Calculus I, Fall 2013 (TA: Yang Ming-Hsuen)

Brief solutions to Final Exam

1. (8 pts) Find the solutions for \frac{dy}{dx} = 3x^{2}e^{−y}.

Ans:

∫ e^{y} dy = ∫ 3x^{2} dx ⇒ e^{y} = x^{3} + C ⇒ y = ln(x^{3} + C).

2. (8 pts) Write down the form of partial fraction expansion for \frac{x^{7}}{(1 − x^{4})^{2}} . Need NOT

find the undetermined coefficients.

Ans:

\frac{x^{4}}{(1 − x^{4})^{2}} = \frac{x^{7}}{(1 + x)^{2} (1 − x)^{2} (1 + x^{2})^{2}}

= \frac{A}{1 + x} + \frac{B}{(1 + x)^{2}} + \frac{C}{1 − x} + \frac{D}{(1 − x)^{2}} + \frac{Ex + F}{1 + x^{2}} + \frac{Gx + H}{(1 + x^{2})^{2}} .

3. (8 pts) Order e^{x}, x^{x}, (ln x)^{x} and x^{e} from slowest to fastest growing rate as x → ∞.

Ans: As x → ∞, e(constant) grows slower than ln x, ln x grows slower than x. Hence e^{x} grows slower than (ln x)^{x}, (ln x)^{x} grows slower than x^{x}. On the other hand, x^{e} grows slower than e^{x}. Thus, the answer (the order of growing rate from the slowest to the fastest) is

x^{e}, e^{x}, (ln x)^{x} , x^{x}.

4. (64 pts) Evaluate

(1) ∫ \frac{1}{2 + sin x}dx (2) ∫ e^{x} sin x dx (3) ∫ \frac{1}{\sqrt{4x − x^{2}}} dx (4) ∫_{0}^{π/4} tan^{3} x sec^{3} x dx

(5) ∫_{1}^{2} \frac{1}{e^{x} − e^{−x}} dx (6) ∫_{0}^{∞} x^{2}e^{−x}dx (7) ∫_{0}^{1} \frac{1}{\sqrt{1 + e^{x}}} dx (8) ∫_{0}^{π} \sqrt{1 + sin x} dx

Ans:

(1) With the substitutions z = tan \frac{x}{2}, z + \frac{1}{2} = \frac{\sqrt{3}}{2} tan θ,

∫ \frac{1}{2 + sin x} dx = ∫ \frac{1}{2 + \frac{2z}{1+z^{2}}} \frac{2}{1 + z^{2}} dz = ∫ \frac{1}{z^{2} + z + 1} dz = ∫ \frac{1}{(z + \frac{1}{2})^{2} + \frac{3}{4}} dz

= ∫ \frac{\frac{\sqrt{3}}{2} sec^{2} θ}{\frac{3}{4} (tan^{2} θ + 1)} dθ = ∫ \frac{2}{3}\sqrt{3} dθ = \frac{2}{3}\sqrt{3} θ + C

= \frac{2}{3}\sqrt{3} tan^{−1} (\frac{2 tan \frac{x}{2} + 1}{\sqrt{3}}) + C.

1 (2) With integration by parts,

∫ e^{x} sin x dx = e^{x} sin x − ∫ e^{x} cos x dx = e^{x} sin x − (e^{x} cos x + ∫ e^{x} sin x dx) . Hence

∫ e^{x} sin x dx = \frac{1}{2} e^{x} sin x − \frac{1}{2} e^{x} cos x + C.

(3) With the substitution x − 2 = 2 sin θ

∫ \frac{1}{\sqrt{4x − x^{2}}} dx = ∫ \frac{1}{\sqrt{− (x − 2)^{2} + 4}} dx = ∫ \frac{2 cos θ}{\sqrt{4 − 4 sin^{2} θ}} dθ

= ∫ dθ = θ + C = sin^{−1} (\frac{x − 2}{2}) + C.

(4) With the substitution u = sec x,

∫0^{\frac{π}{4}} tan^{3} x sec^{3} x dx = ∫0^{\frac{π}{4}} (sec^{2} x − 1) sec^{2} x tan x sec x dx = ∫_{1}^{√\overline{2}} (u^{2} − 1) u^{2} du

= \frac{1}{5}u^{5} − \frac{1}{3}u^{3} |_{1}^{√\overline{2}} = \frac{1}{5}2^{\frac{5}{2}} − \frac{1}{3}2^{\frac{3}{2}} − \frac{2}{15} = \frac{2}{15} (\sqrt{2} + 1) .

(5) With the substitution u = e^{x},

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

= \frac{1}{2} ln |u − 1| − \frac{1}{2} ln |u + 1| |_{e}^{e^{2}} = \frac{1}{2} ln |\frac{e^{2} − 1}{e^{2} + 1}| − \frac{1}{2} ln |\frac{e − 1}{e + 1}|.

(6) With integration by parts,

∫_{0}^{∞} x^{2}e^{−x} dx = lim_{A→∞} ∫_{0}^{A} x^{2}e^{−x} dx = lim_{A→∞} (−x^{2}e^{−x} − 2xe^{−x} − 2e^{−x} |_{0}^{A}) = 2.

