Calculus I, Fall 2015

Final Exam

Jan 12, 2016, 10:10AM

1. (12 pts) Find the solutions for

(a) \frac{dy}{dx} = 2x\sqrt{1 − y^{2}}, −1 < y < 1
(b) \frac{dy}{dx} − y = −y^{2}

Ans:

(1) ∫ \frac{dy}{\sqrt{1−y^{2}}} = ∫ 2x dx ⇒ sin^{−1} y = x^{2} + C (6 pts) for some constant C.

(2) \frac{du}{dx} + u = 1 (2 pts) ⇒ (e^{x}u)′ = e^{x} ⇒ e^{x}u = e^{x} + C ⇒ u = 1 + Ce^{−x} ⇒ y = \frac{1}{1+Ce^{−x}} (4 pts) for some constant C.

2. (12 pts) Find the volume and surface area of the object obtained by rotating the region {(x − 3)^{2} + y^{2} ≤ 1, x ≥ 3} 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((3+\sqrt{1 − y^{2}})^{2}−3^{2}) dy (3pts) = 6π ∫_{−1}^{1} \sqrt{1 − y^{2}} dy+π ∫_{−1}^{1}(1−y^{2}) dy = 3π^{2} + \frac{4}{3}π. ∫_{−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}. (2 pts) ∫_{−1}^{1}(1 − y^{2}) dy = 2 − \frac{2}{3} = \frac{4}{3}. (1 pt)

Surface = ∫_{−1}^{1} 2π(3 + \sqrt{1 − y^{2}})\sqrt{\frac{1}{1−y^{2}}} dy (3pts) + 2π · 3 · 2 (1pt) = 6π ∫_{−1}^{1} \frac{1}{\sqrt{1−y^{2}}} dy +

2π ∫_{−1}^{1} dy + 12π = 6π^{2} + 16π. ∫_{−1}^{1} \frac{1}{\sqrt{1−y^{2}}} dy = sin^{−1} y|_{−1}^{1} = π. (2 pts)

3. (10 8pts) 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}} = lim_{x→∞} \frac{e(e − 1)x^{e−2}}{e^{x}} = lim_{x→∞} \frac{e(e − 1)(e − 2)x^{e−3}}{e^{x}} = 0. (since e < 3)

(3 pts)

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 is large enough)

(3 pts)

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. (since ln x > ln ln x for x is large enough) (3 pts) From slowest to fastest is x^{e}, e^{x}, (ln x)^{x}, x^{x}. (1 pts)

4. (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^{7}}{(1−x^{4})^{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}} . where A ∼ H are undetermined coefficients.(1 pts) for each term

5. (50 pts)

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

(4) ∫_{1}^{2} \frac{1}{e^{2x} − e^{−x}} dx (5) ∫ \frac{1}{\sqrt{4x − x^{2}}} dx

Ans:

(1) Let u^{2} = 2 − x

∫_{1}^{2} \frac{x}{\sqrt{2 − x}} dx = ∫_{1}^{0} \frac{2 − u^{2}}{u} (−2u) du(5 pts) = ∫_{0}^{1} 4 − 2u^{2} dx = (4u − \frac{2}{3}u^{3})|_{0}^{1} = \frac{10}{3}.(5 pts)

(2)

∫_{1}^{e^{π}} sin(ln x) dx = (x sin(ln x))|_{1}^{e^{π}} − ∫_{1}^{e^{π}} cos(ln x) dx(4 pts) = − ∫_{1}^{e^{π}} cos(ln x) dx

= −[(x cos(ln x))|_{1}^{e^{π}} + ∫_{1}^{e^{π}} sin(ln x) dx](4 pts) = e^{π} + 1 − ∫_{1}^{e^{π}} sin(ln x) dx
⇒ ∫_{1}^{e^{π}} sin(ln x) dx = \frac{e^{π} + 1}{2}(2 pts)

(3) Let u = sec x

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

(4) Let u = e^{x}

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

= ∫_{e}^{e^{2}} \frac{\frac{1}{3}}{u − 1} + \frac{−\frac{1}{3}u − \frac{2}{3}}{u^{2} + u + 1} du(2 pts)
= ∫_{e}^{e^{2}} \frac{\frac{1}{3}}{u − 1} du + ∫_{e}^{e^{2}} \frac{−\frac{1}{3}u − \frac{1}{6}}{u^{2} + u + 1} du + ∫_{e}^{e^{2}} \frac{−\frac{1}{2}}{u^{2} + u + 1} du
= [\frac{1}{3} ln |u − 1| − \frac{1}{6} ln |u^{2} + u + 1| − \frac{1}{\sqrt{3}} tan^{−1}(\frac{2}{\sqrt{3}}(u + \frac{1}{2}))]|_{e}^{e^{2}}(2 pts for each term)
= \frac{1}{3} ln(\frac{e^{2} − 1}{e − 1}) − \frac{1}{6} ln(\frac{e^{4} + e^{2} + 1}{e^{2} + e + 1}) − \frac{1}{\sqrt{3}}(tan^{−1}(\frac{2}{\sqrt{3}}(e^{2} + \frac{1}{2})) − tan^{−1}(\frac{2}{\sqrt{3}}(e + \frac{1}{2})))

