Mathematics 136 – AP Calculus Practice Exam 2 Solutions October 29, 2003

I. (Note: On second thought this one is probably more complicated than a reasonable exam question would be; although the algebra of solving for the critical point is not complicated, thinking through the set-up could take some time. It’s good practice though!)

Here’s one way to set it up. Let x be the number of chairs sold. The revenue will be R(x) = x · P (x), where P (x) = price per chair. If x ≤ 300, then R(x) = 90x, so it is increasing (no critical points). For x > 300, the price per chair depends on the number of chairs in the following way:

P (x) = 90 − (.25)(x − 300),

so the revenue is

R(x) = x(90 − .25(x − 300)) = −.25x^{2} − 165x,

So to find the critical points, we take

R′(x) = −.5x − 165 = 0,

and hence x = 330. This is a local maximum by the 2nd derivative test: R′′(x) = −.5 < 0. The revenue at x = 330 is R(330) = (330)(90 − (.25)(30)) = $27225. The revenues at the endpoints of the interval are R(0) = $0 and R(400) = (400)(90 − (.25)(100)) = $26000. Hence the global maximum revenue is $27225, and the minimum is $0.

II. A) If f is continuous on [a, b], then the definite integral of f is the limit of the Riemann

sums for f:

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

(it would also be OK to give the LHS or RHS here). B) The Fundamental Theorem of Calculus says: part 1: If f is continuous on [a, b] and F is an antiderivative of f, then

∫_{a}^{b} f(x) dx = F (b) − F (a)

part 2: The function

F (x) = ∫_{a}^{x} f(t) dt

is an antiderivative of f:

\frac{d}{dx} ∫_{a}^{x} f(t) dt = f(x).

A) (Should have said IV.17 – oops!) Let u = sin^{n−1}(x) and dv = sin x dx, so du = (n − 1) sin^{n−2}(x) cos(x) dx and v = − cos(x). Then the integration by parts formula says

∫ sin^{n}(x) dx = − sin^{n−1}(x) cos(x) + (n − 1) ∫ sin^{n−2}(x) cos^{2}(x) dx

In the remaining integral, use the trig identity cos^{2}(x) = 1 − sin^{2}(x), so

∫ sin^{n}(x) dx = − sin^{n−1}(x) cos(x) + (n − 1) ∫ sin^{n−2}(x)(1 − sin^{2}(x)) dx

= − sin^{n−1}(x) cos(x) + (n − 1) ∫ sin^{n−2}(x) dx − (n − 1) ∫ sin^{n}(x) dx

This is one of the cases where we get the same integral back, up to a constant. Add (n − 1) ∫ sin^{n}(x) dx to both sides to yield

n ∫ sin^{n}(x) dx = − sin^{n−1}(x) cos(x) + (n − 1) ∫ sin^{n−2}(x) dx

Then divide by n to get the table entry:

∫ sin^{n}(x) dx = \frac{−1}{n} sin^{n−1}(x) cos(x) + \frac{(n − 1)}{n} ∫ sin^{n−2}(x) dx

B) There are several ways to do this. I think the most direct is to make a substitution first for the function in the exponent. Let u = x^{2}, then du = 2x dx. The x^{5} dx is \frac{1}{2}u^{2} du so the integrand is

\frac{1}{2} ∫ u^{2}e^{3u} du

Now use table entry III.14 (twice):

\begin{aligned}\frac{1}{2} ∫ u^{2}e^{3u} du & = \frac{1}{2} (\frac{u^{2}}{3} e^{3u} − \frac{2u}{9}e^{3u} + \frac{2}{27}e^{3u}) + C \\ & = \frac{x^{4}}{6} e^{3x^{2}} − \frac{x^{2}}{9} e^{3x^{2}} + \frac{1}{27}e^{3x^{2}} + C \end{aligned}

C) Let u = sin(2x) + 4, so du = 2 cos(2x). The form is

\frac{1}{2} ∫ u^{\frac{1}{3}} du = \frac{1}{2} · \frac{3}{4}u^{\frac{4}{3}} + C = \frac{3}{8}(sin(2x) + 4)^{\frac{4}{3}} + C

D) The denominator is 4x^{2} + 4x − 3 = (2x + 1)^{2} − 4 so there are real roots 2x + 1 = 2, 2x + 1 = −2, or x = 1/2, −3/2. To put this in the form to use number 27 in the table, we need to multiply top and bottom by 1/4 to make the leading coefficient 1:

∫ \frac{x + 3}{4x^{2} + 4x − 3} dx = ∫ \frac{x/4 + 3/4}{x^{2} + x − 3/4} dx = ∫ \frac{x/4 + 3/4}{(x − 1/2)(x + 3/2)} dx

So we use 27 with c = 1/4, d = 3/4, a = 1/2, b = −1/2. The antiderivative is

F (x) = \frac{7}{16} ln(x − 1/2) − \frac{3}{16} ln(x + 3/2)

For the definite integral, we take F (4) − F (2) ≐ .2859.