Q. Fundamental Theorem of Calculus (FTC)

Transcription

Q. Fundamental Theorem of Calculus (FTC) / Accumulation Function
b
What you are finding: Students should certainly know that the FTC says that " f ( x ) dx = F (b) # F ( a)
a
b
where F ( x ) is an antiderivative of f ( x ) . This leads to the fact that F ( a) +
" f ( x ) dx . Students should be
a
prepared to use substitution methods to integrate and be able to change
! the limits of integration using this
b
! substitution. Students should
! also know that
"
a
a
f ( x ) dx = # " f ( x ) dx .
!b
x
The accumulation function looks like this: F ( x ) =
" f (t ) dt . It represents the accumulated area under the
0
curve f starting at zero (or some value)
! and going out to the value of x. The variable t is a dummy variable. It
is important to believe that this is a function of x.
!
"
92.
#e
cosx
sin x dx =
0
A.
1
e
B. e "
!
!
1
e
C. e "1
D. 1"
1
e
E. e
B. u = cos x,du = "sin x dx
x = 0,u = 1, x = #,u = "1
!
!
#
"1
1
cosx
u
"1
1
!
$ e sin x dx = " $ e du = "(e " e ) = e " e
0
1
!
3
93.
#
20 " 4 x dx is equivalent to
A.
"1
!
u du
#
4 "1 5
1
15
B.
"1 3
# u du
4 1
C.
1 16
" u du
4 8
16
16
D. 4 " u du
8
E.
"
u du
8
!
3
#
C.
20 " 4 x dx
!
u = 20 " 4 x,du = "4 dx
!
!
3
#
20 " 4 x dx =
1
x
94.
"
1
"1 8
1 16
u
du
=
#
# u du
4 16
4 8
ln t
dt =
t
x 2 "1 !
A.
2
x2
B.
2
!
!
!
!
x = 1,u = 16, x = 3,u = 8
1
C.
(ln x )
2
2
D.
(ln x )
2
"1
2
E. ln x
1
C. u = ln t,du = dt
t = 1,u = 0,t = x,u = ln x!
t
!
!
ln x
2
x
ln x
ln x )
ln t
u2
(
" t dt = " u du = 2 = 2
1
0
0
!
© 2011 www.mastermathmentor.com
!
- 41 -
Demystifying the MC AB Calculus Exam
95. The graph of the piecewise linear function f is shown in the
x
figure to right. If g( x ) =
" f (t ) dt , which of the following is the
1
greatest?
A. g("4 )
!
D. g( 4 )
B. g("3)
E. g(5)
C. g(1)
-3
!
!
! of the triangle
# f (t) dt : the area
!)=
B. g("3
!
1
4
g( 4 ) =
# f (t ) dt : the area of the triangle
=
1
96. (Calc) Let F ( x ) be an antiderivative of
!
1
("4)("3) = 6
2
=
A. -15.879
B. -11.629
!
4
"
E.
1
( 3)(2) = 3 all others will be less.
2
x 3 + x + 1 . If F (1) = "2.125, then F ( 4 ) =
C. 7.274
D. 15.879
!
!
x + x + 1 dx = F ( 4 ) # F (1) $ F ( 4 ) = F (1) +
3
1
E. 11.629
4
"
x 3 + x + 1 dx
1
4
F ( 4 ) = #2.125 +
"
x 3 + x + 1 dx = #2.125 + 13.754 = 11.629
1
2
97. If f is a continuous function and F "( x ) = f ( x ) for all real numbers x, then
!
1
1
A. 3F ("2) " 3F (2)
B. F ("2) " 3F (2)
3
3
1
1
!
D. 3F ("5) " 3F ( 7)
E. F ( 7) " F ("5)
!
3
3
!
! dx
E. u = 1" 3x,du = "3
!
"
x = "2,u = 7, x = 2,u = "5
C.
1
1
F (2) " 3F ("2)
3
3
!
7
1 2
1 "5
1
1
1
"3 f (1" 3x!) dx = " # f ( u) du = F ( u) "5 = F ( 7) " F ("5)
#
3 "2
37
3
3
3
4
6
98. If f is continuous for all real numbers x and " f ( x ) dx = 10, then
1
!
A. 37
B. 39
C. 35
D. 57
6
A.