(7) With the substitution u = \sqrt{1 + e^{x}},

∫_{0}^{1} \frac{1}{\sqrt{1 + e^{x}}} = ∫\begin{matrix}√ & \overline{1+e} \\ √ & \overline{2} \end{matrix} \frac{1}{u} \frac{2u}{u^{2} − 1} du = ∫\begin{matrix}√ & \overline{1+e} \\ √ & \overline{2} \end{matrix} (\frac{1}{u − 1} − \frac{1}{u + 1}) du

= ln |u − 1| − ln |u + 1| |\begin{matrix}√ & \overline{1+e} \\ √ & \overline{2} \end{matrix} = ln |\frac{\sqrt{e + 1} − 1}{\sqrt{e + 1} + 1}| − ln |\frac{\sqrt{2} − 1}{\sqrt{2} + 1}|.

(8) With the substitution u = sin x,

∫_{0}^{π} \sqrt{1 + sin x} dx = ∫_{0}^{π} \sqrt{1 + sin x}\sqrt{1 − sin x} ; \sqrt{1 − sin x} dx = ∫_{0}^{π} \frac{|cos x|}{\sqrt{1 − sin x}} dx

= ∫_{0}^{1} \frac{1}{\sqrt{1 − u}} du − ∫_{1}^{0} \frac{1}{\sqrt{1 − u}} du = 2 (−2 \sqrt{1 − u} |_{0}^{1}) = 4.

5. (12 pts) Does the improper integral ∫_{1}^{∞} \frac{1}{\sqrt{x^{3} − x}}dx converge or diverge? Explain (DO NOT try to integrate explicitly). Ans: Note that \frac{1}{\sqrt{x^{3}−x}} has two ‘improper’ parts: x → ∞ and x = 1^{+}. For the x = 1^{+} part, let v = x − 1,

∫_{1}^{2} \frac{1}{\sqrt{x^{3} − x}} dx = ∫_{0}^{1} \frac{1}{\sqrt{v(v + 1)(v + 2)}} dv

Since

lim_{v→0^{+}} \frac{\frac{1}{\sqrt{v(v+1)(v+2)}}}{\frac{1}{v^{1/2}}} = \frac{1}{\sqrt{2}}

and ∫_{0}^{1} \frac{1}{v^{1/2}} dv converges, by the limit comparison test,

∫_{1}^{2} \frac{1}{\sqrt{x^{3} − x}} dx = ∫_{0}^{1} \frac{1}{\sqrt{v(v + 1)(v + 2)}} dv

converges. Similarly, since

lim_{x→∞} \frac{\frac{1}{\sqrt{x^{3}−x}}}{\frac{1}{x^{3/2}}} = 1

and ∫_{2}^{∞} \frac{1}{x^{3/2}} dx converges, by the limit comparison test, ∫_{2}^{∞} \frac{1}{\sqrt{x^{3}−x}} dx converges. Therefore,

∫_{1}^{∞} \frac{1}{\sqrt{x^{3} − x}} dx = ∫_{1}^{2} \frac{1}{\sqrt{x^{3} − x}} dx + ∫_{2}^{∞} \frac{1}{\sqrt{x^{3} − x}} dx

converges.

6. (8 pts) Express ∫_{1}^{2} sin x dx as a Riemann sum. That is, lim_{···} ∑_{···}^{···} · · ·.

Ans: Let P = {1 = x_{0} < x_{1} < ⋯ < x_{n} = 2} be a partition of [1, 2] and Δx_{k} = x_{k} − x_{k} − 1. Then

∫_{1}^{2} sin x dx = lim_{∥P ∥→0} ∑_{k=1}^{n} sin c_{k} Δx_{k}, where c_{k} ∈ [x_{k−1}, x_{k}] .

Or consider the partition P = \{1, 1 + \frac{1}{n}, …, 1 + \frac{n−1}{n}, 2\}. Then

∫_{1}^{2} sin x dx = lim_{n→∞} ∑_{k=1}^{n} sin (1 + \frac{k}{n}) \frac{1}{n} = lim_{n→∞} ∑_{k=n+1}^{2n} sin (\frac{k}{n}) \frac{1}{n}.

7. (8 pts) Evaluate lim_{x→0^{+}} x^{x ln x}.

Ans: Since

lim_{x→0^{+}} x (ln x)^{2} = lim_{x→0^{+}} \frac{(ln x)^{2}}{\frac{1}{x}} = lim_{x→0^{+}} \frac{2 ln x \frac{1}{x}}{−\frac{1}{x^{2}}} = lim_{x→0^{+}} \frac{−2 ln x}{\frac{1}{x}} = lim_{x→0^{+}} \frac{−2 \frac{1}{x}}{−\frac{1}{x^{2}}} = 0,

we have

lim_{x→0^{+}} x^{x ln x} = lim_{x→0^{+}} e^{x(ln x)^{2}} = e^{0} = 1.

8. (16 pts) Find the volume and surface area of the object obtained by rotating the region {(x − 2)^{2} + y^{2} ≤ 1, x ≥ 2} around the y axis. Note the surface area consists of two parts, one generated by a half circle, the other generated by a line segment. Ans: It is the same as Problem 7 of the second midterm exam.

V = \frac{4}{3}π + 2π^{2}, A = 4π^{2} + 12π.

9. (.100 pts) Schedule recitation change for next semester.

Ans: The schedule for recitation is changed to Wed evening next semester.