Student (Print)
Section
Last, First Middle
Student (Sign)
Student ID
Instructor
MATH 152,
Fall 2007
Common Exam 2
Test Form A
Solutions
Instructions:
You may not use notes, books, calculator or computer.
For Dept use Only:
1-10
/50
11
/10
12
/10
after 90 minutes and may not be returned.
13
/10
Part II is work out. Show all your work. Partial credit will be given.
14
/10
15
/10
Part I is multiple choice. There is no partial credit.
Mark the Scantron with a #2 pencil. For your own records,
also circle your choices in this exam. Scantrons will be collected
THE AGGIE CODE OF HONOR:
An Aggie does not lie, cheat or steal, or tolerate those who do.
TOTAL
1
Part I: Multiple Choice (5 points each)
There is no partial credit.
1. Use the Trapezoid Rule with 3 intervals to approximate
∫ 1 x 2 dx.
7
a. 168
correct choice
b. 118
c. 114
d. 84
e. 59
Δx = 7 − 1 = 2
fx = x 2
3
T 3 = Δx 1 f1 + f3 + f5 + 1 f7
2
2
= 2 1 + 9 + 25 + 1 49
2
2
2. If you use the trapezoid rule with 47 intervals to estimate
a. too large because e x
3
is concave up.
x3
is concave up.
d. too small because e
x3
is increasing.
e. too small because e
x3
is decreasing.
c. too large because e
Since e x
3
∫0
94
3
e x dx your answer will be
is concave down.
x3
b. too small because e
= 118
correct choice
is concave up, the secant lines are all above the curve.
So the area under the trapezoids is larger than the area under the curve.
π/2
3. Compute
∫0
sec 2 θ dθ
a. 2
b. 1
c. 0
d. −1
e. diverges
π/2
∫0
correct choice
sec 2 θ dθ = tan θ
π/2
0
= lim
tan θ − tan 0 = ∞ − 0 = ∞
π −
x
2
2
4. Compute
∫ −1
4
1 dx
x2
a. 5
4
b. 3
4
c. 0
d. − 5
4
e. diverges
∫ −1
4
1 dx =
x2
correct choice
0
4
= lim− − 1x
x0
− − 1
−1
0
x −1
−1
∫ −1 x −2 dx + ∫ 0 x −2 dx =
+ −1
4
−1
+
− lim+ − 1x
x0
x −1
−1
4
0
0
= − 1x
= “ − 1−
0
−1
+ − 1x
4
0
” − 1 − 1 − “ − 1+
4
0
”
= ∞− 5 +∞ = ∞
4
5. The integral
3 + sinx
dx is
x
∞
∫π
∞
a. divergent by comparison with
∫π
b. divergent by comparison with
∫π
c. divergent by comparison with
∫π
∞
∞
2 dx.
x
3 dx.
x
4 dx.
x
∞
d. convergent by comparison with
∫π
e. convergent by comparison with
∫π
The integrals
∞
∫π
4 dx and
x
∞
∫π
∞
2 dx.
x
4 dx.
x
2 dx both diverge because
x
Since
3 + sinx
≥ 2x , we have
x
∫π
Since
3 + sinx
≤ 4x , we have
x
∫π
6. Find the integrating factor for
correct choice
∞
∫π
k dx = k ln x
x
∞
π
=∞
∞
3 + sinx
dx ≥ ∞ and so diverges.
x
∞
3 + sinx
dx ≤ ∞ and cannot conclude anything.
x
dy
− 4x y = x 2
dx
a. x 4
b. 4 ln x
c. x −4
correct choice
d. −4 ln x
e. − 1 x −3
3
P = − 4x
∫ P dx = −4 ln x
I = e −4 ln x = x −4
3
7. Which of the following diagrams is the direction field of the differential equation
d.
a.
dy
= x2y ?
dx
correct choice
e.
b.
The slope is m = x 2 y.
This is positive when y > 0,
negative when y < 0 and
c.
zero when y = 0. Only plot
d has positive slope in the
second quadrant.
8. Suppose the function y = f x is the solution of
f 1 = 2. Find f ′ 1.
dy
= x 2 − y 2 satisfying the initial condition
dx
a. −1
b. −2
c. −3
correct choice
d. 1
e. 2
As a solution, y = f x satisfies
df
= x 2 − fx 2 . So f ′ 1 = 1 2 − f1 2 = 1 2 − 2 2 = −3
dx
4
9. Find the length of the curve x = 2t 2 , and y = t 3 for 0 ≤ t ≤ 1.
a. 732
b. 108
c. 7
d. 61
correct choice
27
e. 1
3
As a t-integral, the differential of arc length is
2
dx
dt
ds =
2
dy
dt
+
2
dt =
4t 2 + 3t 2  dt =
16t 2 + 9t 4 dt = t 16 + 9t 2 dt
So the arc length is
L=
∫ ds = ∫ 0 t
1
L= 1
18
∫ 16
25
16 + 9t 2 dt
u = 16 + 9t 2
1 2u 3/2
18 3
u du =
25
16
du = 18t dt
= 1 125 − 64 = 61
27
27
10. The curve y = sin x for 0 ≤ x ≤ π is rotated about the x-axis.
Which formula gives the area of the resulting surface?
π
a.
∫ 0 2π cos x
1 + sin 2 x dx
b.
∫ 0 2π sin x
1 + cos 2 x dx
c.
∫ 0 π cos x
1 + sin 2 x dx
d.
∫ 0 π sin x
1 + cos 2 x dx
e.
∫ 0 π sin x dx
π
π
π
correct choice
π
As an x-integral, the differential of arc length is
ds =
1+
dy
dx
2
dx =
1 + cos 2 x dx
The radius of revolution is r = y = sin x. So the surface area is
A=
π
π
∫ 0 2πr ds = ∫ 0 2π sin x
1 + cos 2 x dx
5
Part II: Work Out (10 points each)
Show all your work. Partial credit will be given.