# f (1" 3x ) dx =
"2
6
" [ f ( x # 2) + 2x ] dx =
3
E. 25
6
# [ f ( x " 2) + 2] dx = # !f ( x " 2) dx + # 2 dx
3
3
3
u = x " 2,du = dx x = 3,u = 1, x = 6,u = 4
6
#
4
f ( x " 2) dx =
3
!
#
1
6
f ( x ) dx = 10
# 2x dx = x
3
- 42 -
2 6
3
= 36 " 9 = 27 $ 10 + 27 = 37
R. Derivative of Accumulation Function (2nd FTC)
d x
What you are finding: You are looking at problems in the form of
" f (t) dt . This is asking for the rate
dx a
of change with respect to x of the accumulation function starting at some constant (which is irrelevant) and
ending at that variable x. It is important to understand that this expression is a function of x, not the variable
t. In fact, the variable t in this expression could be any variable (except x).
!
d x
nd
How to find it: You are using the 2 Fundamental Theorem of Calculus that says:
" f (t) dt = f ( x ).
dx a
Occasionally you may have to use the chain rule that says
d
dx
g ( x)
" f (t) dt = f (g( x )) # g$( x ) .
a
!
"2
99. Let f be the function given by f ( x ) =
# t sin2t dt for $ " % x % ". In what interval(s) is f ( x )
x
decreasing?
!
$ # #'
$ "# ' $ # '
B. &" , )
C. &"#, ) * & ,#)
!
% 2 2(
%
2 ( %2 (
!
E. ("#,# )
#
2
(
d%
+# #
B. f "( x ) = '' $ t sin2t dt ** = +t sin2t = 0 , t = 0, ,
dx & !
2 !
2
)
x
!
f "( x ) : + + + + + + 0 + + + + + + + 0 + + + + + + + +0 + + + + + +
A. (0,")
D. ("#,0)
!
!
-#
2
+#
0
#
2
#
% # ( % #(
% # #(
f "( x ) < 0 : ' + ,0* - ' 0, * and f decreasing on ' + , *
& 2 ) & 2)
& 2 2)
100. The graph of the function f shown to the right has a horizontal
tangent at x = 4. Let g be the continuous function defined by
!t
g( x ) = " f ( t ) dt . For what value(s) of x does the graph of g have a
0
point of inflection?
!
A. 3 and 5
B. 0, 3, and 5
D. 2 and 4
E. 4 only
C. 2 only
t
D. g( x ) =
" f (t ) dt # g$( x ) = f ( x ) # g$$( x ) = f $( x )
0
%
&
%
g$$( x ) : + + + + + ' ' ' ' ' ' ' + + + + + g changes concavity at x = 2, x = 4
!!!!!
!
2
4
- 43 -
d
101.
dx
!
x3
# tan(t
4
1
"1) dt =
A. sec 2 ( x12 "1)
B. tan( x 4 "1)
D. 3x 2 tan( x12 "1)
E. 12x11 tan( x12 "1)
d
D.
dx
!
!
x3
# tan(t
1
C. tan( x12 "1)
4
!
!
4
"1) dt = tan ( x 3 ) "1 ( 3x 2 ) = 3x 2 tan( x12 "1)
!
[
]
x
102. (Calc) Let f be the function given by f ( x ) = " cos( t 2 + t ) dt for # 2 $ x $ 2. Approximately, for what
0
!
percentage of values of x for " 2 # x # 2 is f ( x ) decreasing?
A. 30%
B. 26%
!
d
B. f "( x ) =
dx
C. 44%
!
D. 50%
E. 59%
!
# cos(t 2 + t) dt = cos( x 2 + x )
x
0
By the graphs below, f "( x ) < 0 for $ 2 < x < 1.849 and 0.849 < x < 1.728
So f "( x ) < 0 for 1.849 + 2 + 1.728 $ 0.849 = 1.030 values out of 4 = 26%
!
x
103. Let f ( x ) =
" g(t ) dt where g has the graph shown to the right. Which
a
of the following could be the graph of f ?
!
!
A.
B.
C.
D.
E.
A. f ( x ) will have a horizontal tangent at x = b and x = c. The accumulated area
will be negative until slightly to the right of the origin when it becomes positive.
!
