(1%) for the left side of the equation

1091!D13-17í ®M2 ãTŒU–

1. (6%) Find f (x) so that∫

x 0

f (t)dt =∫

x³ sin x

e^−t2dt.

Solution:

By the Fundamental Theorem of Calculus Part 1 (FTC#1), we know that d

dx∫

x 0

f (t) dt = f (x) The right hand side of the given equation can be written as

∫

x³ sin x

e^−t2 dt =∫

x³ 0

e^−t2 dt −∫

sin x 0

e^−t2 dt By Chain Rule we can find the derivative of the right hand side

d dx[∫

x³ 0

e^−t2 dt −∫

sin x 0

e^−t2 dt] = e^−(x3)2(3x²) −e^{−(sin x)2}(cos x) Simplify to get our conclusion

f (x) = 3x²e^−x6 −e^{− sin2x}cos x

● Grading:

This problem is designed to see if the students know how to use Part 1 of the Fundamental Theorem of Calculus.

(1%) for the left side of the equation. (5%) for the right side of the equation.

List of expected mistakes:

ˆ d dx∫

x 0

f (t) dt = f (t) is -1%

ˆ d dx∫

x

0 f (t) dt = f (x) − f (0) is -1%

ˆ Missing negative sign is -1%

ˆ Didn’t apply Chain Rule is -4%

ˆ Taking the derivative of x³ and sin x but got it wrong, -1% each

ˆ e^−x2, e^−(3x2)2, and e^{−(cos x)2} are -2% each

2. Evaluate the following integrals.

(a) (5%) ∫ tan x sec⁴x dx.

(b) (8%) ∫

3x + 2

x(x²+2x + 2)dx.

(c) (7%) ∫

4 0

x³+1

√

16 − x² dx Solution:

(a) Solution 1:

∫ tan x sec⁴x dx =∫ tan x ⋅ sec²x sec²x dx =∫ tan x(tan x + 1) sec²x dx (1 pt)

u=tan x

ÔÔÔÔÔÔ

du=sec²xdx ∫ u(u²+1)du (2 pts for substitution)

= 1 4u⁴+

1

2u²+c = 1

4tan⁴x +1

2tan²x + c (2 pts)

Solution 2:

∫ tan x sec⁴x dx =∫ sec³x sec x tan x dx (1 pt)

u=sec x

ÔÔÔÔÔÔÔ

du=sec x tan xdx ∫ u³du = 1

4u⁴+c = 1

4sec⁴x + c (2 pts) (b) 3x + 2

x(x²+2x + 2) = a x+

bx + c

x²+2x + 2 (1 pt)

⇒ a = 1, b = −1, c = 1 3x + 2

x(x²+2x + 2) = 1 x+

−x + 1

x²+2x + 2 (2 pts)

∫

3x + 2

x(x²+2x + 2)dx =∫ 1 x+

−x + 1 x²+2x + 2dx

=ln ∣x∣ +∫

−x + 1

x²+2x + 2dx (1 pt for∫

1

x =ln ∣x∣)

u=x+1

ÔÔÔ

du=dx ln ∣x∣ +∫

−u + 2

u²+1du (2 pts for completing the square and substitution)

=ln ∣x∣ −1

2ln(u²+1) + 2 tan⁻¹u + c (1 pt for ∫

u u²+1du

=ln ∣x∣ −1

2ln(x²+2x + 2) + 2 tan⁻¹(x + 1) + c 1 pt for ∫ 1 u²+1du) (c)

∫

4 0

x³+1

√

16 − x²dx ^{x=4 sin θ,−}

π 2≤θ≤^π2

ÔÔÔÔÔÔÔÔ

dx=4 cos θdθ ∫

π 2

0

64 sin³θ + 1

4 cos θ 4 cos θ dθ

(4 pts. 1 pt for x = 4 sin θ. 2 pts for integrand and differentials.

1 pt for the upper and lower bound)

= ∫

π 2

0

64 sin³θ + 1 dθ

=64∫

π 2

0

sin²θ sin θ dθ + π

2 (1 pt for ∫

π 2

0

1 dθ)

u=cos θ

ÔÔÔÔ64∫

0 1

(1 − u²)(−du) +π 2

= 128

+

π (2 pts for

π

2 sin³θ dθ)

3. Determine whether the integral is convergent or divergent. Evaluate convergent integral(s).

(a) (5%) ∫

∞ 0

dx

x²+a², where a > 0 is a constant.