11. Compute
∞
∫1
2 ln x dx
x2
Integrate by parts:
∞
∫1
dv = 12 dx
x
v = − 1x
u = 2 ln x
du = 2x dx
2 ln x dx = − 2 ln x + ∫ 2 dx = − 2 ln x − 2
x
x
x
x2
x2
2
x + 2 l=′ H − lim x
= − x∞
lim 2 ln
+2 = 2
x
x∞
1
∞
1
x
2
= x∞
lim − 2 ln
x − x
− − 2 ln 1 − 2
1
1
12. Simpson’s Error Formula says:
The error E S in approximating
∫ a fx dx
b
with n intervals satisfies
Kb − a 5
where K ≥ max f 4 x for a ≤ x ≤ b.
180n 4
4
Use this formula to estimate the error in approximating ∫ 1x dx using n = 30 intervals.
|E S | ≤
1
b−a = 4−1 = 3
f = 1x = x −1
f ′ = −x −2
f ′′ = 2x −3
f ′′′ = −6x −4
f 4 = 24x −5
The largest value of 24x −5 occurs when x = 1. So the smallest K is K = 24.
Kb − a 5
243 5
8
4
=
= 24 ⋅ 3 4 =
=
= 4 × 10 −5
180n 4
18030 4
18010
2010 4
1010 4
|E S | <
13. Find the centroid of the region in the first quadrant bounded by y =
A=
∫ 0 fx dx = ∫ 0
My =
4
4
x dx =
∫ 0 x fx dx = ∫ 0 x
4
4
2x 3/2
3
x dx =
4
0
3/2
= 2⋅4
= 16
3
3
4
2x 5/2
5
0
5/2
= 2⋅4
= 64
5
5
y
My
x̄ =
= 3 64 = 12
5
A
16 5
Mx =
∫0
4
1 fx 2 dx =
2
∫0
4
x , the x-axis and x = 4.
2
1
1 x dx =
2
x2
4
4
0
=4
0
0
2
4
x
ȳ = M x = 3 4 = 3
A
16
4
x, y =
12 , 3
5 4
6
dy
xy
= 2
dx
x +9
14. Consider the differential equation:
a. Find the general solution:
METHOD 1:
The equation is separable.
Separate and integrate.
ln|y| = 1 lnx 2 + 9 + C = ln x 2 + 9 + C
∫ dyy = ∫ x 2 x+ 9 dx
2
Solve for y:
|y| = e C x 2 + 9
METHOD 2:
Px = −
y = A x2 + 9
where A = ±e C
The equation is linear:
x
x2 + 9
∫ Px dx = − ∫
I=e
Find the integrating factor.
x dx = − 1 lnx 2 + 9
2
x2 + 9
∫ Px dx = e − 12
Multiply the standard equation by It.
ln x 2 +9
1
y=A
x +9
2
= x 2 + 9 −1/2 =
1
x +9
2
dy
1
x
−
y=0
3/2
2
dx
x + 9
x +9
2
Recognize the left side as a derivative of a product.
Integrate and solve.
dy
− 2x y = 0
dx
x +9
Put in standard form.
d
dx
1
y
x +9
2
=0
y = A x2 + 9
b. Find the solution satisfying the initial condition:
y0 = 6
Substitute x = 0 and y = 6:
6 = A 0 + 9 = 3A
A=2
y = 2 x2 + 9
c. Find y4:
y4 = 2 4 2 + 9 = 10
7
15. A salt water fish tank contains 20 liters of water with 700 grams of salt. In order to reduce
the salt concentration, you pour in salt water with a concentration of 15 grams of salt per liter
at 2 liters per minute. You keep the tank well mixed and drain the mixture at 2 liters per
minute. Let St be the amount of salt (in grams) in the tank at time t (in minutes).
a. Write the differential equation and initial condition for St.
dS = 15 g ⋅ 2 l − S ⋅ 2 l
min
min
20
dt
l
dS = 30 − 1 S
10
dt
or
S0 = 700
b. Solve the initial value problem for St.
METHOD 1:
The equation is separable.
Separate and integrate.
− 10 ln 30 − 1 S = t + C
∫ dS1 = ∫ dt
10
30 −
S
10
Solve.
ln 30 − 1 S = − t − C
30 − 1 S = e −C/10 e −t/10
10
10
10
10
1 S = 30 − Ae −t/10
30 − 1 S = Ae −t/10
where A = ±e −C/10
10
10
S = 300 − 10Ae −t/10
Use the initial condition. At t = 0, we have S = 700
700 = 300 − 10A
10A = −400
Substitute back.
S = 300 + 400 e −t/10
METHOD 2:
The equation is linear:
A = −40
Put in standard form.
I = exp ∫ 1 dt
10
Find the integrating factor.
dS + 1 S = 30
10
dt
= e t/10
e t/10 dS + 1 e t/10 S = 30e t/10
10
dt
d e t/10 S = 30e t/10
Recognize the left side as a derivative of a product.
dt
Multiply the standard equation by It.
Integrate and solve.
e t/10 S =
∫ 30e t/10 dt = 300e t/10 + C
S = 300 + Ce −t/10
Use the initial condition. At t = 0, we have S = 700.
700 = 300 + C
Substitute back.
C = 400
S = 300 + 400e −t/10
c. After how many minutes will the amount of salt in the tank drop to 400 grams?
Set S = 400 and solve for t:
400 = 300 + 400e −t/10
t = −10 ln 1
4
400e −t/10 = 100
e −t/10 = 1
4
−t = ln 1
10
4
t = 10 ln 4
8