- 44 -
S. Interpretation of a Derivative as a Rate of Change
What you are finding: As mentioned in section I (Related Rates), a quantity that is given as a rate of change
needs to be interpreted as a derivative of some function. Typical problems ask for the value of the function at
a given time. These problems can be handled several ways:
a) solving a Differential Equation with initial condition (although DEQ’s may not have even been
formally mentioned yet)
b) Integral of the rate of change to give accumulated change. This uses the fact that:
b
b
# R"(t) dt = R(b) $ R(a) or R(b) = R(a) +
# R"(t) dt
a
a
104. The table below gives values of a function f and its derivative at selected values of x. If f " is
2
! continuous on the interval [-4, 4], what is the value of
$ f "(2x ) dx ?
#2
x
f ( x)
f "( x )
A. -2
-4
8
-6
B. -8
!
A. !
u = 2x,du = 2dx
2
-2
-1
-6
-4
-4 ! -2
C. 5
2
-4
2
4
4
4
!
D. 20
E. 1
x = "2,u = "4, x = 2,u = 4
4
1
1
$ f #(2x ) dx = 2 $ f #(u) du = 2 f (u)
"2
1
-2
0
"4
4
"4
=
1
1
f ( 4 ) " f ("4 )] = ( 4 " 8) = "2
[
2
2
105. The graph of y = f "( x ) , the derivative of f , is shown in the
!
figure to the right. If f (0) = "1, then f (1) =
A. 1
!
B. 2
!
D. 4
C. 3
E. 5
1
B. Method 1:
#
1
f "( x ) dx = f (1) $ f (0) % f (1) = f (0) +
0
# f "( x ) dx
0
1
1
# f "( x ) dx = area of trapezoid = 2 (5 + 1) = 3 % f (1) = $1+ 3 = 2
0
Method 2 : f "( x ) = $4 x + 5 % f ( x ) = $2x 2 + 5x + C
f (0) = 0 + C = $1 % f ( x ) = $2x 2 + 5x $1 % f ( x ) = $2 + 5 $1 = 2
!
- 45 -
106. (Calc) The rate of change of people waiting in line to buy tickets to a concert is given by
w ( t ) = 100( t 3 " 4t 2 " t + 7) for 0 # t # 4. 800 people are already waiting in line when the box office
opens at t = 0. Which of the following expressions give the change in people waiting in line when the
line is getting shorter?
!
3.773
3.773
# w"(t) dt
A.
B.
1.480
!
C. 800 "
1.480
2.786
D.
3.773
" w(t) dt
$ w#(t) dt
1.480
2.786
# w"(t) dt
E.
!
0
!
" w(t) dt
!
0
!
t= b
B. Since w ( t ) represents the rate of people waiting,
" w(t) dt represents the change of
t= a
the number of people in line between 2 times. The line is getting shorter when its
rate of change is negative. The 800 people already in line doesn't enter into it.
107. (Calc) A cup of coffee is heated to boiling ( 212°F ) and taken out of a microwave and placed in a
72°F room at time t = 0 minutes. The coffee cools at the rate of 16e"0.112t degrees Fahrenheit per
! minute. To the nearest degree, what is the temperature of the coffee at time t = 5 minutes?
A. 105°F
!
B. 133°F
!
C. 166°F
5
!
!
D. T (5) " T (0) =
!
!
D. 151°F
E. 203°F
5
# "16e"0.112t dt $ T (5) = T (0) " # 16e"0.112t dt % 151
!
0
0 !
!
108. (Calc) The Cheesesteak Factory at a Philadelphia stadium has 25 steak sandwiches ready to sell when
it opens for business. It cooks cheesesteaks at the rate of 4 steaks per minute and sells cheesesteaks at
# 2"t &
the rate of 1+ 6sin%
( steaks per minute. For 0 ≤ t ≤ 20, at what time is the number of steaks ready
$ 41 '
to sell at a minimum (nearest tenth of a minute)?
A. 0
!
B. 3.4
C. 10.3
D. 17.1
E. 20
t
$
$ 2#x ' '
$ 2#t '0
D. C( t) = 25+ * 4 " &1+ 6sin&
) ) dt + C ,( t) = 4 " /1+ 6sin&
)2 = 0
% 41 ( (
% 41 (1
.
