Calculus I, Fall 2021

Brief solutions to Final Exam

Jan 13, 2022.

1. (12 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: Volume = ∫_{−1}^{1} π((2 + \sqrt{1 − y^{2}})^{2} − 2^{2})dy = 2π^{2} + \frac{4}{3}π. Here we have used ∫_{−1}^{1} \sqrt{1 − y^{2}} dy = ∫_{−π/2}^{π/2} cos^{2} θ dθ = ∫_{−π/2}^{π/2} \frac{cos 2θ+1}{2} dθ = \frac{π}{2}. and ∫_{−1}^{1}(1− y^{2}) dy = 2 − \frac{2}{3} = \frac{4}{3}. Surface Area = ∫_{−1}^{1} 2π(2 + \sqrt{1 − y^{2}})\sqrt{\frac{1}{1−y^{2}}} dy + 2π · 2 · 2 = 4π^{2} + 12π.

Here we have used ∫_{−1}^{1} \frac{1}{\sqrt{1−y^{2}}} dy = sin^{−1} y|_{−1}^{1} = π.

2. (12 pts) Find the solutions for (a): \frac{dy}{dx} = e^{x−y}, y(0) = 1, and (b): x\frac{dy}{dx} + y = sin x, y(π) = 0, x > 0, respectively. Ans: (a): ∫ e^{y}dy = ∫ e^{x}dx. Integration gives e^{y} = e^{x} + C_{1} for some constant C_{1}. y = ln(e^{x} + C_{1}), y(0) = 1, so C_{1} = e − 1, Answer: y = ln(e^{x} + e − 1). (b): Find the integration factor through standard procedure to obtain (xy)′ = sin x. Integration gives y = \frac{1}{x}(− cos x + C_{2}). Since y(π) = 0, we have C_{2} = −1. Answer: y = \frac{−1}{x}(1 + cos x).

3. (6 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}} .

4. (64 pts) Evaluate (a) : ∫_{0}^{1} x^{2}e^{−x}dx (b) : ∫ cosh^{4} x dx (c) : ∫_{0}^{1} sin^{−1} x dx (d) : ∫_{0}^{π/4} tan^{3} x sec^{3} x dx

(e) : ∫_{1}^{2} \frac{1}{\sqrt{4x − x^{2}}} dx (f) : ∫_{0}^{1} \frac{2x + 1}{(1 + x^{2})^{2}} dx (g) : ∫_{0}^{1} \frac{1}{\sqrt{1 + e^{x}}} dx (h) : ∫_{0}^{π/2} \frac{1}{1 + sin x}dx

Ans:

(a) Use integration by parts:

∫_{0}^{1} x^{2}e^{−x} dx = −x^{2}e^{−x} − 2xe^{−x} − 2e^{−x} |_{0}^{1} = 2 − \frac{5}{e}

(b)

(cosh^{2} x)^{2} = (\frac{cosh 2x + 1}{2})^{2} = \frac{1}{4} cosh^{2} 2x+\frac{1}{2} cosh 2x+\frac{1}{4} = \frac{1}{8}(cosh 4x+1)+\frac{1}{2} cosh 2x+\frac{1}{4}

∫ cosh^{4} x dx = \frac{1}{32} sinh 4x + \frac{1}{4} sinh 2x + \frac{3}{8}x + C

(c)

∫_{0}^{1} sin^{−1} x dx = x sin^{−1} x|_{0}^{1} − ∫_{0}^{1} \frac{x}{\sqrt{1 − x^{2}}} dx = (x sin^{−1} x + \sqrt{1 − x^{2}})|_{0}^{1} = \frac{π}{2} − 1

(d) Let u = sec x

∫_{0}^{π/4} tan^{3} x sec^{3} x dx = ∫_{1}^{√\overline{2}}(u^{2} − 1)u^{2} du = (\frac{u^{5}}{5} − \frac{u^{3}}{3} )|_{1}^{√\overline{2}} = \frac{2 + 2\sqrt{2}}{15}

(e) Let x − 2 = 2 sin θ

∫_{1}^{2} \frac{1}{\sqrt{4x − x^{2}}} dx = ∫_{x=1}^{2} \frac{1}{\sqrt{4 − (x − 2)^{2}}} dx = ∫_{θ=− \frac{π}{6}}^{0} \frac{1}{2 cos θ}2 cos θ dθ = ∫_{− \frac{π}{6}}^{0} 1 dθ = \frac{π}{6}

