This is a 50 in 50 Challenge: 50 questions in 50 minutes

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THIS IS A 60 IN 60 CHALLENGE:
60 QUESTIONS IN 60 MINUTES
•
Some are designated as Calc active.
•
The expectation is that you’ll be able to
get a handle on the solution within a
minute, not actually do the question.
•
The 60 Focused Questions should create
a wealth of other questions. Some of the
questions are answered in the next slide.
#1
4

2
x 1
x
dx
#1 SOLUTION
4

2
x 1
x
4
dx 

4
1 dx 
2
1
x
dx
2
 ( 4  2 )  (ln 4  ln 2 )
 2  ln( 4 / 2 )
 2  ln 2
#2

Find dy/dx for:
3
x  2 xy  y
2
 y
#2 SOLUTION
3
x  2 xy  y
3x
dy
dx
dy
dx
2
2
 y
dy
dy
dy


2 x
 y  2y



dx
dx
dx


2 y  2 x  1
 2 y  3x

2
2 y  3x
2
2 y  2x 1
#3
x  2x  x
3
lim
x   0
2

2x
Doesn’t this allow you to consider all limit

questions
a) at a point or b) at the infinities as
well as any Indeterminate forms (of which this
is one)
#3 SOLUTION

Use L’Hopital – answer is 1/2
#4 LET’S TRY IMPLICIT DIFF. AGAIN !!!
y x  3y  2 x
2
2
The above is non-functional. Could you find
dy/dx ? dx/dy? The value(s) of dy/dx @ y=1?
 of any horizontal/vertical tangents, if
The name
they exist?
#4 SOLUTION

dy/dx=(4x – y2)/(2xy – 3)

dy/dx @ y = 1 would require finding the value(s)
for x. When y = 1, then x = 3/2 or -1. Note
that for (3/2, 1) the value of dy/dx is
undefined (vertical tangency) and for (-1, 1)
the value of dy/dx is 1.
#5
4 

dx
d

x
There is no mystery to the question and the
key to its solution is to use the Change of Base
Formula. Never forget that all derivatives can
be checked in some way or another.
#5 SOLUTION
x
4 (ln 4 )
#6 (C ACTIVE)
Euler and the linear approximation model go
hand in hand…….
 Find the approximated value of f(4) for
f(x) = ln(2x+1), starting @ x = 1 w/ a step
size of 1.5

#6 SOLUTION









f(x + dx)
≈ f(x) + f’ (x) dx
f(1 + 1.5) ≈ f(1) + f’ (1) (1.5)
f(1 + 1.5) ≈ f(1) + f’ (1) (1.5)
f(2.5)
≈ ln(3) + (2/3)(1.5)
f(2.5)
≈ ln(3) + 1
f(2.5 + 1.5) ≈ f(2.5) + f ‘ (2.5)(1.5)
f( 4 )
≈ [ ln 3 + 1] + (2/6)(1.5)
f( 4 )
≈ [ln 3 + 1] + ½
f( 4 )
≈ ln 3 + 1.5
#7

Given a graph of the derivative of a function
and an initial condition, then sketch the graph
of the function…………
#7 SOLUTION
f ‘ (x)
f(x)
#8 (C ACTIVE)

Calculate the area of the region formed by the
function f ( x )  e sin x  3 cos x , the x-axis, the y-axis,
and the line x = 2π.

#8 SOLUTION
2
Y
0
1
dx  13 .009
#9
x

The function g(x) =

3
Find g(4)
 Find g ‘ 
(4)

2t  3
t
2
 3t  5
dt
#9 SOLUTION


g(4) = ln 9 – ln 5 by u-substitution
g ‘ (4) =

2 4  3
4  3 4  5
2

5
9
#10
x

If f(x)= 
0
1
t 2
3

A)
B)

C)
D)

E) f (-1) > 0



dt
then which of the following is FALSE ?
f(0) = 0
f is continuous at x for all x ≥ 0.
f(1) > 0
1
f ‘ (1) =
3

#10 SOLUTON

Choice (E)
1
x
f (x ) 