%
0
$ 2#t '
4 = 1+ 6sin&
) + Critical values : t = 3.417,t = 17.083
% 41 (
t
C( t)
!
0 3.417 17.083 20
25
30
3.2
6.8
- 46 -
T. Average Value of a Function
What you are finding: You are given a continuous function f ( x ) , either an algebraic formula or a graph, as
well as an interval [ a,b] . You wish to find the average value of the function on that interval.
b
How to find it: f avg =
!
" f ( x ) dx
a
b#a
. The units will be!whatever the function f is measured in.
Again, be careful. The average rate of change of a function F on [ a,b] is not the same as the average value of
the function F on [ a,b] .
!
b
Average value of the function:
!
" f ( x ) dx
!
a
b#a
b
Average rate of change of the function (average value of the rate of change):
!
# F "( x ) dx
a
b$a
F (b) $ F ( a)
b$a
=
#" 3" &
109. The average value of sin 2 x cos x on the interval % , ( is
$2 2 '
!
"2
2
A.
B.
C. 0
D. -1
E. 1
3#
3"
!
!
3! 2
2
" sin x cos x dx
#
3#
!
!
A. !
f avg = 2
u = sin x,du = cos x dx x = ,u = 1 x = ,u = $1
3# #
2
2
$
2 2
-1
"u
f avg =
2
du
1
#
$1
u3
$2
=
=
3 1 3#
2
110. The velocity of a particle moving along the x-axis is given by the function v ( t ) = te t . What is the
! average velocity of the particle from time t = 1 to t = 3 ?
A.
19e 9 " 3e
2
B. 4e 9
C.
3
" te
!
!
D. v avg =
v avg =
!
1
1
2
3e 9 " e
2
D.
t2
dt
!
3 #1
3
u = t ,du = 2t !
dt
2
2
" 2te t dt
1
2
=
1
2
9
"e
u
du
1
2
- 47 -
e9 " e
!
4
E. e 9 " e
!
t = 1,u = 1 t = 3,u = 9
1 u9
e
e9 # e
1
=2
=
2
4
sin(2x + 3)
on the interval [-2, 3] ?
x 2 + 2x + 3
111. (Calc) What is the average value of y =
A. -0.057
B. 0.051
C. 0.061
!
3
B. f avg =
D. 0.256
E. 0.303
sin(2x + 3)
dx
2
+ 2x + 3
0.256
=
= 0.051
3+2
5
"x
-2
112. Newton the cat begins to walk along a ledge at time t = 0. His velocity at
time t, 0 ≤ t ≤ 8, is given by the function whose graph is given in the
! What is Newton’s average speed from t = 0 to t = 8?
figure to the right.
A. 0
B. 2
9
4
E. 5
D.
C. 3
8
2
4
6
8
" v (t) dt " v (t ) dt + " v (t ) dt + " v (t ) dt # " v (t ) dt
!
= 0
8#0
6 + 10 + 5 + 3 24
!!!!!Average Speed =
=
=3
8
8
C. Average Speed =
0
2
4
6
8#0
$ "( x # 9) '
113. (Calc) Barstow, California has the daily summer temperature modeled by T ( t ) = 85 + 27sin&
)
% 12 (
!
where t is measured in hours and t = 0 corresponds to 12:00 midnight. What is the average
temperature in Barstow during the period from 1:30 PM to 6 PM?
A. 60°
B. 89°
!
D. 110°
C. 104°
18
!
!
!
!
D. Tavg =
E. 115°
$ " ( x # 9) '
) dt
(
+ 110°
4.5
* 85+ 27sin&% ! 12
13.5
114. Let f ( x ) = 3x 2 " 6x . For how many positive values of b does the average value of f ( x ) on the
!
interval [0, b] equal the average rate of change of f ( x ) on the interval [0, b] ?
!
A. 4
B. 3
C. 2
# (3x 2 " 6x ) dx
0
E. none
b"0
f (b) " f (0)
b 3 " 3b 2 3b 2 " 6b
6 ± 12
=
$
=
$ b 2 " 6b + 6 = 0 $ b =
b
b
b
2
!
!
!
b
C.
D. 1
- 48 -
U. Straight-Line Motion - Integrals
What you are finding: In section J, we looked at straight-line motion by the derivative process. You were
typically given a position function x ( t ) and took its derivative to find the velocity function v ( t ) and took the
velocity’s derivative to get the acceleration function a( t ) . Questions like finding maximum velocity or
minimum acceleration could then be answered.