(5) Let x − 2 = 2 sin θ(5 pts)

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

= ∫ 1 dθ = θ + C = sin^{−1}(\frac{x − 2}{2}) + C(5 pts)

for some constant C.

6. (10 pts) Express ∫_{0}^{π} cos^{6} x sin^{4} x dx in terms of ∫_{0}^{π} cos^{4} x sin^{4} x dx.

Ans:

∫_{0}^{π} cos^{6} x sin^{4} x dx = ∫_{0}^{π} cos^{5} x sin^{4} x cos x dx
= [cos^{5} x(\frac{1}{5} sin^{5} x)]_{0}^{π} − ∫_{0}^{π}(5 cos^{4} x(− sin x))(\frac{1}{5} sin^{5} x) dx(4 pts)
= ∫_{0}^{π} cos^{4} x sin^{6} x dx = ∫_{0}^{π} cos^{4} x sin^{4} x(1 − cos^{2} x) dx
= ∫_{0}^{π} cos^{4} x sin^{4} x dx − ∫_{0}^{π} cos^{6} x sin^{4} x dx(4 pts)
⇒ ∫_{0}^{π} cos^{6} x sin^{4} x dx = \frac{1}{2}∫_{0}^{π} cos^{4} x sin^{4} x dx(2 pts)

7. (10 pts) Start with domain and range for sech and sech^{−1}, derive the formula for the derivative of sech^{−1}. Ans:

sech : ℝ → (0, 1] (2 pts)
sech^{−1} : (0, 1] → [0, ∞) (2 pts)

\frac{d(sech^{−1}x)}{dx} = −\frac{1}{sech(sech^{−1}x)tanh(sech^{−1}x)} (2 pts)
= −\frac{1}{sech(sech^{−1}x)\sqrt{1 − sech^{2}(sech^{−1}x)}} (2 pts)
(∵ x ∈ (0, 1) ⇒ sech^{−1}x > 0 ⇒ tanh(sech^{−1}x) > 0)
= −\frac{1}{x\sqrt{1 − x^{2}}} (2 pts).

8. (8 pts) For what values of p ∈ ℝ, is the function f(x) = |x|^{p} differentiable at x = 0? Explain.

Ans:

For p > 1,

0 ≤ |\frac{|x|^{p}}{x} | = |x|^{p−1} → 0

as x → 0. Thus, |x|^{p} is differentiable at x = 0. (3 pts)

For p = 1,

lim_{x→0+} \frac{|x|}{x} = 1 ≠ −1 = lim_{x→0−} \frac{|x|}{x}.

Thus, |x|^{p} is not differentiable at x = 0. (2 pts)

For p < 1,

lim_{x→0+} \frac{|x|^{p}}{x} = lim_{x→0+} \frac{1}{x^{1−p}} = +∞.

Thus, |x|^{p} is not differentiable at x = 0. (3 pts)

9. (12 pts) State both parts of Fundamental Theorem of Calculus, prove that part 1 implies part 2.

Ans: \underline{Fundamental Theorem of Calculus Part 1}: (3 pts)

Suppose f is continuous on [a, b].

Then F (x) = ∫_{a}^{x} f(t) dt is continuous on [a, b] and differentiable on (a, b)

with derivative

F′(x) = f(x).

\underline{Fundamental Theorem of Calculus Part 1}: (3 pts)

Suppose f is continuous on [a, b] and F is any antiderivative of f on [a, b]. Then

∫_{a}^{b} f(x) dx = F (b) − F (a).

\underline{Part 1 implies Part 2}: (6 pts) Let G(x) = ∫_{a}^{x} f(t) dt. Then by Part 1, G is an antiderivative of f. Given any antiderivative F of f, then

F (x) = G(x) + C, x ∈ (a, b).

Since F and G are both continuous on [a, b],

F (x) = G(x) + C, x ∈ [a, b].

Therefore,

F (b) − F (a) = (G(b) + C) − (G(a) + C) = ∫_{a}^{b} f(x) dx.