S E C T I O N 5.2
The Definite Integral
571
Similarly, if f .x/ is monotone decreasing,
0
1 0
1 0
1
N
!1
N
!1
N
X
X
X
b
!
a
b
!
a
b
!
a
@LN D
f .xk /A > @MN D
f .xk" /A > @RN D
f .xk /A
N
N
N
kD0
kD0
kD1
Thus, if f .x/ is monotonic, then MN always lies in between RN and LN .
Now, as in Figure 6, consider the typical subinterval Œxi !1 ; xi ! and its midpoint xi" . We let A; B; C; D; E, and F be the areas
as shown in Figure 6. Note that, by the fact that xi" is the midpoint of the interval, A D D C E and F D B C C . Let ER represent
the right endpoint approximation error ( D A C B C D), let EL represent the left endpoint approximation error ( D C C F C E)
and let EM represent the midpoint approximation error ( D jB ! Ej).
#
If B > E, then EM D B ! E. In this case,
ER ! EM D A C B C D ! .B ! E/ D A C D C E > 0;
so ER > EM , while
EL ! EM D C C F C E ! .B ! E/ D C C .B C C / C E ! .B ! E/ D 2C C 2E > 0;
#
so EL > EM . Therefore, the midpoint approximation is more accurate than either the left or the right endpoint approximation.
If B < E, then EM D E ! B. In this case,
ER ! EM D A C B C D ! .E ! B/ D D C E C D ! .E ! B/ D 2D C B > 0;
so that ER > EM while
EL ! EM D C C F C E ! .E ! B/ D C C F C B > 0;
#
so EL > EM . Therefore, the midpoint approximation is more accurate than either the right or the left endpoint approximation.
If B D E, the midpoint approximation is exactly equal to the area.
Hence, for B < E, B > E, or B D E, the midpoint approximation is more accurate than either the left endpoint or the right
endpoint approximation.
5.2 The Definite Integral
Preliminary Questions
1. What is
Z
5
dx [the function is f .x/ D 1]?
3
SOLUTION
2. Let I D
Z
Z
5
3
dx D
7
Z
5
1 " dx D 1.5 ! 3/ D 2.
3
f .x/ dx, where f .x/ is continuous. State whether true or false:
2
(a) I is the area between the graph and the x-axis over Œ2; 7!.
(b) If f .x/ # 0, then I is the area between the graph and the x-axis over Œ2; 7!.
(c) If f .x/ $ 0, then !I is the area between the graph of f .x/ and the x-axis over Œ2; 7!.
SOLUTION
(a) False.
(b) True.
(c) True.
Rb
a
f .x/ dx is the signed area between the graph and the x-axis.
3. Explain graphically:
Z
!
0
cos x dx D 0.
!
SOLUTION Because cos." ! x/ D ! cos x, the “negative” area between the graph of y D cos x and the x-axis over Œ 2 ; "!
exactly cancels the “positive” area between the graph and the x-axis over Œ0; !2 !.
Z !5
Z !1
8 dx or
8 dx?
4. Which is negative,
!1
SOLUTION
!5
Because !5 ! .!1/ D !4,
Z
!5
!1
8 dx is negative.
572
CHAPTER 5
THE INTEGRAL
Exercises
In Exercises 1–10, draw a graph of the signed area represented by the integral and compute it using geometry.
1.
Z
3
2x dx
!3
The region bounded by the graph of y D 2x and the x-axis over the interval Œ!3; 3! consists of two right triangles.
One has area
D 9 below the axis, and the other has area 12 .3/.6/ D 9 above the axis. Hence,
SOLUTION
1
2 .3/.6/
Z
3
!3
2x dx D 9 ! 9 D 0:
y
6
4
2
−3
2.
Z
x
−1 −2
−4
−6
−2
1
2
3
3
!2
.2x C 4/ dx
SOLUTION The region bounded by the graph of y D 2x C 4 and the x-axis over the interval Œ!2; 3! consists of a single right
triangle of area 12 .5/.10/ D 25 above the axis. Hence,
Z
3
!2
.2x C 4/ dx D 25:
y
10
8
6
4
2
−2
3.
