{−.08}e^{(−.08)b} − \frac{2}{(.08)^{2}} e^{(−.08)b} + \frac{30}{.08} + \frac{2}{(0.08)^{2}} )

= 0 + 375 + 312.5

= 687.5 (thousand dollars).

IV. A) It’s not necessary to solve the differential equation to do this part. Just check that the

relation is satisfied. For y = 30−10e^{−4t}, \frac{dy}{dt} = 40e^{−4t}. The right side is −4(−10)e^{−4t} = 40e^{−4t} also. Then y(0) = 30 − 20 · 1 = 10.

B) This is a separable differential equation for x: \frac{p}{x} \frac{dx}{dp} = −p(.4 + 2p) so

∫ \frac{dx}{x} = ∫ −.4 − 2p dp
ln |x| = −.4p − p^{2} + c
x = be^{−4p−p^{2}}

The initial condition was supposed to be x(0) = 1 (another typo, D’OH!), which says b = 1. I did not take off credit if the constant was not determined.

V. A) The idea on this part is that a probability density function must satisfy ∫_{−∞}^{∞} f(x) dx =

1. This lets us determine a value for c as follows:

\begin{aligned}∫_{0}^{100} c(100x^{2} − x^{3}) dx & = 1 \\ c · (\frac{100x^{3}}{3} − \frac{x^{4}}{4} ) |_{0}^{100} & = 1 \\ c\frac{10^{8}}{12} & = 1 \\ c & = 12 × 10^{−8} \end{aligned}

B) The expected value is

E(x) = ∫_{0}^{100}(12 × 10^{−8})x(100x^{2} − x^{3}) dx

= (12 × 10^{−8}) (\frac{100x^{4}}{4} − \frac{x^{5}}{5} ) |_{0}^{100}

= (12 × 10^{−8}) · (10^{10}/20) = 60

C) The standard deviation is square root of

V ar(x) = ∫_{0}^{100}(12 × 10^{−8})x^{2}(100x^{2} − x^{3}) dx − 60^{2}

= (12 × 10^{−8}) (\frac{100x^{5}}{5} − \frac{x^{6}}{6} ) |_{0}^{100} − 3600

= 4000 − 3600 = 400

So σ = \sqrt{V ar(x)} = 20.

VI. To maximize the volume, we clearly want the sum of length and girth to be as large as possible, so z + 2x + 2y = 108. Use this to solve for z in terms of x, y: z = 108 − 2x − 2y. Then the volume is V = xyz = xy(108 − 2x − 2y). We have

\begin{aligned}V_{x} & = 108y − 4xy − 2y^{2} = y(108 − 4x − 2y) \\ V_{y} & = 108x − 2x^{2} − 4xy = x(108 − 2x − 4y) \end{aligned}

The critical points satisfy either y = 0 or 4x+2y = 108, and either x = 0 or 2x+4y = 108. The critical points with x = 0 and/or y = 0 will clearly not maximize the volume (because they make V = 0). So we want the critical point with 4x + 2y = 2x + 4y = 108. Solving simultaneously gives x = y = 18, and z = 108 − 36 − 36 = 36. It can be seen that this is a maximum for the volume by the second derivative test.

VII. Using Lagrange Multipliers, let F (x, y, λ) = (x^{2} + y^{2} − xy) + λ(x + 2y − 14). Then our system of equations is

\begin{aligned}F_{x} & = 2x − y + λ = 0 \\ F_{y} & = 2y − x + 2λ = 0 \\ F_{λ} & = x + 2y − 14 = 0 \end{aligned}

From the first two equations

−4x + 2y = 2λ = x − 2y

so 5x = 4y, or y = \frac{5}{4}x. Substitute this in the constraint equation: x + \frac{10}{4}x = 14, so x = 4, y = 5. Since this is the only critical point satisfying the constraint, it must be the minimum we want (f(x, y) = x^{2} + y^{2} − xy = (x − y/2)^{2} + 3y^{2}/4) has no local maximum on the constraint line).