# BC Calculus: Extra Review Problems for Chapter 6

• Class: 5H: BC Calculus
• Author: Peter Atlas
• Text: Calculus Finney, Demana, Waits, Kennedy

1. Suppose that a body of mass $$m$$ is moving in a straight line with velocity $$v$$ encounters a resistance proportional to the velocity and that this is the only force acting on the body. If the body starts with velocity $$v_0$$, how far does it travel in time $$t$$? Assume $$F = \frac{d}{dt} \left( mv \right) .$$
2. Solution $$\frac{d}{dt} \left( mv \right) = kv \implies \displaystyle \int \frac{dv}{v} = \int \frac{k}{m} \, dt \implies \ln{|v|} = \frac{k}{m}t + C \implies |v| = Ce^{\frac{kt}{m}}$$, and $$v = v_0e^{\frac{kt}{m}} \implies \int_0^t v_0 e^{\frac{kx}{m}} \, dx = \left. \frac{v_0 m}{k} e^{\frac{kx}{m}} \right|_{0}^{t} = \frac{v_0m}{k} \left( e^{\frac{kt}{m}} - 1 \right)$$
3. Question two has been intentionally left blank.
4. Solution Really? Get a life.
5. The half-life of polonium is 140 days, but your sample will not be useful to you after 90% of the radioactive nuclei originally present have disintegrated. About how many days can you use the polonium?
6. Solution $$y = y_0 e ^{kt} \implies \frac{1}{2} = e^{140 k} \implies \frac{\ln{\frac{1}{2}}}{140} = k \implies 0.1 = e^{\frac{\ln {0.5}}{140}t} \implies t = 465.0699$$
7. Solve the differential equation $$\frac{dy}{dx} = 10 - 10y$$ subject to the initial condition that $$x = 0$$ when $$y = 2$$.
8. Solution $$\displaystyle \int \frac{dy}{1 - y} = \int 10 \, dx \implies \ - \ln {|1 - y|} = 10x + C \implies \ln{|1 - y|} = -10x + C \implies |1 - y| = Ce^{-10x}.$$ The initial condition is (0, 2), so $$C = 1$$ and $$y - 1 = e^{-10x} \implies y = e^{-10x} + 1$$
9. The velocity of a particle moving along the x-axis is proportional to $$x$$. This is to say the particle's velocity is proportional to its displacement. At time $$t = 0$$, the particle is located at $$x = 2$$, and at time $$t = 10$$, it is at $$x = 4$$. Find its position at $$t = 5$$.
10. Solution $$\frac{dx}{dt} = kx \implies x = x_0e^{kt} \implies x = 2e^{kt} \implies 4 = 2e^{10k} \implies k = \frac{\ln{2}}{10} \implies x = 2e^{\frac{\ln{2}}{10}t} \implies x(5) = 2e^{\frac{\ln{2}}{2}} = 2\sqrt{2}.$$
11. Evaluate the following integrals:
1. $$\displaystyle \int x^2 \cos{x} \, dx$$
2. Hint Tabular with $$x^2$$ going down.
Solution $$x^2 \sin{x} + 2x \cos{x} - 2\sin{x} + C$$
3. $$\displaystyle \int x^2 \ln{x} \, dx$$
4. Hint By parts with $$ln$$ going down.
Solution $$\frac{x^3 \ln{x}}{3} - \frac{x^3}{9} + C$$
5. $$\displaystyle \int x^3 e^{4x} \, dx$$
6. Hint Tabular with $$x^3$$ going down.
Solution $$\frac{x^3 e^{4x}}{4} - \frac{3x^2 e^{4x}}{16} + \frac{6x e^{4x}}{64} - \frac{6 e^{4x}}{256} + C$$
7. $$\displaystyle \int 4x \sec^2{x} \, dx$$
8. Hint By parts with $$\sec^2{x}$$ going up.
Solution $$4x \tan{x} - 4\ln{|\sec{x}|} + C$$
9. $$\displaystyle \int e^{-2x} \sin{(2x)} \, dx$$
10. Hint Solve for unknown integral. Have $$\sin {(2x)}$$ go down.
Solution $$-\frac{1}{4}e^{-2x}\sin{(2x)} -\frac{1}{4}e^{-2x}\cos{(2x)} + C$$
11. $$\displaystyle \int x \sqrt{1 - x} \, dx$$
12. Hint Substitution. Let $$u = 1 - x$$.
Solution $$-\frac{2}{3}(1 - x)^{\frac{3}{2}} + \frac{2}{5}(1 - x)^{\frac{5}{2}} + C$$