0
1
dt , f (1) 
t 2
3
f (1)  0. So ( E ) is false

0
0
1
t 2
3
dt  f (x )  

1
1
t 2
3
dt  0
#11
8

Find
lim
h   0

8
1
 1 
8   h   8  
2
 2 
h

A question
that demands your understanding of the definition
of the derivative. Of course, L’Hopital’s Rule would also be
appropriate !
#11 SOLUTION
f(x) = 8x8
 This limit is the definition of the derivative of
the above f(x) function evaluated @ x = ½
7
 The answer is f ‘(1/2) = 64 (1/2) = 1/2

#12
If h (x )  f (x )  g (x ), f ' (x )  g(x ), & g ' (x )  f (x ),
2
2
then h ' (x )  ______________________________

#12 SOLUTION
If
 Then

h ( x )  f ( x )  g ( x ),
2
2
f ' ( x )   g( x ), & g ' ( x )  f ( x ),
h'(x)  2 f (x)  f '(x)  2 g(x)  g'(x)
h ' ( x )  2 g ' ( x )   g ( x )  2 g ( x )  g ' ( x )
h'(x )  4 g(x )  g'(x )
h'(x )  4 f (x )  g(x )
OR
#13

Let’s see what you learned from the last question………
If
d
dx
then
 f (x )  g(x ) &
d
dx
d
dx
g(x )  f ( x ),
2
 f (x ) g(x )  ______________________________
#13 SOLUTION
If
d
dx
then
then
 f ( x )  g( x ) &
d
dx
d
dx
d
dx
g( x )  f ( x ),
2
 f ( x )  g( x )  f ( x ) 
d
dx
 g( x )  g( x ) f ' ( x )

 f ( x )  g( x )  f ( x )  f ( x )
2

 f ' (x) f (x )
2
 f (x ) f (x )  f (x )
2
2

#14 (C ACTIVE)

What is the volume of the solid formed when the
region bounded by y = ex, y = ln(1/x), x = .1, &
y = 0 is rotated about the x-axis?
Sketch the region and store the intersection point. This will require
two integrations. Be efficient with your use of FnInt.
#14 SOLUTION
.26987

.1
  e
x
 dx
2
1

   ln( 1 / x ) dx
2
 1.688
.26987
NOTE: ln(1/x) is easier to “see” and work with by simplifying
to ln (x-1) which is equivalent to - ln(x)
NOTE: A boundary line of x = .1 is necessary because if the
y-axis is used (w/out the x-axis boundary), then an
improper integral results !!
#15

This looks harmless………
 /2

Find

 /4

cos x
sin x
dx
#15 SOLUTION

(ln2)/2
#16

What is the average value of the function
3t3 – t2 over the interval -1 ≤ t ≤ 2 ?
#16 SOLUTION

2.75
#17

A region in the plane is bounded by the graph of y = 1/x,
the x-axis, the line x = m, and the line x = 2m, m > 0. The
area of this region

A) is independent of m
B) increases as m increases
C) decreases as m increases
D) decreases as m increases when m < ½;
increases as m increases when m > ½
E) increases as m increases when m < ½;
decreases as m increases when m > ½




#17 SOLUTION

A)
2m

m
1
x
dx  ln x | m  ln( 2 m )  ln( m )  ln 2
2m
#18

The 4th degree Taylor polynomial for f about
x = 1 is given by
T(x) = 5 - 10(x -1) + 15(x – 1)2 – 20(x – 1)3 + 25(x – 1)4

iv
Calculate f(1) + f ‘ (1) + f (1)
#18 SOLUTION
T(x) = 5 - 10(x -1) + 15(x – 1)2 – 20(x – 1)3 +
25(x – 1)4
 f(1) = 5
 f ‘ (1) = -10
 f ‘’ (1) = 30
 f ‘’’ (1) = -120
iv
 f (1) = 600
iv
 So, f(1) + f ‘ (1) + f (1) = 595

#19
Create the 3rd degree Maclaurin polynomial for
f(x) = ln(1 – x)
 Use this polynomial to approximate ln(3)

#19 SOLUTION

From previous experience (or memorization) we
know ln (1 + x) = x  x  x  x  ....(  1) x  ...
2
3
4
n
n 1
2


So, ln (1 – x) =

x 
x
3
2
2

x
4
3
3

x
4
 NOTE:
 ....(  1)
4
And ln (1 – (-2) ) = ln (3) ≈

n
x
n
 ...
n
 ( 2 ) 
( 2 )
2
2

( 2 )
3
3

8
3
ln 3 ≈1.0986,
so the 3rd degree polynomial is

not a great approximation

#20

Which of the following series converge?