!
!
Using integrals, we can work backwards. Typical questions involve knowing the velocity function, the
!
position of the particle at the start of the problem,
and integrating to find the position function.
How to find it: x ( t ) =
" v (t ) dt + C and v (t ) = " a(t ) dt + C . Using these equations is essentially solving
DEQ’s with an initial condition, usually the position at time t = 0 or velocity at t = 0. Two other concepts that
come into play are displacement and distance over some time interval [ t1,t 2 ] .
!
Displacement: the difference in position over [ t1,t 2 ] : x ( t 2 ) " x ( t1 ) =
t2
# v (t ) dt . Displacement can be
t1
positive, negative or zero.
!
t2
Distance: how far the particle traveled over [ t1,t 2 ] :
!
" v (t ) dt . Distance is always positive.
t1
115. A particle moves along the x-axis so that at any time t ≥ 0, its velocity is given by v ( t ) = 5 " 2t " 3t 2 . If
the particle is at position x = "3 at
! time t = 2, where was the particle at t = 1?
A. -5
B. 2
C. 3
!
B. x ( t ) =
D. 4
# (5 " 2t " 3t ) dt = 5t " t
2
2
!
E. 11
" t3 + C
x (2) = 10 " 4 " 8 + C = "3 $ C = "1
x ( t ) = 5t " t 2 " t 3 "1 $ x (1) = 5 "1"1"1 = 2.
116. A spider is walking up a vertical wall. At t = 0, the
spider is 18 inches off the floor. The velocity v of the
! in inches per minute at time t
spider, measured
minutes, 0 ≤ t ≤ 10 is given by the function whose
graph is shown to the right. How many inches off the
floor is the spider at t = 10?
A. 13.5
D. 31.5
B. 19.5
E. 37.5
C. 24
10
D. y ( t ) = 18 +
" v (t ) dt = 18 + 9 + 4.5 # 3 + 3 = 31.5
0
!
- 49 -
117. The table below gives selected values of v ( t ) , of a particle moving along the x-axis. At time t = 0, the
particle is at the origin. Which of the following could be the graph of the position, x ( t ) , of the
particle for 0 ≤ t ≤ 4 ?
!
0
3
t
v(t)
1
0
2
-1
3
1
4
3
!
!
A.
B.
D.
E.
C.
D. Velocity is zero at t = 1 and at 2 < t < 3 so the graph of x ( t ) must
have horizontal tangents at t = 1 and at 2 < t < 3.
118. A particle moves along the y-axis with velocity v ( t ) for 0 ≤ t ≤ 5.
!
The graph of v ( t ) is shown to the right with the positive areas
indicated by P, Q, and R. What is the difference between the
distance that the particle travels and its displacement for
!
0≤t≤5?
!
A. P - R
D. 2R
B. 2(P + Q + R)
E. 2(Q + R)
C. 2Q
5
E. Distance =
" v (t ) dt = P + Q + Q + R = P + 2Q + R
0
5
Displacement =
" v (t) dt = P # Q + Q # R = P # R
0
P + 2Q + R # (P # R ) = 2Q + 2R
!
- 50 -
119. (Calc) A particle moves along the x-axis so that its velocity v ( t ) = 12te"2t " t + 1. At t = 0, the particle
is at position x = 0.5. What is the total distance that the particle traveled from t = 0 to t = 3 ?
A. 1.448
B. 1.948
C. 2.911
!
D. 4.181
E. 4.681
!
1.69
3
3
# v (t ) dt " # v (t) dt or
D. Distance =
0
1.69
# v (t ) dt = 4.181.
0
The starting position has no bearing on the answer.
120. (Calc) A particle moves along the x-axis so that at any time t > 0, its acceleration is given by
a( t ) = cos(1" 2 t ) . If the velocity of the particle is -2 at time t = 0, then the speed of the particle at
!
t = 2 is
A. 0.613
!
B. 0.669
C. 1.331
D. 1.387
E. 1.796
2
D. v ( t ) = v (0) +
$ a(t) dt = "2 + 0.613 = "1.387 # Speed = 1.387
0
!
