MATH 2260 - 43005 MIDTERM EXAM 2 MODEL SOLUTION Write ...

MATH 2260 - 43005 MIDTERM EXAM 2 MODEL SOLUTION FALL 2012 - MOON

Write your answer neatly and show intermediate steps. Calculator, computer and any electronic devices are not allowed. (1) (1pt) Quick survey: Check one of them for each row. really sometimes not I’ve never useful useful useful used them Worksheets Model solutions Recommended problems (Optional) If you have any supplement materials you want to be provided, describe it.

(2) (5 pts) Solve the following differential equation dy ey x−3 = e−y x2 . dx

Z

e2y

x2 ey dy = dx ⇒ e2y dy = x5 dx −y −3 e x Z Z 2y e dy = x5 dx Z 1 2y 1 dy = e + C1 , x5 dx = x6 + C2 2 6 1 2y 1 6 e = x +C 2 6

• Separating variables and getting e2y dy = x5 dx: 2 pts. Z • Evaluating e2y dy and obtaining 21 e2y + C1 : +1 pt. Z • Evaluating x5 dx and getting 61 x6 + C2 : +1 pt. 1 1 • Finding correct answer e2y = x6 + C: 5 pts. 2 6

Date: October 9, 2012. 1

MATH 2260 - 43005 MIDTERM EXAM 2 MODEL SOLUTION

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(3) (6 pts) A fossilized animal bone is unearthed on the ruins of Pompeii. It contains 80% of the carbon-14 found in living matter. About how old is the bone? (It takes 5730 years to be half of the initial amount of carbon-14.) y(t) = amount of carbon-14 at the time t y(t) = y0 e−kt 1 y(5730) = y0 2 ln 1 1 1 1 ⇒ y0 e−5730k = y0 ⇒ e−5730k = ⇒ −5730k = ln ⇒ k = − 2 2 2 2 5730 1 ln 2

y(t) = y0 e 5730 t 1 ln 2

1 ln 2

y(t) = 0.8y0 ⇒ 0.8y0 = y0 e 5730 t ⇒ 0.8 = e 5730 t ln 12 5730 ln 0.8 ⇒ t = ln 0.8 ⇒ t = 5730 ln 12 5730 ln 0.8 answer : years ln 12 • Writing correct general solution of given differential equation y(t) = y0 e−kt : 1 pt. 1 • Stating given initial condition y(5730) = y0 using equality: 2 pts. 2 ln 21 • Finding k = − : 3 pts. 5730 5730 ln 0.8 • Solving the equation and getting the correct value t = : 5 pts. ln 12 5730 ln 0.8 • Writing the answer with appropriate unit years: 6 pts. ln 12


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(4) (5 pts) Evaluate the integral Z x sin x dx. f (x) = x,

g 0 (x) = sin x

f 0 (x) = 1,

g(x) = − cos x Z x sin x dx = x(− cos x) − − cos x dx Z = −x cos x + cos x dx = −x cos x + sin x + C Z

• Finding appropriate f (x) = x, g 0 (x) = sin x: 1 pt. Z • Applying integration by parts formula correctly and getting Z x(− cos x) − − cos x dx: 3 pts.

x sin x dx =

• Obtaining correct answer −x cos x + sin x + C: 5 pts.

(5) (7 pts) Evaluate the integral Z

2

x3 ex dx.

1 u = x2 ⇒ du = 2xdx, du = xdx Z Z 2 1 u 2 ue du x3 ex dx = 2 1 f (u) = u, g 0 (u) = eu 2 1 0 f (u) = , g(u) = eu 2 Z Z 1 u 1 u 1 u ue du = ue − e du 2 2 2 1 1 1 1 2 2 = ueu − eu + C = x2 ex − ex + C 2 2 2 2 Z 1 u • Finding appropriate substitution ue du: 2 pts. 2 1 • Applying integration by parts correctly and getting ueu − 2 1 u 1 u • Obtaining ue − e + C: 6 pts. 2 2 1 1 2 2 2 • Substituting u = x and getting x2 ex − ex + C: 7 pts. 2 2

Z

1 u e du: 4 pts. 2



Z

3

Z

2

2

sin3 x cos2 x dx. Z

2

(1 − cos2 x) cos2 x sin x dx

sin x cos x dx =

sin x cos x sin x dx =

Z

u = cos x ⇒ Z du = − sin xdx

2

2

4

2

(1 − cos x) cos x sin x dx =

2

(1 − u )u (−1) du =

Z

u4 − u2 du

1 1 1 1 = u5 − u3 + C = cos5 x − cos3 x + C 5 3 5 3 Z • Expressing given integral as (1 − cos2 x) cos2 x sin x dx: 3 pts. Z • Applying appropriate substitution (1 − u2 )u2 (−1) du: 5 pts. 1 1 • Evaluating antiderivative u5 − u3 + C: 6 pts. 5 3 1 1 • Substituting u = cos x and getting cos5 x − cos3 x + C: 7 pts. 5 3 (7) (6 pts) Evaluate the integral Z