Z
x
−1
1
2
3
1
!2
.3x C 4/ dx
SOLUTION The region bounded by the graph of y D 3x C 4 and the x-axis over the interval Œ!2; 1! consists of two right triangles.
One has area 12 . 23 /.2/ D 32 below the axis, and the other has area 12 . 73 /.7/ D 49
6 above the axis. Hence,
Z
1
!2
.3x C 4/ dx D
49 2
15
! D
:
6
3
2
y
8
6
4
2
−2
4.
Z
−1
x
−2
1
1
4 dx
!2
SOLUTION The region bounded by the graph of y D 4 and the x-axis over the interval Œ!2; 1! is a rectangle of area .3/.4/ D 12
above the axis. Hence,
Z 1
4 dx D 12:
!2
S E C T I O N 5.2
The Definite Integral
573
y
4
3
2
1
−2
5.
Z
x
−1
1
8
.7 ! x/ dx
6
SOLUTION The region bounded by the graph of y D 7 ! x and the x-axis over the interval Œ6; 8! consists of two right triangles.
One triangle has area 12 .1/.1/ D 12 above the axis, and the other has area 12 .1/.1/ D 12 below the axis. Hence,
Z
8
.7 ! x/ dx D
6
1 1
! D 0:
2 2
y
1
0.5
x
2
−0.5
4
6
8
−1
6.
Z
3!=2
sin x dx
!=2
The region bounded by the graph of y D sin x and the x-axis over the interval Π!2 ; 3!
2 ! consists of two parts of equal
area, one above the axis and the other below the axis. Hence,
Z 3!=2
sin x dx D 0:
SOLUTION
!=2
y
1
0.5
x
1
− 0.5
2
3
4
−1
7.
Z
0
5p
25 ! x 2 dx
p
The region bounded by the graph of y D 25 ! x 2 and the x-axis over the interval Œ0; 5! is one-quarter of a circle of
radius 5. Hence,
Z 5p
1
25"
25 ! x 2 dx D ".5/2 D
:
4
4
0
SOLUTION
y
5
4
3
2
1
x
1
8.
Z
2
3
4
5
3
!2
jxj dx
SOLUTION The region bounded by the graph of y D jxj and the x-axis over the interval Œ!2; 3! consists of two right triangles,
both above the axis. One triangle has area 12 .2/.2/ D 2, and the other has area 12 .3/.3/ D 92 . Hence,
Z
3
!2
jxj dx D
13
9
C2 D
:
2
2
574
CHAPTER 5
THE INTEGRAL
y
3
2
1
−2
9.
Z
x
−1
1
2
3
2
!2
.2 ! jxj/ dx
SOLUTION The region bounded by the graph of y D 2 ! jxj and the x-axis over the interval Œ!2; 2! is a triangle above the axis
with base 4 and height 2. Consequently,
Z 2
1
.2 ! jxj/ dx D .2/.4/ D 4:
2
!2
y
2
1
−2
10.
Z
x
−1
1
2
5
!2
.3 C x ! 2jxj/ dx
SOLUTION The region bounded by the graph of y D 3 C x ! 2jxj and the x-axis over the interval Œ!2; 5! consists of a triangle
below the axis with base 1 and height 3, a triangle above the axis of base 4 and height 3 and a triangle below the axis of base 2 and
height 2. Consequently,
Z 5
1
1
1
5
.3 C x ! 2jxj/ dx D ! .1/.3/ C .4/.3/ ! .2/.2/ D :
2
2
2
2
!2
y
3
2
1
−2
4
x
2
−1
−2
−3
11. Calculate
Z
10
0
.8 ! x/ dx in two ways:
(a) As the limit lim RN
N !1
(b) By sketching the relevant signed area and using geometry
R
R
Let f .x/ D 8 ! x over Œ0; 10!. Consider the integral 010 f .x/ dx D 010 .8 ! x/ dx.