(b) (5%) ∫

∞ 0

dx x². (c) (5%) ∫

∞ 0

dx

x²−a², where a > 0 is a constant.

Solution:

(a)

∫

t 0

dx x²+a² =

1 a∫

t 0

1 1 + (^x_a)²

dx a

u=^xa

ÔÔÔÔ

du=¹_adx

1 a∫

t a

0

1

1 + u²du = 1

atan⁻¹( t

a) (3 pts)

∫

∞ 0

dx

x²+a² = lim

t→∞∫

t 0

dx

x²+a² =lim

t→∞

1

atan⁻¹(t a) = π

2a (2 pts)

Hence the improper integral converges and the value is π 2a. (b)

∫

∞ 0

1

x²dx =∫

1 0

dx x² + ∫

∞ 1

dx

x² (2 pts for decomposing it into two improper integrals)

∵ ∫

1 0

1

x²dx = lim

t→0⁺∫

1 t

1

x²dx = lim

t→0⁺( 1

t −1) = ∞ diverges

∴ ∫

∞ 0

1

x²dx diverges (2 pts for the divergence of ∫

1 0

dx

x². 1 pt for the conclusion) (c)

1

x²−a² → −∞ as x → a⁻ (1 pt)

∫

a 0

1

x²−a²dx = 1 2a∫

a 0

1 x − a−

1

x + adx = 1 2a lim

t→a⁻∫

t 0

1 x − a−

1

x + adx = 1 2a lim

t→a⁻ln ∣t − a

t + a∣ = −∞.

Hence ∫

a 0

1

x²−a²dx diverges. (3 pts) Therefore ∫

∞ 0

dx x²−a² = ∫

a 0

dx x²−a² + ∫

2a a

dx x²−a² + ∫

∞ 2a

dx

x²−a²diverges (1 pt)

4. Let S be the region enclosed by y = ln x, y = 1, and x = 4. Let R be the solid of the revolution by rotating S around the x-axis.

(a) (7%) Find the volume of R by the disc method.

(b) (7%) Find the volume of R by the cylindrical shell method.

Solution:

(a) We use the disc method. We first evaluate V =π˜ ∫

4 e

(ln x)²dx = π∫

4 e

ln xd(x ln x − x) (2% for the correct integral)

=π(x ln x − x) ln x∣⁴_e−π∫

4 e

(x ln x − x)(1 x)dx

=π(x ln x − x) ln x∣⁴_e−π[x ln x − x − x]∣⁴_e

=π(x(ln x)²−2x ln x + 2x)∣⁴_e (2 % for correct antiderivative)

=π(4(ln 4)²−8 ln 4 + 8) − π(e − 2e + 2e) = 16π((ln 2)²−ln 2) + π(8 − e). (2%) So the required volume is

V = ˜V − π(4 − e) = 16π((ln 2)²−ln 2) + 4π. (1%) Some people may do the following way:

∫ (ln x)²dx =x(ln x)²− ∫ xd(ln x)²

=x(ln x)²−2∫ x(ln x)(1 x)dx

=x(ln x)²−2(x ln x − x) = x(ln x)²−2x ln x + 2x.

(b) Now we use the cylindrical shell method. Here we write x = e^y and the domain of integration becomes y ∈ [1, ln 4]. So the volume is

V =2π∫

ln 4 1

y[4 − e^y]dy = 2π(2y²−ye^y+e^y)∣^{ln 4}₁

(3% for the correct integral, 2% for the correct antiderivative)

=2π[(2(ln 4)²−4 ln 4 + 4) − (2 − e + e)]

=16π((ln 2)²−ln 2) + 4π. (2%)

5. The consumer’s surplus CS is obtained from the demand curve p = D(q) = 1000

(0.2q + 1)³ and the current unit price p with the formula

CS(p) = [∫

q 0

D(x) dx] − pq = [∫

D⁻¹(p) 0

D(x) dx] − pD⁻¹(p) where q = D⁻¹(p) is the quantity demanded obtained from p = D(q).

(a) (4%) Compute D⁻¹(p).

(b) (6%) Find the consumer surplus for the demand function D(x) when p = 8.

(c) (5%) Compute CS^′(p).

(d) (3%) Explain why CS(p) is a decreasing function on its domain.