Z b

2

2

1 √ dx. x 0 Z 2 Z 2 1 1 √ dx = lim √ dx + b→0 x x 0 b Z 2 h 1 i2 √ √ 1 1 1 1 √ dx = x− 2 dx = 2x 2 = 2 · 2 2 − 2 · b 2 = 2 2 − 2 b x b b √ √ √ lim+ 2 2 − 2 b = 2 2 b→0

2 1 √ dx: 2 pts. • Writing the definition of given improper integral lim+ b→0 x b Z 2 √ √ 1 √ dx and obtaining 2 2 − 2 b: 4 pts. • Evaluating the ordinary integral x b √ • Taking limit and getting 2 2: 6 pts. • If one computes the integral without using limit: 2 pts.

Z



5

1 dx, (x > 1). x2 − 1 If you need, use the derivative table below. Z 0 f (x) f (x) f (x) f (x) dx x2

√

sec x sec x tan x csc x − csc x cot x cot x − csc2 x

sec x ln | sec x + tan x| + C csc x − ln | csc x + cot x| + C cot x ln | sin x| + C

Derivative table for some trigonometric functions √ 2 x = sec t ⇒ Zx − 1 = tan t, dx = sec t tanZtdt Z 1 1 1 √ dx = sec t tan t dt = dt 2 sec t tan t sec t x2 x2 − 1 Z = cos t dt = sin t + C x = sec t ⇒ t = sec−1 x ⇒ sin t + C = sin(sec−1 x) + C Another solution: √ sin t/ cos t x2 − 1 tan t sin t + C = +C = +C = +C 1/ cos t sec t x Z • Finding appropriate trigonometric substitution x = sec t and getting

1 dt: sec t

3 pts. • Simplifying given integrand by using the definition of secant and obtaining Z cos t dt: 5 pts. • Finding antiderivative sin t + C: 6 pts. • Getting sin(sec−1 x) + C from x = sec t: 8 pts.



3x2 + 3x + 1 dx. x(x + 1)2

B C A 3x2 + 3x + 1 + + = x(x + 1)2 x x + 1 (x + 1)2 B C A A(x + 1)2 Bx(x + 1) Cx + + = + + 2 2 2 x x + 1 (x + 1) x(x + 1) x(x + 1) x(x + 1)2 A(x2 + 2x + 1) + B(x2 + x) + Cx (A + B)x2 + (2A + B + C)x + A = = x(x + 1)2 x(x + 1)2 ⇒ A + B = 3, 2A + B + C = 3, A = 1 ⇒ A = 1, B = 2, C = −1 3x2 + 3x + 1 1 2 1 = + − 2 x(x + 1) x x + 1 (x + 1)2 Z Z 2 1 3x2 + 3x + 1 1 dx = + − dx x(x + 1)2 x x + 1 (x + 1)2 1 = ln |x| + 2 ln |x + 1| + +C x+1 A B C Setting up a general form of partial fraction + + : 3 pts. x x + 1 (x + 1)2 Finding three relations A + B = 3, 2A + B + C = 3, A = 1: 5 pts. 1 2 1 Solving linear equations and obtaining + − : 6 pts. x x + 1 (x + 1)2 1 Evaluating correct antiderivative ln |x| + 2 ln |x + 1| + + C: 8 pts. x+1 ⇒

• • • •

6


7

(10) (7 pts) Determine whether the following integral Z ∞ 1 dx x e + e−x 0 converges or not.

•

• • • •

1 1 ≤ x e−x > 0 ⇒ ex + e−x ≥ ex ⇒ x −x e +e e Z ∞ Z ∞ 1 1 ⇒ dx ≤ dx x −x e +e ex 0 0 Z b Z ∞ 1 1 dx = lim dx b→∞ 0 ex ex 0 Z b Z b 1 1 −x −x b −b 0 dx = e dx = [−e ] = −e + e = 1 − 0 x eb 0 e 0 Z b 1 1 dx = lim 1 − =1 lim b→∞ b→∞ 0 ex eb Z ∞ Z ∞ 1 1 dx converges ⇒ dx converges x x e e + e−x 0 0 1 Finding a function x with convergent integral and making an inequality e 1 1 ≤ x : 3 pts. If there is no justification of convergence, you can’t ex + e−x e get the point. Z b Z ∞ 1 1 dx as a limit lim dx: 4 pts. Expressing the improper integral x b→∞ 0 ex e 0 1 Evaluating the integral and getting 1 − b : 5 pts. e Taking limit and obtaining 1: 6 pts. Concluding appropriate result: 7 pts.