Mathematics 136, section 2 – AP Calculus Solutions for Final Exam Practice Questions December 14, 2009

I. A) The graph y = 2f(x − 2) + 1 would be obtained by shifting the graph y = f(x) to

the right by 2 units, stretching vertically by a factor of 2, then shifting the resulting graph up by 1 unit.

B) The graph y = −f(−x) would be obtained by reflecting the graph y = f(x) across the y-axis, then reflecting the resulting graph across the x-axis.

II. A) For problems of this kind, look for where the given graph has horizontal tangents

(zeroes of f′), and positive and negative slopes. I will look for qualitative information rather than any quantitative estimates of actual slope values.

B) These are the reverse of question A. When f(x) is > 0, the antiderivative g(x) is increasing. When f(x) < 0, the antiderivative g(x) is decreasing. g(x) will have local maxima at x-values where f changes from positive to negative, and local minima where f changes from negative to positive (First Derivative Test).

III. A) The derivative f′(a) is:

f′(a) = lim_{h→0} \frac{f(a + h) − f(a)}{h},

provided that the limit exists. B) The definite integral of f(x) over [a, b] is the limit of the Riemann sums

∫_{a}^{b} f(x) dx = lim_{n→∞} ∑_{i=1}^{n} f(x_{i}^{∗})Δx,

provided that the limit exists (independently of the choice of the x_{i}^{∗}). C) Let f(x) be continuous on [a, b]. Then

1) the function

G(x) = ∫_{a}^{x} f(t) dt

is an antiderivative of f(x) on (a, b) (that is G′(x) = f(x) for all x in (a, b)).

2) If F (x) is any antiderivative of f(x), then

∫_{a}^{b} f(x) dx = F (b) − F (a).

IV.

A) The plate is a 6 × 4 rectangle. Subdivide it into 6 vertical strips of width Δx = 1 at the given x-values. Since the density of the metal does not depend on y, we can estimate the density in any vertical strip by a density value from the table on that strip times the area of the strip. Using the left endpoints, for instance, this gives

M ≐ 200 · 1 · 4 + 190 · 1 · 4 + 170 · 1 · 4 + 140 · 1 · 4 + 100 · 1 · 4 + 90 · 1 · 4 = 3560

grams. B) Since we used the left endpoints, and the density appears to be decreasing left-to-right

across the plate, this estimate is probably too large.

V. A) Use a u-substitution on each term:

∫_{0}^{1} cos(πx) − x^{2}e^{x^{3}} dx = \frac{1}{π} sin(πx) − \frac{1}{3}e^{x^{3}} |_{0}^{1}

= 0 − \frac{1}{3}e + \frac{1}{3}

= \frac{1}{3}(1 − e).

B) We integrate by parts twice letting u be the power of x each time:

∫ x^{2} sin(3x) dx = \frac{−x^{2}}{cos(3x)} + \frac{2}{3} ∫ x cos(3x) dx
= \frac{−x^{2}}{3} cos(3x) + \frac{2}{3} (\frac{x}{3} sin(3x) − \frac{1}{3} ∫ sin(3x) dx)
= \frac{−x^{2}}{3} cos(3x) + \frac{2x}{9} sin(3x) + \frac{2}{27} cos(3x) + C

C) The partial fractions look like

\frac{x}{(x + 3)(x^{2} + 1)} = \frac{A}{x + 3} + \frac{Bx + C}{x^{2} + 1} .

Clearing denominators gives:

x = A(x^{2} + 1) + (Bx + C)(x + 3),

or

x = (A + B)x^{2} + (3B + C)x + (3C + A).