121. A particle moves along the x-axis so that its velocity
v ( t ) = 2e1"t " 2t , as shown in the figure to the right. Find the
total distance that the particle traveled from t = 0 to t = 2.
2
"2
e
2
C. 2e " + 4
e
2
E. 2e + + 2
e
2
B. 2e " " 4
e
A. 2e +
!
!
D. 2e " 2
!
!
!
A. Distance =
!
1
2
# (2e1"t " 2t) dt "
# (2e
0
1"t
1
1
" 2t ) dt
2
Distance = ["2e1"t " t 2 ] 0 " ["2e1"t " t 2 ]1 = "2 "1" ("2e) + 2e"1 + 4 " (2 + 1)
Distance = "3 + 2e +
!
2
2
+ 4 " 2 "1 = 2e + " 2
e
e
- 51 -
V. Area/Volume Problems
What you are finding: Typically, these are problems with which students feel more comfortable because
they are told exactly what to do or to find. Area and volume are lumped together because, almost always,
they are both tested within the confines of a single A.P. free response question. Usually, but not always, they
are on the calculator section of the free-response section.
Area problems usually involve finding the area of a region under a curve or the area between two curves
between two values of x. Volume problems usually involve finding the volume of a solid when rotating a
curve about a line.
How to find it: Area: Given two curves f ( x ) and g( x ) with f ( x ) " g( x ) on an interval [a, b], the area
x= b
between f and g on [a, b] is given by A =
# [ f ( x ) " g( x )] dx . While integration is usually done with respect
x= a
!
!
to the x-axis, these problems sometimes show up in terms of y: A =
y= d
# [m( y ) " n( y )] dy .
y= c
Volume: Disks and Washers:
! The method I recommend is to establish the outside Radius R, the distance
from the line of rotation to the outside curve, and, if it exists, the inside radius r, the distance from the line of
rotation to the inside curve. The formula when rotating these curves about a line on an interval is given by:
!
y= d
x= b
2
2
2
2
V = " $ [ R( x )] # [ r( x )] dx or V = " $ [ R( y )] # [ r( y )] dy
(
)
x= a
(
)
y= c
A favorite type of problem is creating a solid with the region R being the base of the solid. Cross sections
perpendicular to an axis are typically squares, equilateral triangles, right triangles, or semi-circles. Rather
than give formulas for this, it is suggested that you draw the figure, establish its area in terms of x or y, and
!expression on the given interval. !
integrate that
122. What is the area of the region enclosed by the graphs of y = x " 4 x 2 and y = "7x ?
A.
4
3
16
68
80
C. 8
D.
E.
3
3
3
! be able to sketch the line and parabola.
B. Even without a calculator, students should
B.
x " 4 x 2 = "7x # 4 x 2 " 8x = 0 # 4 x ( x " 2) = 0 # x = 0, x = 2
!2
!
!
2
% 2 4 x 3 (2
32 16
2
2
A = $ ( x " 4 x + 7x ) dx = $ (8x " 4 x ) dx = '4 x "
=
* = 16 "
3 )0
3
3
&
0
0
!
123. (Calc) What is the area of the region in the first quadrant enclosed by the graph of y = 2cos x, y = x,
and the x-axis?
!
A. 0.816
B. 1.184
C. 1.529
1.03
A. A =
#2
!
1.03
" x dx + " 2cos x dx
0
D. 1.794
1.03
or A =
+
" -,cos
0
E. 1.999
% y( .
' * $ y0 dx = 0.816
& 2) /
$1
!
- 52 -
124. The three regions A, B, and C in the figure to the right
are bounded by the graph of the function f and the
x-axis. If the areas of A, B, and C are 5, 2, and 1
1
# [ f ( x ) + 2] dx ?
respectively, what is the value of
"4
A. 28
C. 6
E. -12
!
1
C.
B. 10
D. -4
1
1
# [ f ( x ) + 2] dx = # f ( x ) dx + # 2 dx
"4
"4
"4
1
" 5 + 2 "1+ [2x ]"4 = "4 + [2 " ("8)] = "4 + 10 = 6
!
x"3
2
is shown in the figure to the right. Which of the following
calculations would accurately compute the area of region R?
!
9
!
