(kah2852) huang HW05 Radin (54915)
This print-out should have 21 questions.
Multiple-choice questions may continue on
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before answering.
001
10.0 points
Find the bounded area enclosed by the
graph of
f ( x) = 3 x x 2
and the x-axis.
keywords: AreaBetween, AreaBetweenExam,
10.0 points
002
Find the area enclosed by the graphs of
f (x) = sin x ,
g (x) = cos x
on [0, ].
1. area = 4( 2 + 1)
2. area = 2 + 1
3. area = 2 2 correct
4. area = 2( 2 + 1)
5. area = 2
6. area = 4 2
9
1. Area = sq.units correct
2
2. Area = 4 sq.units
3. Area = 3 sq.units
4. Area =
1
7
sq.units
2
5. Area = 5 sq.units
Explanation:
The graph of f is a parabola opening downwards and crossing the x-axis at x = 0 and
x = 3. Thus the required area is similar to
the shaded region in
Explanation:
The area between the graphs of y = f (x)
and y = g (x) on the interval [a, b] is expressed
by the integral
b
A=
y
a
|f (x) g (x)| dx ,
which for the given functions is the integral
A=
0
3
|sin x cos x| dx .
But, as the graphs
x
y
As a denite integral, therefore, the required
area is given by
3
2
0
(3x x ) dx =
32 13
x x
2
3
Consequently,
9
Area =
sq.units .
2
3
0
/2
.
cos :
sin :
huang (kah2852) HW05 Radin (54915)
of y = cos x and y = sin x on [0, ] show,
cos sin
0,
on [0, /4],
0,
on [/4, ].
Explanation:
The graph of f is a parabola opening upwards and having y -intercept at y = 9. In
addition, since
f (x) = (5x + 3)(x 3),
Thus
3
the graph has x-intercepts at x = 5 and at
x = 3. Thus the graph of f and the vertical
line are given by
/4
A=
{cos sin } d
0
2
/4
{cos sin } d = A1 A2 .
But by the Fundamental Theorem of Calculus,
A1 =
sin + cos
/4
0
=
21,
while
A2 =
sin + cos
/4
= (1 +
2) .
Consequently,
area = A1 A2 = 2 2 .
003
10.0 points
In terms of denite integrals, therefore, the
total area A is given by
3
A=
3
140
2. A =
sq.units correct
3
137
sq.units
3. A =
3
4. A =
277
sq.units
6
143
sq.units
5. A =
3
5x2 12x 9 dx
f (x) dx =
0
0
53
x 6 x2 9 x
3
=
3
0
= 36 ,
while
4
283
1. A =
sq.units
6
f (x) dx.
3
3
2
the x-axis, and the line x = 4.
0
But
Find the total area, A, of the bounded region in the rst and fourth quadrants enclosed
by the graphs of
f (x) = 5x 12x 9,
4
f (x) dx +
4
f (x) dx =
3
3
5x2 12x 9 dx
53
x 6 x2 9 x
3
=
4
3
=
32
.
3
Consequently,
A=
140
sq.units.
3
004 (part 1 of 3) 10.0 points
The shaded region in
huang (kah2852) HW05 Radin (54915)
3
1
|f (x) g (x)| dx
2.
3
1/2
|f (x) g (x)| dx
3.
1
1
|f (x) g (x)| dx correct
4.
2
2
is bounded by the graphs of
f ( x) = 1 + x x 2 x3
and
g (x) = 1 x.
|f (x) g (x)| dx
5.
3
Explanation:
Because g (x) f (x) on [2, 0] and f (x)
g (x) on [0, 1], the area of the shaded region is
given by
(i) Find the x-coordinates of all the points of
intersection of the graphs of f and g .
1
2
1. x = 1, 0, 1/2
2. x = 2, 0, 1 correct
|f (x) g (x)| dx .
006 (part 3 of 3) 10.0 points
(iii) Calculate the area of this shaded region.
3. x = 3, 0, 2
1. area =
5. x = 2, 0, 3/2
Explanation:
The x-coordinates of the points of intersection of the two graphs are the solutions of the
equation g (x) = f (x). Now
3
2
g ( x) f ( x) = x + x 2 x
= x(x + 2)(x 1).
9
sq. units
4
2. area =
4. x = 3, 0, 1
37
sq. units correct
12
3. area =
35
sq. units
12
4. area =
10
sq. units
3
5. area =
17
sq. units
6
Thus the graphs intersect when
x = 2, 0, 1 .
Explanation:
From (ii) we see that
0
005 (part 2 of 3) 10.0 points
(ii) Set up the denite integral determining the
area of this shaded region.
