Math 101 A04 Midterm 2 Name:

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Midterm 2. Name: 1. (12 points) Compute the following antiderivatives. (a) (6 points) ∫ sin3 x dx. ∫ (sin x)(sin2 x) dx = ∫ sin x(1 − cos2 x) dx = ∫ sin x dx ...
Math 101 A04

Midterm 2

Name:

1. (12 points) Compute the following antiderivatives Z (a) (6 points) sin3 x dx Z

2

(sin x)(sin x) dx =

Z

2

sin x(1 − cos x) dx =

Z

sin x dx −

2

2

Z

cos2 x sin x dx

The second integral is done by a u-substitution with u = cos x. cos3 x + C. The final answer is − cos x + 3 Z (b) (6 points) tan2 x sec4 x dx Z

2

2

2

(tan x)(sec x)(sec x) dx =

Z

2

tan x(1 + tan x) sec x dx =

Z

(tan2 x + tan4 x) sec2 x dx

Now this integral can be done by u-substitution with u = tan x. tan3 x tan5 x The answer is + + C. 3 5 2. (10 points) Compute the following antiderivative: Z

2x3 + x2 + 1 dx (x − 1)2 (x2 + 1)

Since the degree of the top is less than the bottom, no long division is needed and the theorem on partial fraction says there is a decomposition of the form:

2x3 + x2 + 1 A B Cx + D = + + 2 2 2 2 (x − 1) (x + 1) x − 1 (x − 1) x +1 Multiplying both sides by the common denominator (x − 1)2 (x2 + 1) gives the equation (after collecting “like” terms):

2x3 + x2 + 1 = (A + C)x3 + (−A + B − 2C + D)x2 + (A + C − 2D)x − A + B + D Equating the coefficients of same power terms from each side gives the linear system:

2=A+C 1 = −A + B − 2C + D 0 = A + C − 2D 1 = −A + B + D

A simple way to solve this system is to use the first equation in the third to solve for D. Then use the D value in the fourth equation to solve for −A + B and use that in the second equation where −A + B (and also D) appears. That lets you solve for C and the rest is easy. The solution is A = 2, B = 2, C = 0, and D = 1.

So the antiderivative we want is the sum of the following: Z

2 dx + x−1

Z

2 dx + (x − 1)2

Z

  1 1 + tan−1 x + C dx = 2 ln(x − 1) − 2 2 x +1 x−1

3. (8 points) Compute the following integral: Z

−3

−4



9x2 − 16 dx x



u 2 − 42 du. u −12 Now we make a secant trig substitution, u = 4 sec θ, du = 4 sec θ tan θ dθ. Z −9 Z −9 p 2 Z −9 √ 4 (sec2 θ − 1) 2 | tan θ| tan θ dθ 4 sec θ tan θ dθ = 4 tan θ tan θ dθ = 4 4 sec θ u=−12 u=−12 u=−12 Now we need to figure out the sign of tan θ so we can get rid of the absolute value. u u Since the substitution is u = 4 sec θ (really sec−1 = θ), graph sec−1 and note that 4 4 for u ≤ −4, θ is between π/2 and π while for u ≥ 4, θ is between 0 and π/2. In our case −12 ≤ u ≤ −9, so θ is between π/2 and π. But for those values of θ, tan θ ≤ 0. So | tan θ| = − tan θ. Z −9 Z −9 2 So the integral is, 4 − tan θ dθ = 4 1 − sec2 θ dθ = 4 [θ − tan θ]−9 u=−12 First set u = 3x, du = 3 dx. Then the integral becomes

u=−12

Z

−9

u=−12

"

i−9 h √ u −1 u 2 + sec θ − 1 = 4 sec−1 + = 4 sec 4 4 u=−12 √ Notice that we again used − tan θ = sec2 θ − 1.

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r

u2 −1 16

#−9

−12

4. (16 points) For each improper integral, determine if it converges or diverges. If it converges, find the value. If it diverges, explain why: Z π/2 cos x √ dx (a) (4 points) sin x 0 π/2

cos x √ dx t→0 sin x t Using u = sin x, we have: Z π/2 Z π/2 √ 1 cos x √ dx = u−1/2 du = 2u1/2 u=sin t = 2 − 2 sin t sin x x=t t √ Now taking the limit, limt→0+ 2 − 2 sin t = 2, so the improper integral converges to 2.

= lim+

Z

Z



(b) (7 points) e−x cos x dx Z t 0 = lim e−x cos x dx t→∞

0

This integral can be done by two integration by parts after which we notice it repeats. u = e−x , du = −e−x dx and dv = cos x dx, v = sin x Z t Z t t −x −x e−x sin x dx e cos x dx = e sin x 0 + 0

0

u = e−x , du = −e−x dx and dv = sin x dx, v = − cos x Z t Z t t −x −x e sin x dx = e cos x 0 − e−x cos x dx 0

0

So putting this all together: Z t Z t t t −x −x −x e sin x dx = e sin x 0 + e cos x 0 − e−x cos x dx 0 0 R t −x So solving for 0 e sin x dx: Z t t t e−x sin x dx = 1/2( e−x sin x 0 + e−x cos x 0 ) = 1/2[e−t (sin t − cos t) + 1] 0 Z t lim e−x cos x dx = lim 1/2[e−t (sin t − cos t) + 1] = 1/2

t→∞

0

t→∞

So it converges to 1/2. To compute the limit, note that sin t−cos t is always between -2 and 2 so you can use the squeeze theorem by squeezing e−t (sin t − cos t) between 2e−t and −2e−t . Since e−t goes to 0, this means e−t (sin t − cos t) does so also.