I.

n 0
2
n 1
2

II .

n1
2 1

n
3 1
n
III .

n 0
4
n
n
A) I only B) II only
C) III only
 D) II and III
E) I and II


#20 SOLUTION
Choice E
 Item I may be broken up into to summations; one
from 0 to 1 which is a finite series that must
converge and the other from 1 to ∞ which when
compared to the p-series with p = 2 must
converge.
 Item II closely resembles a geometric series with r
= 2/3. Convergence must occur
 Item III is a p-series with p = ¾ < 1, so it diverges.

#21

Use the definition of the derivative
f ' ( x )  lim
h   0
f (x  h)  f (x)
h
to find the instantaneous
rate of change for f(x) = sin(2x)
#21 SOLUTION
f (x  h)  f (x)
f ' ( x )  lim
h
h   0
sin( 2 ( x  h ))  sin( 2 x )
f ' ( x )  lim
h
h   0
sin( 2 x  2 h )  sin( 2 x )
f ' ( x )  lim
h
h   0
sin( 2 x )  cos( 2 h )  cos( 2 x )  sin( 2 h )  sin( 2 x )
f ' ( x )  lim
h
h   0
sin( 2 x ) cos( 2 h ) 1
f ' ( x )  lim
h   0
h
f ' ( x )  lim sin( 2 x )  lim
h   0
f ' (x) 
f ' (x) 
f ( x )  sin( 2 x )
with
sin( 2 x ) 
2 cos( 2 x )
h   0
 lim
cos( 2 x )  sin( 2 h )
h
h   0
 cos( 2 h )  1
h
0
 lim cos( 2 x )  lim
h   0

h   0
cos( 2 x )  2
sin( 2 h )
h
#22

Which of the following is an equation of the
normal line to y = sin x + cos x at x = π ?
A) y = - x +π – 1
 C) y = x – π – 1
 E) y = x + π - 1

B) y = x – π + 1
D) y = x + π + 1
#23

Which of the following gives dy/dx for the
parametric curve y = 3 sin t, x = 2 cos t ?
A) -3/2 cot t
 C) -2/3 tan t
 E) tan t

B) 3/2 cot t
D) 2/3 tan t

Choice A
#24

Which of the following gives dy/dx if
-1
y = sin (2x) ?
A)
2
1 4 x
D)
B)
1 4 x
1 4 x
2
1
E)
2
1
C)
2
2x
1 4x
2
2
1 4 x
2
#24 SOLUTION


Choice C
Consider all the inverse trig derivatives, including tan -1 (f (x) ) and the
associated antiderivatives that are sure to come up such as:
6


36  4 x
1
6
2
dx (  each
1
 1 x

2
term
by 36 )
dx
/9
Let u 
2
x
2
  u
9
x
3
du  (1 / 3)dx
3 du  dx
AND
1
6
1
 1 x
2
dx
/9

1
6
3 du
 1 u
2

1
tan
1
u C
2

1
2
tan
1
x 
   C
3 
#25

A)
B)
C)
D)
E)
A slope field for the differential equation
dy/dx = 42 – y will show
A line with slope -1 and y-intercept of 42
A vertical asymptote at x = 42
A horizontal asymptote at y = 42
A family of parabolas opening downward
A family of parabolas opening to the left.
#25 SOLUTION
A family of exponential functions result that by
transformational geometry result in horizontal
asymptotes at y = 42, regardless of the
constant C that you use in the general solution
of the differential equation.
 So, choice C