2
(g (x) f (x)) dx
0
=
2
(x3 + x2 2x) dx
3/2
1.
2
|f (x) g (x)| dx
=
14 13
x + x x2
4
3
0
2
=
8
.
3
huang (kah2852) HW05 Radin (54915)
4
On the other hand,
y
graph of g
1
0
(f (x) g (x)) dx
1
x
(x3 x2 + 2x) dx
=
0
=
1
1
x4 x3 + x 2
4
3
1
0
=
5
.
12
Consequently, the shaded region has
area =
37
sq. units .
12
graph of f
This shaded region has
a
Area =
|f (y ) g (y )| dy
a
keywords: AreaBetween,
007
10.0 points
Find the area of the bounded region enclosed by the graphs of
f (y ) = 1 y 2 ,
g (y ) = y 2 1 .
where a and a are the points where the
graphs of f and g intersect, i.e. at the y intercepts y = 1, 1 of the graphs. On the
other hand, the graphs are symmetric about
the x-axis, so
1
Area = 2
1
1
(1 y 2 ) dy = 4 y
=4
0
1. Area = 3
2. Area =
8
correct
3
Area =
keywords:
parabola,
4. Area = 2
5. Area =
10
3
Explanation:
The graph of f is a parabola opening to the
left, while the graph of g is a parabola opening to the right. Thus the bounded region
enclosed by the graphs of f and g is similar to
the shaded region shown in
y3
3
1
0
.
Consequently,
7
3
3. Area =
(f (y ) g (y )) dy
area,
008
8
.
3
area between curves,
10.0 points
Find the area of the bounded region enclosed by the graphs of
f (x) = 2 sin x,
on [ ,
3
3 ].
Correct answer: 0.613706.
Explanation:
g (x) = tan x
huang (kah2852) HW05 Radin (54915)
5
We have to nd the shaded region in
009
y
10.0 points
Find the area enclosed by the curves
f(x) = 2sin x
x + y = 3 , x y = 0 , y + 2x = 3 .
g(x) = tan x
-/3
1. Area = 0 sq. units
x
/3
2. Area =
3
sq. units correct
4
3. Area =
1
sq. units
2
To nd the points of intersection, note that
4. Area = 1 sq. units
2 sin x = tan x
2 sin x tan x = 0
sin x
2 sin x
=0
cos x
2 sin x cos x sin x
=0
cos x
sin x(2 cos x 1) = 0
1
sin x = 0 cos x =
2
x=0 x=
3
5. Area =
1
sq. units
4
Explanation:
Each of the curves is a straight line, so the
enclosed region is the shaded triangle in
y
Thus
0
[g (x) f (x)] dx
A=
3
3
+
0
[f (x) g (x)] dx
3
=2
x
[f (x) g (x)] dx
0
3
=2
(2 sin x tan x) dx
0
3
=4
0
3
sin x dx 2
0
sin x
dx
cos x
3
= 4( cos x)
0
3
2( ln | cos x|)
= 4 cos + cos 0
3
+ ln |cos 0|
2 ln cos
3
1
= 2 + 2 ln = 2 2 ln 2 .
2
0
To express the area as a denite integral we
need to nd where the lines intersect. Now
the line x y = 0 intersects
x+ y = 3,
y + 2x = 3
at the points
( 1, 1 ) ,
33
,
22
huang (kah2852) HW05 Radin (54915)
respectively. Thus the area of the shaded
triangle is given by
1
0
((3 x) (3 2x)) dx
When
f ( x) =
6
=
x dx +
1
0
2
Consequently,
3
2
1
(3 2x) dx .
V =
Consequently, the shaded triangle has
Area =
3
sq.units .
4
keywords: area between curves, linear functions
010
10.0 points
Find the volume, V , of the solid obtained
by rotating the region bounded by
1
y = , x = 2, x = 6, y = 0
x
about the x-axis.
1. V =
1
12
2. V =
1
12
3. V =
1
6
4. V =
1
6
5. V =
011
1
x
6
=
2
1
.
3
10.0 points
Find the volume, V , of the solid generated
by rotating about the x-axis the region enclosed by the graphs of
y = sec x, x = 0, y = 0, x =
.
3
1. V = cu. units
2. V = 3 cu. units correct
8
cu. units
3
4. V = cu. units
3
1
5. V = ln
+3
2
3. V =
1
3
6. V =
1
dx .
x2
V x) =
((3 (x)) dx
1
1
, a = 2, b = 6 ,
x
therefore,
3
2
+
6
1
correct
3
cu. units
Explanation:
The solid is generated by rotating the region
Explanation:
The volume of the solid of revolution obtained by rotating the graph of y = f (x) on
[a, b] about the x-axis is given by
b
volume =
2
f (x) dx .
a
/3
huang (kah2852) HW05 Radin (54915)
about the x-axis. The cross-section perpendicular to the x-axis has area
A(x) = y 2 = sec2 x .