(a) Let N be a positive integer and set a D 0, b D 10, #x D .b ! a/ =N D 10=N . Also, let xk D a C k#x D 10k=N ,
k D 1; 2; : : : ; N be the right endpoints of the N subintervals of Œ0; 10!. Then
0 0
1
0
11
"
N
N !
N
N
X
10 X
10k
10 @ @ X A 10 @ X AA
RN D #x
f .xk / D
8!
D
8
1 !
k
N
N
N
N
SOLUTION
kD1
D
N2
N
C
2
2
kD1
!!
D 30 !
kD1
50
:
N
"
50
D 30.
N
N !1
N !1
(b) The region bounded by the graph of y D 8 ! x and the x-axis over the interval Œ0; 10! consists of two right triangles. One
triangle has area 21 .8/.8/ D 32 above the axis, and the other has area 12 .2/.2/ D 2 below the axis. Hence,
Hence lim RN D lim
!
10
10
8N !
N
N
kD1
30 !
Z
10
0
.8 ! x/ dx D 32 ! 2 D 30:
S E C T I O N 5.2
The Definite Integral
575
y
8
6
4
2
10
2
12. Calculate
Z
#
6
x
8
4
!1
SOLUTION
4
.4x ! 8/ dx in two ways: As the limit lim RN and using geometry.
N !1
Let f .x/ D 4x ! 8 over Œ!1; 4!. Consider the integral
Z
4
!1
f .x/ dx D
Z
4
!1
.4x ! 8/ dx.
Let N be a positive integer and set a D !1, b D 4, #x D .b ! a/ =N D 5=N . Then xk D a C k#x D !1 C 5k=N ,
k D 1; 2; : : : ; N are the right endpoints of the N subintervals of Œ!1; 4!. Then
0
1
0
1
"
N
N !
N
N
X
5 X
20k
60 @ X A 100 @ X A
RN D #x
f .xk / D
!4 C
!8 D!
1 C 2
k
N
N
N
N
kD1
D!
kD1
kD1
N2
N
C
2
2
60
100
.N / C 2
N
N
kD1
!
50
50
D !10 C :
N
N
!
"
50
Hence lim RN D lim !10 C
D !10.
N
N !1
N !1
The region bounded by the graph of y D 4x ! 8 and the x-axis over the interval Œ!1; 4! consists of a triangle below the axis
with base 3 and height 12 and a triangle above the axis with base 2 and height 8. Hence,
Z 4
1
1
.4x ! 8/ dx D ! .3/.12/ C .2/.8/ D !10:
2
2
!1
D !60 C 50 C
#
y
−1
5
x
1
−5
2
3
4
−10
In Exercises 13 and 14, refer to Figure 1.
y
y = f(x)
2
4
6
x
FIGURE 1 The two parts of the graph are semicircles.
13. Evaluate: (a)
Z
2
f .x/ dx
0
(b)
Z
6
f .x/ dx
0
Let f .x/ be given by Figure 1.
R
(a) The definite integral 02 f .x/ dx is the signed area of a semicircle of radius 1 which lies below the x-axis. Therefore,
SOLUTION
Z
2
0
R6
"
1
f .x/ dx D ! " .1/2 D ! :
2
2
(b) The definite integral 0 f .x/ dx is the signed area of a semicircle of radius 1 which lies below the x-axis and a semicircle of
radius 2 which lies above the x-axis. Therefore,
Z 6
1
1
3"
f .x/ dx D " .2/2 ! " .1/2 D
:
2
2
2
0
576
CHAPTER 5
14. Evaluate: (a)
Z
THE INTEGRAL
Z
4
(b)
f .x/ dx
1
6
1
jf .x/j dx
Let f .x/ be given by Figure 1.