#26
 If
5
5 a
3
3 a
 f ( x  a) dx  7, then
A) 7 + a
 D) a – 7


B) 7
E) -7
 f ( x ) dx 
C) 7 – a
#26 SOLUTION
If you shift the limits of integration a units to
the left (5 to 5 – a and 3 to 3 – a), then that
will compensate for the integrand’s shift from
f(x – a) to f(x) which also would have been a
units to the left…….CHOICE B
 A good numerical example might help:

5
 ( x  1)
3
4
dx

 (x)
2
dx
#27




The graph of y = f(x) conforms to the slope field for the
differential equation dy/dx = 4x ln x, as shown below. Which of
the following could be f(x)?
A) 2x2(ln x)2 + 3
B) x3ln x + 3
C) 2x2ln x – x2 + 3
D) (2x2 + 3)ln x – 1
E) 2x(ln x)2 – (4/3)(ln x)3 + 3
[0,2] by [0,5]
#27 SOLUTION
Choice C done through Integration by Parts
 Obviously, this is tied to the initial condition
which is imbedded within the window of [0,2]
by [0,5]
 You need the general solution after IBP and
“window” observation to see this solution !!!

#28 A GOOD SUBSTITUTION PROBLEM

If the substitution
1/2
integrand of  x
1 x
0
is made in the
, the resulting integral is
x  sin y
dx

1/2
A)

 sin
1/2
2
y
dy
B)
0
0
 /4
C )
2  sin
y
dy
0
D)
y
dy
cos y
 sin
0
 /6
2  sin
2
 /4
2
0
E)
2 
sin
2
y
dy
2
y
dy
#28 SOLUTION (THIS “TRIG SUB” IS BEYOND WHAT YOU
NEED TO STUDY, BUT AN INTERESTING BY PRODUCT OF
“U-SUB” WORK ! )
1/2

0
x
1 x
dx  is to be converted to a “y” integral

y  sin
dy


dx=

0
1
dx
x
x
and
1

 
1
1 x
 /4
x

1
1

dy 
1/2
x  sin y
Let
2
dx

2
1
2
x
, so
x
 2
x
 2 sin y  sin y  dy
0
x  dy 
dx
1 x
 2 sin y  dy
#29 (C ACTIVE)

Let v(t) be the velocity, in feet per second, of a
skydiver at time t seconds, t ≥ 0. After her
parachute opens, her velocity satisfies the
differential equation dv/dt = -2(v + 17), with
the initial condition v(0) = -47.

It is safe to land when her speed is 20 feet per
second. At what time t does she reach this
speed?
#29 SOLUTION

1.15
#30 (C ACTIVE)

Pollution is being removed from a lake at a rate
-0.5t
modeled by the function y = 20e
tons/yr,
where t is the number of years since 1995.
Estimate the amount of pollution removed from
the lake between 1995 and 2005. Round your
answer to the nearest ton.
A) 40
 D) 61

B) 47
E) 71
C) 56
#30 SOLUTION

Choice A) 40
#31

Let f and g be the functions given by f(x) = ex
and g(x) = 1/x. Which of the following gives the
area of the region enclosed by the graphs of f
and g between x = 1 and x = 2?
A) e2 – e – ln 2
 C) e2 – ½
 E) 1/e – ln 2

B) ln 2 - e2 + e
D) e2 - e – ½
#31 SOLUTION
A simple setup once you know that x = 1 is the
right of the point of intersection of the two
curves
 CHOICE A

#32 (C ACTIVE)

Let R be the region enclosed by the graphs of
y = e –x, y = ex , and the line x = 1. Which of the
following gives the volume of the solid
generated when R is revolved about the x-axis?
1
x
A )  (e  e ) dx
x
0
B )  (e  e
2x
2 x
1
x
  (e  e
2x
0
2 x
0
1
) dx
x 2
E )   (e  e )
x
0
x 2
C )  (e  e )
) dx
0
1
D)
1
dx
dx
#32 SOLUTION
This question, while calculator active, is
calculator neutral.
 CHOICE D