Explanation:
Since the graphs of y = 1 + x and x = 1
intersect at (1, 2), the shaded region consists
of the points
Consequently,
3
V=
sec2 x dx =
tan x
0
= tan
012
(x, y ) : g (x) y f (x) , 0 x 1
3
0
= 3 cu. units
3
7
.
in the plane where
g ( x) = 1 ,
f ( x) = 1 +
x.
Thus the solid obtained by rotating the region
around the x-axis has
10.0 points
1
The shaded region in
volume =
0
y
(f (x)2 g (x)2) dx .
But
f (x)2 g (x)2 = (1 +
x) 2 1 = 2 x + x .
On the other hand,
1
(2 x + x) dx =
0
x
is bounded by the graphs of
y = 1 + x, x = 1,
1
0
=
Consequently, the solid has
volume =
11
6
.
y = 1.
Find the volume of the solid obtained by rotating this region around the x-axis.
1. volume =
4 3/2 1 2
x +x
3
2
5
cu. units
6
4
cu. units
2. volume =
3
3. volume =
5
cu. units
3
10.0 points
The shaded region in
y
7
cu. units
6
5. volume =
013
11
cu. units correct
6
4. volume =
keywords:
6. volume =
13
cu. units
6
x
11
.
6
huang (kah2852) HW05 Radin (54915)
is bounded by the graphs of
y = x+ 4, y = 4,
x = 1,
(not drawn to scale). Find the volume of the
solid obtained by rotating this shaded region
about the dotted line y = 2.
1. volume =
7
cu. units
2
2. volume =
19
cu. units correct
6
8
keywords: denite integral, volume of revolution, hollow solid, square root function,
014
10.0 points
Find the volume, V , of the solid obtained
by rotating the region bounded by the graphs
of
x = y2,
x=y
about the line x = 1.
4. volume =
2. V =
11
5. volume =
cu. units
3
Explanation:
When a region
9
10
4. V =
23
cu. units
6
3
4
3. V =
10
cu. units
3
3
5
2
5. V =
3. volume =
1. V =
2
5
6. V =
29
30
7. V =
7
correct
15
8. V =
31
30
(x, y ) : g (x) y f (x) , a x b
is rotated about a line y = c to form a hollow
solid of revolution, then the volume of this
solid is given by the integral
b
a
{(f (x) c)2 (g (x) c)2 } dx .
In the gure above,
f ( x) = x + 4 ,
g ( x) = 4 ,
c = 2,
where 0 x 1. Thus the solid has volume
1
0
=
x = y,
x = y2,
is the shaded region in
( x + 2)2 4 dx
1
Explanation:
The region enclosed by the graphs of
y
x + 4 x dx .
(1, 1)
0
Consequently,
volume =
1 2 8 3/2
x+ x
2
3
1
0
=
19
.
6
x
x=y:
x = y2 :
huang (kah2852) HW05 Radin (54915)
9
To determine the volume of the solid generated when this region is rotated about the
line x = 1 lets see rst what a horizontal
cross-section of the solid looks like. As shown
in
d1
d0
1
y
x
the base is the circle
1
d0 = 1 + y 2 ,
(1, 1)
z
x2 + y 2 = 4
d1 = 1 + y
and the cross-section perpendicular to the y axis is a square.
it is a washer having
inner radius = d0 (y ) = 1 + y 2 ,
Set up a denite integral expressing the
volume, V , of the solid.
and
2
1. volume =
outer radius = d1 (y ) = 1 + y ,
2
(4 y 2 ) dy cu.units
2
where 0 y 1. Thus
2
1
d 1 ( y ) 2 d 0 ( y ) 2 dy .
V =
0
(4 + y 2 ) dy cu.units
2. volume =
2
2(4 y 2 ) dy cu.units
3. volume =
2
For the given values of d0 and d1 , therefore,
2
4. volume =
1
y 2 + 2y y 4 2y 2 dy
V =
0
correct
4
0
2y y y
2
4(4 + y 2 ) dy cu.units
5. volume =
dy
1
1
= y2 y5 y3
5
3
2
1
0
.
Explanation:
At the point P = (x, y )
Consequently,
V=
z
1
015
cu.units
2
1
=
2
4(4 y 2 ) dy
7
11
=
.
53
15
10.0 points
The volume of a solid can often be determined by integration even when the crosssection of the solid is not necessarily a circle.
