Mathematics 134 – Intensive Calculus for Science 2 Practice Final Exam Solutions

I. A) The statement of the Fundamental Theorem of Calculus is: If f is continuous on [a, b] and F is an antiderivative of f, then

∫_{a}^{b} f(x) dx = F (b) − F (a)

B) The function F is decreasing for −2 < x < −1 (since f is < 0 there), increasing for −1 < x < 1 (since f is > 0 on that interval, constant for 2 < x < 3, and decreasing for 3 < x < 4. For the intervals where f is horizontal, the slope of F is constant (so that part of the graph of F is a part of a straight line). The values of F at the points −2, −1, 0, 1, 2, 3, 4 can be found from the Fundamental Theorem. Since F (0) = 0 is given, by considering areas we see:

F (0) − F (−2) = ∫_{−2}^{0} f(x) dx = 0 so F (−2) = 0
F (0) − F (−1) = ∫_{−2}^{0} f(x) dx = 1 so F (−1) = −1
F (1) − F (0) = ∫_{0}^{1} f(x) dx = 1/2 so F (1) = 1/2
F (2) − F (0) = ∫_{0}^{2} f(x) dx = 1/2 so F (2) = 1/2
F (3) − F (0) = ∫_{0}^{3} f(x) dx = 3/2 so F (3) = 3/2
F (4) − F (0) = ∫_{0}^{4} f(x) dx = 1/2 so F (4) = 1/2

Here is a sketch of the plot:

II. A) By the power (and sum and constant multiple) rules

∫ 5x^{4} − \sqrt{x} + \frac{3}{x} dx = x^{5} − \frac{2}{3}x^{3/2} + 3 ln |x| + c

B) Let u = x^{4} + 1, so du = 4x^{3} + 1. The given integral is

∫ x^{3} sin(x^{4} + 1) dx = \frac{1}{4} ∫ sin(u) du

= \frac{−1}{4} cos(u) + c

= \frac{−1}{4} cos(x^{4} + 1) + c

C) Integrate by parts twice, letting u be the power of x and dv = e^{−5x}dx:

∫ x^{2}e^{−5x} dx = \frac{−x^{2}}{5} e^{−5x} + \frac{2}{5} ∫ xe^{−5x} dx

= \frac{−x^{2}}{5} e^{−5x} + \frac{2}{5} (\frac{−x}{5}e^{−5x} + \frac{1}{5} ∫ e^{−5x} dx)

= \frac{−x^{2}}{5} e^{−5x} + \frac{2}{5} (\frac{−x}{5}e^{−5x} − \frac{1}{25}e^{−5x}) + c

= \frac{−x^{2}}{5} e^{−5x} − \frac{2x}{25}e^{−5x} − \frac{2}{125}e^{−5x} + c

(This can also be derived from entry 14 in the table of integrals, taking a = −5.) D) Use entry 24 in the table, with a = 3, u = 2x. We have

∫ \frac{1}{u^{2} + a^{2}} = \frac{1}{a} arctan (\frac{u}{a}) + c

so

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

= \frac{1}{2} · \frac{1}{3} arctan (\frac{x}{3}) + c

= \frac{1}{6} arctan (\frac{x}{3}) + c

E) Let u = ln(x). Then du = \frac{1}{x}dx, so the given integral is

∫ \frac{(ln(x))^{4}}{x} dx = ∫ u^{4} du

= \frac{u^{5}}{5} + c

= \frac{(ln(x))^{5}}{5} + c

III. A) Rotating about the x-axis, we get disk cross-sections, so the volume is

V = ∫_{0}^{4} π(x^{3/2})^{2} dx = ∫_{0}^{4} πx^{3} dx = π \frac{x^{4}}{4} |_{0}^{4} = 64π

B) Rotating about the line y = −4, we get “washer” cross-sections with inner radius 4 and outer radius 4 + x^{3/2}. The volume is

V = ∫_{0}^{4} π(x^{3/2} + 4)^{2} − π4^{2} dx

= π ∫_{0}^{4} x^{3} + 8x^{3/2} + 16 − 16 dx

= π (\frac{x^{4}}{4} + \frac{16}{5}x^{5/2}) |_{0}^{4}

= \frac{832π}{5} ≐ 522.76

IV. A) Since the density is 0 for all x < 0, we have

∫_{−∞}^{∞} p(x) dx = ∫_{0}^{∞} .1e^{−.1x} dx = lim_{b−>∞} −e^{−.1x}|_{0}^{b} = lim_{b−>∞}(1 − e^{−.1b}) = 1

B) The median time between cars is the solution T of the equation ∫_{−∞}^{T} p(x) = 1/2. As in part A, since p(x) = 0 if x < 0, this says

1/2 = ∫_{0}^{T} .1e^{−.1x} dx = 1 − e^{−.1T} .