#33 (C ACTIVE)

The base of a solid S is the region enclosed by
the graph of y = ln x, the line x = e, and the
x-axis. If the cross sections of S perpendicular to
the x-axis are squares, which of the following
gives the best approximation of the volume of
S?
A) 0.718
 D) 3.171

B) 1.718
E) 7.338
C) 2.718
#33 SOLUTION
e
 ln x  dx
2
1
CHOICE A
#34 (C ACTIVE)

A developing country consumes oil at a rate
given by r(t) = 20e 0.2t million barrels per year,
where t is the time measured in years, for
0 ≤ t ≤ 10. Which of the following expressions
gives the amount of oil consumed by the
country during the time interval 0 ≤ t ≤ 10.
10
A ) r (t )
B ) r (10 )  r (0)
C )  r ' ( t ) dt
0
10
D )  r ( t ) dt
0
E ) 10  r (10 )
#34 SOLUTION

CHOICE D
#35 (C ACTIVE)

Which of the following gives the length of the
path described by the parametric equations
4
x = (1/4)t and y = t3, where 0 ≤ t ≤ 2 ?
2
A )  t  9t
6
2
4
B )  t  1 dt
6
dt
0
0
2
D )  t  9t
6
0
dt
E)
C )  1  9t
0
2
4
2
3
2
 t  3t dt
0
4
dt
#35 SOLUTION

CHOICE D – Length is

dx  dy 
    
 dt   dt 
2
2
dt
#36 (BC)

Which of the following is the limit of the
3  
a

n
sin

 ?
sequence with nth term
 n 
n
A) 1
B) 
C ) 2

D) 3 
E) 4 

#36 SOLUTION
lim
an 
n 
lim
n 

3  
n sin 

 n 
lim
n 
3  
sin 

 n 
( apply
1
L ' Hopital
b / c this is 0 / 0 )
n
3 

lim
n
3  

cos


2
 n 
n

1
1
n
2
CHOICE D
lim
n 
3  
3   cos 

 n 
1
 3
#37

Which of the following gives the value of
x
 sin t dt
lim
x   2

A)
 sin 2

4
B)
sin 2
4
2
x 4
2
C) 
sin 2
2
?
D)
sin 2
2
E ) DNE
#37 SOLUTION

Choice B; This is a L’Hopital Rule example
#38

Which of the following gives the area under the
curve y = 1/(x2 + 1) in the 1st quadrant ?
A)

4
B) 1
C)

2
D) 
E ) diverges
#38 SOLUTION

Choice C
#39


Which of the following is the coefficient of x4 in
the Maclaurin series generated by cos(3x) ?
A) 27/8
D) 0
B) 9
E) -27/8
C) 1/24
#39 SOLUTION

Choice A
#40

Which of the following is the fourth order
Taylor polynomial generated by f(x) at x =
π/2 ?
A ) ( x   / 2) 
(x   / 2)
2

(x   / 2)
2!
B)
1
(x   / 2)
2

2!
C)
1
(x   / 2)
4!
( x   / 2)
4
4!
2

2!
(x   / 2)
4
4!
D)
1  (x   / 2)
2
 (x   / 2)
4
E)
1  (x   / 2)
2
 ( x   / 2)
4
4
#40 SOLUTION

Choice C
#41

Which of the following is the sum of the series


n 0

A)
D)
e
e

e
e
2n
?

e
C)

e
2
2
e 
2
B)

n
E ) The series
diverges .
#41 SOLUTION

The sum of an infinite series will only be asked if the
series is geometric or directly related to one of the
basic series. . In this case the ratio is easy to see
n
n
>>>>>




e

2n
  2 
e 
Since r = π/e2 and r < 1 , then


lim
n 
Sn 
a
1 r
 e
0
( 20 )
1

e
2

1
1

e
2

e
2
e 
2
;
choice
D
#42 (C ACTIVE)

Which of the following is equal to the area of
the region inside the curve r = 2 cos θ and
outside the curve r = cos θ ?
 /2
A ) 3  cos  d 
2