Solution:

(a) p = 1000

(0.2q + 1)³, (0.2q + 1)³= 1000

p , 0.2q = ³

√ 1000

p −1, q = 5³

√ 1000

p −5 D⁻¹(p) = 50p^−1/3−5

(b) When p = 8, q = 50(8^−1/3) −5 = 20 CS(8) =∫

20 0

1000

(0.2x + 1)³ dx − 8 ⋅ 20 = [−2500(0.2x + 1)⁻²]

20

0 −160 = −100 + 2500 − 160 = 2240

(c)

CS(p) =∫

50p^−1/3−5 0

1000

(0.2x + 1)³ dx − p (50p^−1/3−5) CS^′(p) = p (−50

3 p^−4/3) − (50p^−1/3−5) + p (−50 3 p^−4/3) CS^′(p) = − (50p^−1/3−5)

or

CS(p) =∫

50p^−1/3−5 0

1000

(0.2x + 1)³ dx − p (50p^−1/3−5)

= [−2500(0.2x + 1)⁻²]

50p^−1/3−5

0 −p (50p^−1/3−5) = −25p^2/3−p (50p^−1/3−5) CS^′(p) = (−50

3 p^−1/3) − (50p^−1/3−5) + p (−50 3 p^−4/3) CS^′(p) = − (50p^−1/3−5)

(d) From (c), CS is a decreasing functions when −(50p^−1/3−5) < 0, in other words, p < 1000.

Notice that when p = 1000. q = D⁻¹(1000) = 0. Therefore p ≤ 1000 is the domain and CS(p) is decreasing on its domain.

● Grading:

(a) Finding the inverse (4%). Algebra mistakes are -1% each. Any mistake here should change their answer in (b) and (c).

(b) Find corresponding q (2%) and evaluate the integral (4%). Full credit if they got (a) wrong but found correct q and integrated correctly using their answer. Algebra mistakes -1% each, integral mistakes -2% each.

(c) They can use FTC or just integrate first, then derivative. (3%) for the derivative of the integral part. (2%) for the derivative of pq. Algebra mistakes -1% each, other bigger mistakes -2% each.

(d) They can explain using the answer from (c) or just explain with words and/or graph (3%).

They don’t need to find the domain. Gaps in argument is -1% and wrong argument is -2%.

Expected mistakes:

ˆ I expect a lot of people to have trouble understanding the problem.

ˆ I expect a lot of mistakes in (c)

ˆ I expect a lot of answers in (d) that don’t use (c)

ˆ When grading, award points for anything in the right direction

6. Solve the following differential equations.

(a) (7%) (1 + x²)y^′+2xy² =0, y(0) = 1.

(b) (8%) (1 + x²)y^′+xy = x(1 + x²), y(0) = 1.

Solution:

(a) We have

1 y²y^′=

−2x

1 + x², (2%) hence

1

y =ln(1 + x²) +C (2%), y = 1

ln(1 + x²) +C (1%).

Now we have

1 = y(0) = 1

ln 1 + C, C = 1, (2%) so

y = 1

ln(1 + x²) +1. (b) We have

y^′+ x

1 + x²y = x, so P (x) = x

1 + x², Q(x) = x, and

I(x) = e^{∫ P (x)dx}=e^12ln(1+x2)=

√

1 + x². (3%) Thus

y = 1

I(x)∫ I(x)Q(x)dx (1%)

= 1

√

1 + x²∫ x

√

1 + x²dx

= 1

3(1 + x²) + C

√

1 + x². (2%) Now

1 = y(0) = 1

3+C, C = 2

3. (2%) Hence

y = 1

3(1 + x²+ 2

√ 1 + x²

).

7. Consider the parametric curve (x(t), y(t)) = (1 − t², t − t³).

(a) (2%) The curve passes through the origin twice. Find t₁ and t₂ so that (x(t1), y(t1)) = (x(t2), y(t2)) = (0, 0).

(b) (4%) Find the tangent lines of the curve at (0, 0).

(c) (6%) (x(t), y(t)) = (1 − t², t − t³), t₁ ≤t ≤ t₂, forms a loop. Find the area enclosed by the loop.

Solution:

(a) t₁= −1 (1%), t₂ =1 (1%).

(b)

dy dx =

y^′(t) x^′(t) =

1 − 3t²

−2t , (2%)

so the tangent line is y = −x when t = −1 (1%), and y = x when t = 1 (1%). These are two tangent lines of the curve at (0, 0).

(c)

Area = ∣∫

1

−1 y(t)x^′(t)dt∣ (2%)

= ∣ ∫

1

−1(t − t³)(−2t)dt∣ (1%)

= ∣ −2∫

1

−1t²−t⁴dt∣

=2 (1 3t³−1

5t⁵) ∣

1

−1 (2%)

= 8

15. (1%)