Mathematics 136, section 2 – Advanced Placement Calculus Solutions for Exam 2 – November 4, 2009

I. Consider the function f(x) = x^{3} + 6x^{2} − 15x.

A) (10) Find all the critical numbers of f and classify each as a local maximum, local minimum, or neither. Solution: We have f′(x) = 3x^{2} + 12x − 15 = 3(x + 5)(x − 1). This is defined for all x and zero only at x = −5, 1 so those are the critical numbers. We have f′′(x) = 6x + 12. Hence f′′(−5) = −18 and f′′(1) = 18. By the Second Derivative Test, x = −5 is a local maximum, and x = 1 is a local minimum. (Note: this can also be determined by the First Derivative Test, finding the intervals where f′ is positive (x < −5 and x > 1) and the interval where f′ is negative (−5 < x < 1).

B) (5) What are the absolute (global) maximum and minimum values of f(x) on the interval [0, 3]? Solution: The critical number x = 1 is in this interval. f(0) = 0, f(1) = −8 and f(3) = 36. So the absolute maximum is f(3) = 36 and the absolute minimum is f(1) = −8.

II. (10) Evaluate the following limit:

lim_{θ→0} \frac{cos(5θ) − 1}{θ^{2}}.

Solution: This is indeterminate of the form 0/0, so we can apply L’Hopital’s Rule (twice):

lim_{θ→0} \frac{cos(5θ) − 1}{θ^{2}} = lim_{θ→0} \frac{−5 sin(5θ)}{2θ} (still 0/0)

= lim_{θ→0} \frac{−25 cos(5θ)}{2}

= \frac{−25}{2}.

III. The velocity of a accelerating car is measured at each of the following times, yielding a table

of values:

t = time(sec) 1 3 5 7 9 11 ; v(t) = velocity(ft/sec) 1 3.5 5.1 7.8 9.0 11.2

A) (5) Give an estimate for the total distance traveled by the car between t = 1 and t = 11 using a left-hand Riemann sum for the velocity function and Δt = 2. Solution: The left-hand Riemann sum estimate is

v(1)Δt + v(3)Δt + v(5)Δt + v(7)Δt + v(9)Δt = 2 + 7 + 10.2 + 15.6 + 18 = 52.8ft.

B) (5) Is your result from A) less than or greater than the actual total distance traveled, assuming v is always increasing between t = 1 and t = 11? Explain. Solution: If v is always increasing on this interval, then each term in the left-hand Riemann sum underestimates the actual distance traveled on that interval. Hence the answer in A is less than the actual distance traveled.

IV. Compute each of the following integrals. Show all work.

A) (5) ∫_{1}^{4} x^{2} − 3\sqrt{x} + 4 dx

Solution: By the Evaluation Theorem, this gives:

\frac{x^{3}}{3} − 2x^{3/2} + 4x|_{1}^{4} = \frac{64}{3} − 16 + 16 − \frac{1}{3} + 2 − 4 = 19.

B) (10) ∫ sec^{2}(2x)\sqrt[3]{tan(2x) + 4} dx

Solution: Let u = tan(2x) + 4. Then du = 2 sec^{2}(2x) dx, and the form is

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

Integrate using the indicated method and the table (say which entry you are using if you use one).

C) (10) Use parts: ∫ x^{2}e^{4x} dx (full credit only for showing all work; you can check your

result by the table)

Solution: Integrate by parts twice making u the power of x both times:

∫ x^{2}e^{4x} dx = \frac{x^{2}e^{4x}}{4} − \frac{2}{4} ∫ xe^{4x} dx

= \frac{x^{2}e^{4x}}{4} − \frac{1}{2} (\frac{xe^{4x}}{4} − \frac{1}{4} ∫ e^{4x} dx)

= \frac{x^{2}e^{4x}}{4} − \frac{xe^{4x}}{8} + \frac{e^{4x}}{32} + C.

D) (12.5) Use a trigonometric substitution: ∫ \frac{\sqrt{25 − x^{2}}}{x} dx.