B ) 3  cos  d 
2
0
0
 /2

D ) 3  cos  d 
0
E ) 3  cos  d 
0
C)
3  /2
2
 cos  d 
0
2
#42 SOLUTION

Choice A
#43


For -1 < x < 1 if f(x) = 
n1 2n1
(1) x
2n  1
n1



A )  (1)
n1
x

C )  (1) x
2n
2n
n1
D )  (1) x
n
n1
n1

n1
x
2n 2


E )  (1)
B )  (1) x
n
2n 2
n1
n1
, then f ' ( x) 
2n
2n
#43 SOLUTION

Choice A
#44

If n is a positive integer, then
2
2

1 1  2 
lim       .... 
n n 
n  n 


1
A) 
0
1
x
2
1
dx
0
3
D )  x dx
2
0
B) 3 
2 
3n 
   can be exp ressed as
 n  
3
1
x
2
dx
0
3
E ) 3  x dx
2
0
C) 
1
x
2
dx
#44 SOLUTION

Choice D

#45

For x > 0,

A)
1
x
3
C
B)
x

1
8
x
 
ln x
D)


1

x

2
2
C
E)
4

du
 dx 
u 

2

x
ln x 
2
2
C
2
C
C ) ln (ln x )  C
#45 SOLUTION

A u-substitution example………




1

 x

1
du 

u 
1 
 ln u

x 

du 
u
dx
dx
u
2
2
 C

1

2
2
x

dx
x
1 
 ln x

x 

ln

x
 C 
#46

Which of the following is the slope of the
inverse of y = ln (x – 3) at x = 2 ?

A) ½
B) 1/e2

D) e2 – 3
E) e2
C) -1
#46 SOLUTION
If
-1
f
(x):
y = ln (x – 3) , then
x = ln (y – 3)
y = ex + 3
x
 Therefore dy/dx = e
2
 and @ x = 2 dy/dx = e
choice E


You’ll have more difficult inverse derivative
questions than this one. A strong suggestion is to
graph, when possible, both the given function and
its inverse.
#47

A)
B)
C)
D)
E)
Which of the following gives y “ for
y = cos x + tan x ?
– cos x + 2 sec 2 x tan x
cos x + 2 sec 2 x tan x
- sin x + sec 2 x
- cos x + sec 2 x tan x
cos x + sec 2 x tan x
#47 SOLUTION

Choice A
#48

If f is a function such that f ‘ (x) exists for all x
and f(x) > 0 for all x, which of the following is
NOT necessarily true?
1
A)

f ( x ) dx  0
1
1
B)

1

2 f ( x ) dx  2
1
1
1
C)

1
f ( x ) dx  2

1

1
f ( x ) dx  
1

1

f ( x ) dx
1
1
E)
f ( x ) dx
0
1
D)
f ( x ) dx
0
f ( x ) dx


1
1
f ( x ) dx


0
f ( x ) dx
#48 SOLUTION

Choice C
#49

What is the volume of the solid generated by
rotating about the x-axis the region enclosed by
the curve y = sec x and the lines x = 0, y = 0,
and x = π/3 ?
A)

B) 
C)  3
3
D)
8
3
1

E )  ln   3 
2

#49 SOLUTION

Choice C
#50 (FINALLY !!!) (C ACTIVE)

A rectangle is inscribed between the parabolas y = 4x2
and y = 30 – x2 as shown.
[-3,3] by [-2, 40]

What is the maximum area of such a rectangle ?
A ) 20
D ) 50
2
B ) 40
E ) 40
C ) 30
2
2

#50 SOLUTION
A(x) = 2x(30 – x2 -4x2 )
 A ‘ (x) = 60 – 30x2
 A ‘ (x) = 0 @ x = 2
 A “( 2 ) < 0, so a maximum occurs
 Maximum
Area is Choice E


#51

If f is a continuous function for all x, the
maximum # of horizontal asymptotes that f can
have is
A) 0
 D) 3

B) 1
C) 2
E) there is no maximum #
51 SOLUTION

Choice C; there can be a horizontal asymptote
at -∞ and a different one at +∞
#52
lim
x    / 4
A) 0
D) 10
tan( 5 x )  1
x   /4
B) 2
E) d.n.e.