In the following gure
y
P
x
huang (kah2852) HW05 Radin (54915)
on the base of the solid the cross-section is a
square having side-length 2x. Thus the area
of cross-section is a square having area 4x2 , so
the volume of the solid is given by the denite
integral
2
10
so long as r = 1. Consequently,
2
xe +1
x2
e +1
1
2
I=
0
= 1
1
.
1 + e2
It follows that
2
V=
4x dy.
2
I=
But
x2 = 4 y 2 ,
Hence
e2
.
1 + e2
10.0 points
017
Evaluate the denite integral
1
2
2
volume =
2
(ex + 2ex )2 dx .
I=
4(4 y ) dy cu.units .
0
1. I = e2 7 + 2e2
keywords: volume, integral, cross-section
10.0 points
016
Evaluate the denite integral
1
I=
0
2
( 2x xe ) dx .
12
e+
2
1
3. I = e2 +
2
2. I =
11
2e2 correct
2
11
+ 2e2
2
4. I = e2 + 7 2e2
5. I = e2 4 + 2e2
12
e 4 2e2
2
Explanation:
After expansion
6. I =
1. I =
3 + 2e2
1 + e2
2. I =
1 + 2e2
1 + e2
1 + 2e2
3. I =
1 + e2
e2
4. I =
1 + e2
e2
5. I =
2+
1 + e2
6. I =
e2
correct
1 + e2
(ex + 2ex )2 = e2x + 4 + 4e2x ,
in which case
1
e2x + 4 + 4e2x dx
I=
0
1 2x
e + 4x 2e2x
2
=
1
0
Consequently,
Explanation:
Up to an arbitrary constant
xr+1
xr dx =
1+r
I=
1 2 11
e+
2e2 .
2
2
018
10.0 points
Find the value of f (1) when
2
f (x) = 8xe7x ,
f (0) = 8.
.
huang (kah2852) HW05 Radin (54915)
11
1. I =
4
1. f (1) = e7
7
1
5 1 correct
2
2. f (1) =
60 4 7
e
7
7
2. I =
1
51
4
3. f (1) =
4 7 60
e
7
7
3. I =
1
5+1
4
4. f (1) =
60 4 7
+e
7
7
4. I =
3
5+1
4
5. I =
1
5+1
2
6. I =
3
51
4
4
5. f (1) = e7
7
6. f (1) =
60 4 7
e
correct
7
7
Explanation:
By integration
Explanation:
Set u = 4ex 3. Then du = 4ex dx, while
2
8xe7x dx.
f ( x) =
After substitution u = x2 the integral becomes
4
4
e7u du = e7u + C
7
x=0
=
u=1
x = ln 2
=
u = 5.
In this case,
I=
5
1
4
1
with C an arbitrary constant. Thus
2
4
f (x) = e7x + C.
7
=
C=
60
,
7
i.e.,
10.0 points
019
Evaluate the denite integral
ln 2
I=
0
ex
dx .
4ex 3
.
10.0 points
020
Find the value of the integral
60 4 7x2
e
.
7
7
At x = 1, therefore,
60 4 7
e.
7
7
1 1/2
u
2
1
51
2
I=
4
f ( x) =
f (1) =
u1/2
du =
Consequently,
The condition f (0) = 8 determines C since
f (0) = 8
1
I=
0
1. I = 5e
2. I =
5
e+4
2
3. I =
5
e
2
4. I = 5e + 1
5etan x + 4
dx .
cos2 x
5
1
.
huang (kah2852) HW05 Radin (54915)
5
e4
2
5. I =
5.
1 5x
e
e5x
20
Explanation:
Since
d
tan x = sec2 x ,
dx
so
e5x + e5x
dx
(e5x e5x )5
set u = tan x. Then
du = sec2 x dx ,
=
while
x=0
x=
4
=
=
1
5
e5x + e5x
dx
(e5x e5x )5
u = 1.
=
1
(5eu + 4) dx =
5eu + 4u
0
Consequently,
I = (5e + 4) 5 = 5e 1 .
keywords:
021
10.0 points
Find the indenite integral
e5x + e5x
dx.
(e5x e5x )5
1 5x
e
e5x
25
4
+C
2.
1 5x
e
+ e5x
20
4
3.
1 5x
e
e5x
20
4
1 5x
e
e5x
4
+C
+ C correct
4
+C
1
1
du =
+ C.
5
u
20 u4
Thus
u = 0,
In this case
4.
+C
du = 5(e5x + e5x ) dx,
1
= sec2 x,
cos2 x
1.
4
Explanation:
Set u = e5x e5x . Then
6. I = 5e 1 correct
I=
12
1
0
.
1 5x
(e
e5x )4 + C
20
with C an arbitrary constant.