Solution: From the form of the integrand, let x = 5 sin θ, so dx = 5 cos θ dθ. Then the integral can be evaluated by using the trigonometric identity cos^{2} θ = 1 − sin^{2} θ:

∫ \frac{\sqrt{25 − x^{2}}}{x} dx = ∫ \frac{5 cos θ}{5 sin θ} 5 cos θ dθ

= 5 ∫ \frac{cos^{2} θ}{sin θ} dθ

= 5 ∫ \frac{1 − sin^{2} θ}{sin θ} dθ

= 5 ∫ csc θ − sin θ dθ

= 5 (ln | csc θ − cot θ| + cos θ) + C # 15 in table

= 5 ln |\frac{5}{x} − \frac{\sqrt{25 − x^{2}}}{x} | + \sqrt{25 − x^{2}} + C.

E) (12.5) Use partial fractions:

∫ \frac{x + 4}{(x + 1)(x^{2} + 16)} dx.

Solution: From the form of the denominator we set up the partial fractions:

\frac{2x + 19}{(x + 1)(x^{2} + 16)} = \frac{A}{x + 1} + \frac{Bx + C}{x^{2} + 16}
2x + 19 = A(x^{2} + 16) + (Bx + C)(x + 1).

Equating coefficients gives A + B = 0, B + C = 2 and 16A + C = 19. So A = 1, B = −1, and C = 3. Integrating with #15 in table, we obtain:

∫ \frac{dx}{x + 1} + ∫ \frac{−x + 3}{x^{2} + 16} dx = ln |x + 1| − \frac{1}{2} ln |x^{2} + 16| + \frac{3}{4} tan^{−1}(x/4) + C.

V. A) (15) A fish swimming against a current of u ft/sec (constant) for a distance L (constant) will expend energy

E(v) = \frac{Lv^{3}}{v − u}

if it swims at v ft/sec (u < v). What v minimizes the total energy expended? Solution: By the quotient rule,

E′(v) = \frac{(v − u) · 3Lv^{2} − Lv^{3} · (1)}{(v − u)^{2}},

and E′(v) = 0 when the numerator is zero, so 3L(v − u)v^{2} = Lv^{3}. Dividing through by Lv^{2}, 3(v − u) = v, so v = \frac{3u}{2}. This a minimum by the first derivative test.

B) (15) Using the definition (not the Evaluation Theorem) compute

∫_{2}^{3} x^{2} + 4x dx.

The following formulas may be useful:

∑_{i=1}^{n} i = \frac{n(n + 1)}{2}  ∑_{i=1}^{n} i^{2} = \frac{n(n + 1)(2n + 1)}{6}.

Solution: For the right-hand Riemann sum, we have Δx = \frac{3−2}{n} = \frac{1}{n} and x_{i} = 2 + \frac{i}{n}. Then

∫_{2}^{3} x^{2} + 4x dx = lim_{n→∞} ∑_{i=1}^{n} ((2 + \frac{i}{n})^{2} + 4 (2 + \frac{i}{n})) \frac{1}{n}

= lim_{n→∞} ∑_{i=1}^{n} (\frac{i^{2}}{n^{2}} + \frac{8i}{n} + 12) \frac{1}{n}

= lim_{n→∞} (\frac{1}{n^{3}} ∑_{i=1}^{n} i^{2} + \frac{8}{n^{2}} ∑_{i=1}^{n} i + 12)

= lim_{n→∞} \frac{n(n + 1)(2n + 1)}{n^{3}} + \frac{8n(n + 1)}{2n^{2}} + 12

= \frac{1}{3} + \frac{8}{2} + 12 = \frac{49}{3}.

(Check by Evaluation Theorem:

∫_{2}^{3} x^{2} + 4x dx = \frac{x^{3}}{3} + 2x^{2}|_{2}^{3} = 9 + 18 − \frac{8}{3} − 8 = \frac{49}{3}.)