Hence A + B = 0, 3B + C = 1, and 3C + A = 0. From the last equation, A = −3C. So the first equation becomes B − 3C = 0 and the second is 3B + C = 1. Then (B − 3C) + 3(3B + C) = 10B = 3, so B = \frac{3}{10}, C = \frac{1}{10}, and A = \frac{−3}{10}. Integrating the partial fractions gives:

\begin{aligned}∫ \frac{\frac{−3}{10}}{x + 3} + \frac{\frac{3}{10}x + \frac{1}{10}}{x^{2} + 1} dx & = \frac{−3}{10} ∫ \frac{1}{x + 3} dx + \frac{3}{10} ∫ \frac{x}{x^{2} + 1} dx + \frac{1}{10} ∫ \frac{1}{x^{2} + 1} dx \\ & = \frac{−3}{10} ln |x + 3| + \frac{3}{20} ln(x^{2} + 1) + \frac{1}{10} tan^{−1}(x) + C. \end{aligned}

D) Completing the square gives

∫ \frac{dx}{\sqrt{(x − 8)^{2} + 1}}

We let x − 8 = tan θ, so dx = sec^{2} θ dθ, and the integral becomes

∫ \frac{sec^{2} θ dθ}{\sqrt{tan^{2} θ + 1}} = ∫ sec θ dθ

Using the appropriate entry from the integral table, this gives

= ln | sec θ + tan θ| + C = ln |\sqrt{(x − 8)^{2} + 1} + x − 8| + C.

VI. A) The cross-sections by planes x = constant are washers, so the volume is computed by

the integral

V = ∫_{0}^{2} π(4 − x^{2})^{2} dx = \frac{256π}{15} .

B) The cross-sections by planes y = constant are washers with outer radius 3 and inner radius 3 − \sqrt{4 − y}. So the volume is

V = ∫_{0}^{4} π(3)^{2} − π(3 − \sqrt{4 − y})^{2} dy = 24.

C) The coordinates of the centroid are

\overline{x} = \frac{∫_{0}^{2} x(4 − x^{2}) dx}{∫_{0}^{2} 4 − x^{2} dx} = \frac{3}{4},

and

\overline{y} = \frac{∫_{0}^{2} \frac{1}{2}(4 − x^{2})^{2} dx}{∫_{0}^{2} 4 − x^{2} dx} = \frac{8}{5}.

VII.

A) Graph omitted – the curves xy = c are rectangular hyperbolas with asymptotes along

the coordinate axes. B) Separating variables and integrating:

\begin{aligned}∫ \frac{dy}{y} & = ∫ x dx \\ ln |y| & = \frac{x^{2}}{2} + C \\ y & = ke^{x^{2}/2} \end{aligned}

where k = ±e^{C}. To get y(0) = 4, we want k = 4, so the solution is

y = 4e^{x^{2}/2}.

VIII. A) The differential equation is \frac{dT}{dt} = k(T − A). B) The solution, after separating and integrating is

T = A + ke^{kt},

C) Using the result from B with A = 70, we have T (0) = 200, so k = 130. Then at t = 5, 170 = 70 + 130e^{5k}, so

k = \frac{1}{5} ln (\frac{10}{13}) = −.0525.

Then we want the time t when 120 = 70 + 130e^{(−.0525)t}, so

t = \frac{ln (\frac{50}{130})}{−.0525} ≐ 18.2

minutes.

IX. Put the top of the punch bowl at y = 0. The work done is

W = ∫_{−0.5}^{0} π(\sqrt{(.5)^{2} − y^{2}})^{2} · 950 · 9.8(0 − y) dy ≐ 457

Joules. (Note: 1J is 1 kg · m^{2}/sec^{2}.)

X. A) Since 0 < \frac{e}{π} < 1, the geometric series converges to

\frac{1}{1 − \frac{e}{π}} = \frac{π}{π − e}.

B) For each k ≥ 0, the kth derivative of f(x) = e^{2x} at a = 0 is f^{(k)}(0) = 2^{k}. So the Taylor series is

∑_{k=0}^{∞} \frac{2^{k}x^{k}}{k!} = 1 + 2x + \frac{4x^{2}}{2!} + \frac{8x^{3}}{6!} + · · · .

C) The fifth degree Taylor polynomial of sin(x) at a = 0 is

p_{5}(x) = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!}

At x = .6, this gives

sin(.6) ≐ p_{5}(.6) = .564648.

The actual value of sin(.6) is about .564642, so the absolute error is about .000006.