#
x " 3&
dx
I. ) % x "
(
$
2 '
0
1
9
$
x # 3'
II. " 2 x dx + " & x #
) dx
%
2 (
0
1
125. Region R, enclosed by the graphs of y = ± x and y =
3
!
III.
# (2x + 3 " x ) dx
2
"1
!
A. I only
!
B. II only
C. III only
D. II and III
E. I, II, and III
D. I does not take into consideration y = " x
II is accurate. y = x is always the top curve but y = " x is the bottom curve
x"3
on [0,1] and y =
is the bottom curve on [1,9].
2
III is accurate. If we integrate with respect to y, we would use right minus left.
3
That gives
# (2y + 3 " y ) dy. y is a dummy variable, You can use any variable.
2
"1
(when you calculate it with your calculator, you use x ).
!
- 53 -
126. (Calc) On the graph to the right, the line y = kx "10 , k a constant, is tangent
to the graph of y = x 2 " 4 . Region R is enclosed by the graphs of
y = kx "10 , y = x 2 " 4 , and the y-axis. Find the area of region R.
A. 4.287
!
!
D. 9.798
!
B. 4.899
!
C. 6.124
E. 12.247
B. If the curves are tangent they intersect so x 2 " 4 = kx "10 and are tangent so 2x = k.
x 2 " 4 = 2x 2 "10 # x = 6 which is the intersection point. Thus k = 2 6.
6
The line is therefore y = 2x 6 "10. A =
$ [x
2
)]
(
" 4 " 2x 6 "10 dx = 4.889.
0
127. If the region bounded by the y-axis, the line y = 4, and the curve y = x is revolved about the x-axis,
!
the volume of the solid generated would be
A.
88"
3
B. 56"
!
D.
128"
3
E.
640"
3
C. Students should be able to sketch the graph without a calculator.
16
16
!
2(
%
x = 4 " x = 16!so V = # + '4 2 $ !
x * = # + (16 $ x )
&
)
0
0
!
!
C. 128"
( )
16
%
,
x2(
256 /
V = #'16x $ * = #. 256 $
1 = 128#
2 )0
2 0
&
!
"
128. (Calc) If the region bounded by the x-axis, the line x = , and the curve y = sin x is revolved about
2
"
the line x = , the volume of the solid generated would be
2
!
A. 1.142
B. 1.468
C.!1.793
D. 3.142
E. 3.586
!
"
"
# x = # sin#1 y
2
2
2
1
$"
#1 '
V = " * & # sin y ) dy = 1.142" = 3.586
%2
(
0
E. R =
" 2
$"
'
by shells (not in curriculum) :V = 2" * & # x ) sin x dx = 3.586
%2
(
0
!
- 54 -
129. A piece of candy is determined by the curves y = ± x 3 , for 1 ≤ x ≤ 2 as shown in
the figure to the right. For this piece of candy, the cross sections perpendicular
to the x-axis are equilateral triangles. What is the volume of the piece of candy
in cubic units?
!
127
7
127 3
D.
21
15 3
2
127 3
E.
7
A.
B.
!
!
!
!
C.
254 3
7
!
1
E. Area = (2) y y 3 = y 2 3 = x 6 3
2
2
2
#x7&
3 7
3
127 3
6
V = 3 " x dx = 3% ( =
2 )1) =
(127) =
(
7
7
7
$ 7 '1
1
(
)
130. A lake shaped in a right triangle is located next to a park as shown in
the figure to the right. The depth of the lake at any point along a strip
! edge is f ( x ) feet. Which of the following
x miles from the park’s
expressions gives the total volume of the lake?
3
A.
#
4 &
) %$ 4 " 3 x (' f ( x ) dx! B.
0
4
D.
" f ( x ) dx
E.
0
!
!
4
3
#
3 &
) %$ 3 " 4 x (' f ( x ) dx
0
3
2
C. 2 " f ( x ) dx
0
4
" f ( x ) dx
0
!
4
A. The hypotenuse is given by y = 4 " x. The dimensions of our
3
!
4
rectangular solid will be height : 4 " x, width : dx and depth : f ( x )
3
3
#
4 &
So the volume is given by V = ) % 4 " x ( f ( x ) dx
$
3 '
0
!
131. (Calc) Let R be the region between the graphs of
y = 2sin x and y = cos x as shown in the figure to the right. The region
!