R
(a) The definite integral 14 f .x/ dx is the signed area of one-quarter of a circle of radius 1 which lies below the x-axis and
one-quarter of a circle of radius 2 which lies above the x-axis. Therefore,
Z 4
1
1
3
f .x/ dx D " .2/2 ! " .1/2 D ":
4
4
4
1
R6
(b) The definite integral 1 jf .x/j dx is the signed area of one-quarter of a circle of radius 1 and a semicircle of radius 2, both of
which lie above the x-axis. Therefore,
Z 6
1
1
9"
:
jf .x/j dx D " .2/2 C " .1/2 D
2
4
4
1
SOLUTION
In Exercises 15 and 16, refer to Figure 2.
y
y = g(t)
2
1
1
−1
2
3
4
5
t
−2
FIGURE 2
15. Evaluate
Z
3
g.t/ dt and
0
Z
5
g.t/ dt .
3
SOLUTION
The region bounded by the curve y D g.t/ and the t-axis over the interval Œ0; 3! is comprised of two right triangles, one with
area 21 below the axis, and one with area 2 above the axis. The definite integral is therefore equal to 2 ! 21 D 32 .
# The region bounded by the curve y D g.t/ and the t-axis over the interval Œ3; 5! is comprised of another two right triangles,
one with area 1 above the axis and one with area 1 below the axis. The definite integral is therefore equal to 0.
Z a
Z c
16. Find a, b, and c such that
g.t/ dt and
g.t/ dt are as large as possible.
#
0
b
a
g.t/ dt as large as possible, we want to include as much positive area as possible. This
Z c
happens when we take a D 4. Now, to make the value of
g.t/ dt as large as possible, we want to make sure to include all of
SOLUTION
To make the value of
Z
0
b
the positive area and only the positive area. This happens when we take b D 1 and c D 4.
17. Describe the partition P and the set of sample points C for the Riemann sum shown in Figure 3. Compute the value of the
Riemann sum.
y
34.25
20
15
8
x
0.5 1
2 2.5 3 3.2
4.5 5
FIGURE 3
SOLUTION
The partition P is defined by
x0 D 0
<
x1 D 1
<
x2 D 2:5
<
x3 D 3:2
<
x4 D 5
The set of sample points is given by C D fc1 D 0:5; c2 D 2; c3 D 3; c4 D 4:5g. Finally, the value of the Riemann sum is
34:25.1 ! 0/ C 20.2:5 ! 1/ C 8.3:2 ! 2:5/ C 15.5 ! 3:2/ D 96:85:
S E C T I O N 5.2
The Definite Integral
577
18. Compute R.f; P; C / for f .x/ D x 2 C x for the partition P and the set of sample points C in Figure 3.
SOLUTION
R.f; P; C / D f .0:5/.1 ! 0/ C f .2/.2:5 ! 1/ C f .3/.3:2 ! 2:5/ C f .4:5/.5 ! 3:2/
D 0:75.1/ C 6.1:5/ C 12.0:7/ C 24:75.1:8/ D 62:7
In Exercises 19–22, calculate the Riemann sum R.f; P; C / for the given function, partition, and choice of sample points. Also,
sketch the graph of f and the rectangles corresponding to R.f; P; C /.
19. f .x/ D x, P D f1; 1:2; 1:5; 2g,
SOLUTION
C D f1:1; 1:4; 1:9g
Let f .x/ D x. With
and
P D fx0 D 1; x1 D 1:2; x2 D 1:5; x3 D 2g
C D fc1 D 1:1; c2 D 1:4; c3 D 1:9g;
we get
R.f; P; C / D #x1 f .c1 / C #x2 f .c2 / C #x3 f .c3 /
D .1:2 ! 1/.1:1/ C .1:5 ! 1:2/.1:4/ C .2 ! 1:5/.1:9/ D 1:59:
Here is a sketch of the graph of f and the rectangles.