C) 5/2
#52 SOLUTION

Choice C
#53

If f is continuous on [2,5] and differentiable on
(2,5) with f(2) = -4 and f(5) =14 which is true?
 I.
f(x) = 6 has a solution in (2, 5)
 II. f ‘ (x) = 6 has a solution in (2, 5)
 III. f ‘’ (x) = 6 has a solution in (2, 5)
 A)
I only
B) II only C) I & II only
 D) I & III only E) I, II, & III
#53 SOLUTION
Both the Intermediate Value Theorem and the
Mean Value Theorem are used in the problem
 Choice C

#54
2

If w = 2x, then 
f (2 x ) dx 
0
2

A)
 f (w) dw

0
4
D)
 f (w) dw
0
B)
1
2
f (w) dw

2
0
1
E ) 2  f '(w) dw
0
C)
1
4
f (w) dw

2
0
#54 SOLUTION

Choice C
#55


If g(x) =
g ‘ (3).
2x

f (t)dt
0
use the table below to compute
X 
0
3
6
f(x)
1
5
7
f ‘ (x)
9
11
-4
A) -4
B) 5
C) 10
D) 11
E) 14
#55 SOLUTION

Choice E
#56

Let F(x) =

A) -3/4
 D) ½

x

1
1
t 1
3
dt .
Then F “ (1) =
B) -1/4
E) 3/4
C) 0
#57

If
3

2
f ( x )dx  6 &
0

f ( x )dx  4
0
2
then

f ( x )dx 
3


A) -10
D) 2
B) -2
E) 10
C) -1
#57 SOLUTION

Choice B
#58

Let g(x) =
x
2

t  64
3
dt
0

A) What
is the domain of g ?
 B) Find the interval(s) on which g is increasing.
 C) What is g “ (0) ?

#58 SOLUTION




A) The domain is must include all values of x such
that the antiderivative exists. Therefore this will
include all reals. [NOTE: If the upper limit had been
just x, then the domain is restricted to be x ≥ -4]
3
2
B) g ' ( x )  2 x x   64 > 0 for x > 0.
Therefore, g(x) increases for x > 0
C) g” (0) = 0 ; NOTE: Without actually find the second
derivative you should sense that each factor of g ‘(x) will
include a factor of x in the 2nd derivative.
#59

A solid has base given by the triangle with vertices (-4,0), (0, 8),
and (4, 0). Cross sections perpendicular to the y-axis are semicircles with diameter in the plane. The volume of the solid is
given by
4
A)
  8  2 x  dx
2
0
8
C)

 8 8  y  dy
0
2
0


2
B)
2
x

8
 dx 
  
2 4
8
D)

 4 8  y  dy
0
2
4

0

8  2 x  dx 

2
8
E)

 2 8  y  dy
0
2
#59 SOLUTION
Choice D
 The lines forming the triangle are y = 2x + 8
and y = -2x + 8, and y = 0. Since the diameter
of the semicircles are perpendicular to the yaxis, then you need to represent the distance in
terms of y………each diameter is y2 8  y 2 8   8  y

8

V 

0

2
8  y 
 
 dy
 2 
…….remember radius is squared, not diameter !!!!

#60
Circle City has a population density of
p(r) = 4  r for 0  r  3, where r is the distance in
miles from the center of the city, and p (r) is in
thousands of people/square mile. Which of the

following gives the total population of Circle City?

2
3
A)
 2 r
3
4  r dr
2
B)
6
 2 r
0
4  r dr
3
0
D)
 2 r
3
2
4  r dr
E

3

3
3
2
C)
4  r dr
2

4  r dr
2
#60
p ( r ) times square miles will give total
population
 For a small amount of area (dA) we still multiply
by the population density function p ( r )
2
 Since A =  r , then dA = 2πr
 Consequently, Total Pop. = the accumulation
from 0 to 3 miles away from the center of the
city
 of p ( r ) times 2πr…………Choice A


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