R is the base of a solid with cross sections perpendicular to the x-axis as
rectangles that are twice as high as wide. Find the volume of the solid.
!
A. 4.472
B. 7.854
D. 15.708
E. 31.416
C. 8.944
D. 2sin x = cos x " x = 0.464,3.605
3.605
2
s = 2sin x # cos x. Area = 2s( s) = 2s . V = 2
$ (2sin x # cos x )2dx = 15.708
0.464
!
- 55 -
W. Differential Equations
dy
= ( Algebraic Expression) . The
dx
goal of solving a DEQ is to work backwards from the derivative to the function; that is to write an equation
in the form of y = f ( x ) + C (since the technique involves integration, the general solution will have a
constant of integration). If the value of the function at some value of x is known (an initial condition), the
!
value of C can be found. This is called a specific solution.
What you are finding: A differential equation (DEQ) is in the form of
How!to find it: In Calculus AB, the only types of DEQ’s studied are called separable. Separable DEQ’s are
those than that can be in the form of f ( y ) dy = g( x ) dx . Once in that form, both sides can be integrated with
the constant of integration on only one side, usually the right.
Word problems involving change with respect to time are usually models of DEQ’s. A favorite type is a
! rate of change of y is proportional to some expression. The equation that
problem using the words: The
dy
describes this statement is:
= k " (expression) . The rate of change is usually with respect to time.
dt
Usually there is a problem that requires you to create a slope field. Simply calculate the slopes using the
given derivative formula and plot them on the given graph. Usually the slopes will be integer values.
132. The slope!field for the equation in the figure to the right could be
A.
dy
= x + y2
dx
B.
dy
= x " y2
dx
D.
dy
=x+y
dx
!
E.
dy
= x2 " y
dx
!
!
C.
dy
= xy
dx
E. Since the slope is 0 at (1,1), choices A and C are eliminated
!
Since the slope
! is 0 at ("1,1), choice B is eliminated
The slope field cannot be D because along the line y = "x,
all slopes should be zero.
133. If
!
A.
!
dy
= 3x 2 y 2 and if y = 1 when x = 2, then when x = -1, y =
dx
1
10
B.
A.
!
!
1
8
dy
= 3x 2 dx "
2
y
!
1
19
D.
"1
8
E. -1
1
= # 3x 2 dx " $ = x 3 + C
y
!
1
$1
1
$1 = 8 + C " C = $9 $ = x 3 $ 9 " y = 3
=
y
x $ 9 9 $ x3
1
1
x = $1 " y =
3 =
9 $ ($1) 10
!
C.
dy
#y
2
- 56 -
134. The rate of change of the volume V of the oil in a tank with respect to time t is inversely proportional
to the square root of V. If the tank is empty at time t = 0 and V = 9 when t = 2, find the equation
which describes the relationship between t and V.
A. V =
9 2
t
4
B. V =
D.
!
!
t2
36
C. V 3 2 =
4
t
9
D. V 3 2 =
27
t
2
E. V = 3t
dV
k
2
=
" V 1 2 dV = kt " V 3 2 = kt + C
!
dt
3
V
!
!
2
t = 0,V = 0 : V 3 2 = kt
3
2 32
2
27
(9) = 2k " k = 9 V 3 2 = 9t " V 3 2 = t
3
3
2
135. (Calc) The Athabasca Glacier in Canada is one of the few glaciers to
! drive. Like many glaciers in the world, it is receding (in
which you can
this case its toe is moving away from the road). In the year 2000, its toe
was 500 feet from the road and in the year 2010, its toe was 600 feet from
the road. The rate the distance that the toe increases from the road is
proportional to that distance. Assuming the glacier recedes at the same
rate, what will be the average distance in feet per year (nearest integer)
the glacier recedes from 2000 to the year 2050?
A. 10
B. 13
C. 15
D. 16
E. 18
ds
= ks " s = Ce kt .
dt
#6&
600 = 500e10k " k = ln% ( 10
$5'
C. s is the distance from the road :
500 = Ce 0 " s = 500e kt
The year 2050 : s = 500e 50k ) 1244.
Average rate of change :
s(50) * s(0) 1244 * 500
=
) 15
50
50
!
- 57 -

Q. Fundamental Theorem of Calculus (FTC)

Transcription

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