Hence e^{−.1T} = 1/2, so T = ln(1/2)/(−.1) ≐ 6.93.

V. A) To compute the derivatives here, you need to recall the product rule from last semester! f(x) = xe^{x}, so f′(x) = xe^{x} + e^{x}, f′′(x) = xe^{x} + e^{x} + e^{x} = xe^{x} + 2e^{x}, and f′′(x) = xe^{x} + e^{x} + e^{x} + e^{x} = xe^{x} + 3e^{x}. Therefore the 3rd degree Taylor polynomial at a = 0 is

p_{3}(x) = f(0) + f′(0)x + \frac{f′′(0)}{2}x^{2} + \frac{f′′(0)}{6}x^{3} = x + (2/2)x^{2} + (3/6)x^{3} = x + x^{2} + x^{3}/2.

B) Both p_{3}(.1) and (.1)e^{.1} equal .1105 to four decimal places, so the agreement is close.

VI. A) From the Taylor series for \frac{1}{1−u} at a = 0, we have \frac{1}{1−u} = 1 + u + u^{2} + u^{3} + · · ·. Letting u = −x^{2}, we get

\frac{1}{1 + x^{2}} = 1 − x^{2} + x^{4} − x^{6} + ⋯

B) We know ∫ \frac{1}{1+x^{2}} dx = arctan(x). Hence to get the Taylor series for arctan we can integrate series for \frac{1}{1+x^{2}} term by term. The result is

arctan(x) = c + x − \frac{x^{3}}{3} + \frac{x^{5}}{5} − \frac{x^{7}}{7} + ⋯

for some constant c. Since arctan(0) = 0, c must be zero, and

arctan(x) = x − \frac{x^{3}}{3} + \frac{x^{5}}{5} − \frac{x^{7}}{7} + ⋯

C) The series for g(x) = \frac{arctan(x)−x}{x^{3}} can be found by algebraic manipulations of the series from part B:

\frac{arctan(x) − x}{x^{3}} = \frac{x − \frac{x^{3}}{3} + \frac{x^{5}}{5} + ⋯ − x}{x^{3}} = \frac{−1}{3} + \frac{1}{5}x^{2} + ⋯

The coefficient of x^{2} is the second derivative of g(x) at zero, divided by 2, so g′′(0) = \frac{2}{5} > 0. Hence g is concave up at 0.

VII. A) The number of calories needed per day to maintain weight is 20W, so if I is the constant calorie intake per day,

\frac{dW}{dt} = k(20W − I)

for some constant of proportionality k. B) Separating variables and integrating:

∫ \frac{dW}{20W − I} = ∫ k dt
\frac{1}{20} ln |20W − I| = kt + c
20W − I = de^{20kt}

W = \frac{I + de^{20kt}}{20}

C) Given I = 3000, and k = \frac{−1}{3500}. When t = 0, W (0) = 160, so we have 160 = \frac{3000+d}{20}, so d = 3200 − 3000 = 200. Then the person’s weight at the end of one year is

W (365) = \frac{3000 + 200e^{−365/3500}}{20} ≐ 151.2

Extra Credit: The idea here is to integrate the equations of the parabolas to find the areas of the given cross-sections, then use those to derive a right-hand sum approximation of the integral ∫_{0}^{20} A(x) dx with Δx = 5 to approximate the volume enclosed by the hull. For example, the area of the cross-section at x = 5 is ∫_{−1}^{1} 0 − (2x^{2} − 2) dx = 8/3. Doing the same for the other given cross-sections, the volume is estimated by

(8/3) · 5 + 4 · 5 + (16/3) · 5 + \frac{32}{3\sqrt{5}} · 5 ≐ 83.9ft^{3}