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(c) (5 points)

Z

∞ 3

x2 e−x dx

−∞

= lim

t→∞

Z

t

2 −x3

xe

0

dx + lim

s→−∞

Z

0

3

x2 e−x dx x

These integrals can be done by u-substitution, u = −x3 . Z 0 Z t 1 u 1 u e du = lim − e du + lim s→−∞ s 3 t→∞ 0 3 t 0 1 −x3 1 −x3 = lim − e + lim − e t→∞ s→−∞ 3  s 0   3 1 1 −s3 1 −t3 1 + lim − + e = lim − e + s→−∞ t→∞ 3 3 3 3 = 1/3 + ∞ Diverges

5. (6 points) Solve the following initial value problem: dy = xy 3 (4x2 + 1) dx

We solve this by separation of variables: Z Z 1 dy = x(4x2 + 1) dx y3 Z Z 1 dy = 4x3 + x dx y3 x2 −1 4 = x + +C 2y 2 2 y2 =

−1 2x4 + x2 + 2C

12 =

−1 2 · 1 + 1 + 2C

3 + 2C = −1, which means C = −2. So y 2 =

2x4

−1 . + x2 − 4 Page 4

y(1) = 1

6. (7 points) A police medical examiner needs to to determine the time of death of a murder victim. She is told by police investigators the room temperature was 75◦ F. When the police first arrive at the crime scene, the body temperature was 92◦F. After one hour examining the crime scene, investigators took the body temperature again and found it was 90◦ F. When was the time of death? Use Newton’s Law of Cooling. [Assume normal body temperature is 98.6◦F. Set t = 0 for the time of police arrival.] Since room temperature is 75, Newton’s law of cooling for this situation is: du = −k(u − 75) dt Z Z 1 du = −k dt u − 75 ln |u − 75| = −kt + C1

Applying the exponential to both sides (and noticing u − 75 > 0) u − 75 = C2 e−kt where C2 = eC1 Now using t = 0, u = 92 92 − 75 = C2 17 = C2

Using t = 1, u = 90 90 − 75 = 17e−k

So k = − ln(15/17) and u = 17et ln(15/17) + 75.

To solve for the time of death, we know at that time, body temperature was 98.6. Plugging this into our solution for u: 98.6 = 17et ln(15/17) + 75 23.6/17 = et ln(15/17) t ln(15/17) = ln(23.6/17) t=

ln(23.6/17) ln(15/17)

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7. (10 points) Determine if the following sequences converges or diverges. Give a sufficient reason/explanation. If it converges, find the limit.   2 5n − 4n + 10 (a) (4 points) an = sin 2n2 + 5 2

First notice that limn→∞ 5n 2n−4n+10 = 5/2. This is because the top and bottom 2 +5 polynomials are of the same degree. For large n the higher degree terms of each 5n2 5n2 − 4n + 10 ≈ . (L’Hospital’s rule also polynomial dominate the others and 2n2 + 5 2n2 works)   5n2 − 4n + 10 = sin 5/2. Now since sin x is a continuous function, lim an = sin lim n→∞ n→∞ 2n2 + 5

(b) (6 points) an =

ln 3n ln 2n

∞ so we can use L’Hospital’s rule. It’s The limit of an is an indeterminate form ∞ easy enough to apply the rule directly, but often with logarithms we can make some simplifications first. ln 3n ln 3 + ln n an = = ln 2n ln 2 + ln n 1/n So lim an = lim = 1 (using L’Hospital’s rule). n→∞ n→∞ 1/n Notice that our simplification also made it intuitively clear that the ratio of the top to bottom would approach 1.

8. (14 points) Determine if the following sequences converges or diverges. Give a sufficient reason/explanation. If it converges, find the limit. ∞ X 5n − 1 (a) (7 points) 6n n=1

The series converges because it is the sum of two convergent series:

∞ X

(5/6)n and

n=1

∞ X n=1

−(1/6)n . These series converge because they are geometric with ratio 5/6 and

1/6 which are both less than 1.

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∞ X

(5/6)n =

n=1

∞ X

5/6 =5 1 − 5/6

−1/6 = −1/5 1 − 1/6 n=1  X ∞ ∞ ∞ ∞  X X 5n − 1 X 5n 1 n So = (5/6) + −(1/6)n = 5 + −1/5 = − n n n 6 6 6 n=1 n=1 n=1 n=1 −(1/6)n =

(b) (4 points)

∞ X n=1

n2 2n2 + e−n

Diverges by the n-th term test for divergence. n2 = 1/2 lim an = lim n→∞ n→∞ 2n2 + e−n To find the limit, you can use L’Hospital directly or squeeze an between

n2 and 2n2

n2 since e−n is between 0 and 1. (Intuitively, since e−n becomes small much 2n2 + 1 2 n2 quicker than any polynomial, 2n2n+e−n ≈ 2n 2 for very large n) ∞ X 3 (c) (3 points) n n=1

∞ X 1 . diverges, since it is a multiple of a divergent series, the harmonic series n n=1

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