(f)

∫_{0}^{1} \frac{2x + 1}{(1 + x^{2})^{2}} dx = ∫_{0}^{1} \frac{2x}{(1 + x^{2})^{2}} dx + ∫_{0}^{1} \frac{1}{(1 + x^{2})^{2}} dx

∫_{0}^{1} \frac{2x}{(1 + x^{2})^{2}} dx = ∫_{0}^{1} \frac{d(1 + x^{2})}{(1 + x^{2})^{2}} = −(1 + x^{2})^{−1}|_{0}^{1} = \frac{1}{2}

∫_{0}^{1} \frac{1}{(1 + x^{2})^{2}} dx =^{x=tan t} ∫t=0^{\frac{π}{4}} cos^{2} t dt = ∫0^{\frac{π}{4}} \frac{cos 2t + 1}{2} dt = \frac{π}{8} + \frac{1}{4}

Answer = \frac{π}{8} + \frac{3}{4}

(g) With the substitution u = \sqrt{1 + e^{x}}, 2u du = e^{x} dx, e^{x} = u^{2} − 1,

∫_{0}^{1} \frac{1}{\sqrt{1 + e^{x}}} dx = ∫\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 |\frac{u − 1}{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}| = 2 ln (\frac{\sqrt{e + 1} − 1}{\sqrt{2} − 1}) − 1

(h) With the substitution t = tan \frac{x}{2} (x = 2 tan^{−1} t), sin x = \frac{2t}{1 + t^{2}} , dx = \frac{2 dt}{1 + t^{2}} ,

∫_{x=0}^{π/2} \frac{1}{1 + \frac{2t}{1 + t^{2}}} \frac{2dt}{1 + t^{2}} = ∫_{t=0}^{1} \frac{2dt}{(1 + t)^{2}} = −2(1 + t)^{−1}|_{0}^{1} = 1.

5. (8 pts) Order e^{x}, x^{x}, (ln x)^{x} and x^{e} from slowest to fastest growing rate as x → ∞. Explain.

Ans:

lim_{x→∞} \frac{x^{e}}{e^{x}} = lim_{x→∞} \frac{ex^{e−1}}{e^{x}} i = lim_{x→∞} \frac{e(e − 1)x^{e−2}}{e^{x}} i = lim_{x→∞} \frac{e(e − 1)(e − 2)x^{e−3}}{e^{x}} = 0 (∵ e < 3).

lim_{x→∞} \frac{e^{x}}{(ln x)^{x}} = lim_{x→∞} \frac{e^{x}}{e^{x ln ln x}} = lim_{x→∞} \frac{1}{e^{x(ln ln x−1)}} = 0 (since ln ln x > 1 for x large enough).

lim_{x→∞} \frac{(ln x)^{x}}{x^{x}} = lim_{x→∞} \frac{e^{x ln ln x}}{e^{x ln x}} = lim_{x→∞} \frac{1}{e^{x(ln x−ln ln x)}} = 0 (∵ ln x > ln ln x for x large enough).

Answer: From slowest to fastest: x^{e} ≪ e^{x} ≪ (ln x)^{x} ≪ x^{x}.

6. (10 pts) let f^{−1} be the inverse function of f. Suppose f and f^{−1} are twice differentiable

with

f(1) = 2, f′(1) = 3, f′′(1) = 4,
f(2) = 5, f′(2) = 6, f′′(2) = 7,
f(3) = 8, f′(3) = 9, f′′(3) = 10,

Find (f^{−1})′′(2).

Ans:

(f^{−1})′(y) = \frac{1}{f′(f^{−1}(y))}

\frac{d}{dy} ⟹ (f^{−1})′′(y) = \frac{−f′′(f^{−1}(y))}{(f′(f^{−1}(y)))^{3}} ,

f(1) = 2 ⟹ f^{−1}(2) = 1.

(f^{−1})′′(2) = \frac{−f′′(f^{−1}(2))}{(f′(f^{−1}(2)))^{3}} = \frac{−f′′(1)}{(f′(1))^{3}} = \frac{−4}{27}

7. (10 pts) State both parts of Fundamental Theorem of Calculus. Then prove that ’part 1 implies part 2’. Ans: See the textbook or Lecture 19 note.