y
2
1.5
1
0.5
x
0.5
20. f .x/ D 2x C 3,
SOLUTION
P D f!4; !1; 1; 4; 8g,
Let f .x/ D 2x C 3. With
1
1.5
2
2.5
C D f!3; 0; 2; 5g
and
P D fx0 D !4; x1 D !1; x2 D 1; x3 D 4; x4 D 8g
C D fc1 D !3; c2 D 0; c3 D 2; c4 D 5g;
we get
R.f; P; C / D #x1 f .c1 / C #x2 f .c2 / C #x3 f .c3 / C #x4 f .c4 /
D .!1 ! .!4//.!3/ C .1 ! .!1//.3/ C .4 ! 1/.7/ C .8 ! 4/.13/ D 70:
Here is a sketch of the graph of f and the rectangles.
y
20
15
10
−4
5
−2
x
2
−5
21. f .x/ D x 2 C x, P D f2; 3; 4:5; 5g,
SOLUTION
Let f .x/ D
x2
C x. With
4
6
8
C D f2; 3:5; 5g
and
P D fx0 D 2; x1 D 3; x3 D 4:5; x4 D 5g
C D fc1 D 2; c2 D 3:5; c3 D 5g;
we get
R.f; P; C / D #x1 f .c1 / C #x2 f .c2 / C #x3 f .c3 /
D .3 ! 2/.6/ C .4:5 ! 3/.15:75/ C .5 ! 4:5/.30/ D 44:625:
Here is a sketch of the graph of f and the rectangles.
y
30
25
20
15
10
5
1
2
3
4
5
x
578
CHAPTER 5
THE INTEGRAL
˚
22. f .x/ D sin x, P D 0; !6 ;
SOLUTION
we get
! !
3; 2
#
, C D f0:4; 0:7; 1:2g
Let f .x/ D sin x. With
n
"
"
"o
P D x0 D 0; x1 D ; x3 D ; x4 D
6
3
2
and
C D fc1 D 0:4; c2 D 0:7; c3 D 1:2g;
R.f; P; C / D #x1 f .c1 / C #x2 f .c2 / C #x3 f .c3 /
$"
%
$"
$"
"%
"%
D
! 0 .sin 0:4/ C
!
.sin 0:7/ C
!
.sin 1:2/ D 1:029225:
6
3
6
2
3
Here is a sketch of the graph of f and the rectangles.
y
1
0.8
0.6
0.4
0.2
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
x
In Exercises 23–28, sketch the signed area represented by the integral. Indicate the regions of positive and negative area.
Z 5
23.
.4x ! x 2 / dx
0
SOLUTION
Here is a sketch of the signed area represented by the integral
y
R5
0
.4x ! x 2 / dx.
4
2
5
1
−2
2
3
x
4
−4
24.
Z
!=4
tan x dx
!!=4
SOLUTION
Here is a sketch of the signed area represented by the integral
y
R !=4
!!=4
tan x dx.
1.0
0.5
−0.6
+
−0.2
x
0.2 0.4 0.6
−
−0.5
−1.0
25.
Z
2!
sin x dx
!
SOLUTION
Here is a sketch of the signed area represented by the integral
y
R 2!
!
0.4
x
−0.4
−0.8
−1.2
26.
Z
0
3!
sin x dx
1
2
3
4
5
−
6
7
sin x dx.
S E C T I O N 5.2
Here is a sketch of the signed area represented by the integral
SOLUTION
y
R 3!
0
The Definite Integral
579
sin x dx.
1
+
0.5
+
x
2
−0.5
4
6
−
8
10
−1
27.
Z
2
ln x dx
1=2
Here is a sketch of the signed area represented by the integral
SOLUTION
R2
1=2 ln x dx.
0.6
0.4
+
0.2
0.5
–0.2
–
1
1.5
2
–0.4
–0.6
28.
Z
1
!1
tan!1 x dx
Here is a sketch of the signed area represented by the integral
SOLUTION
y
0.5
−1
!1 tan
!1 x dx.
+
−0.5
−
R1
0.5
1
x
−0.5
In Exercises 29–32, determine the sign of the integral without calculating it. Draw a graph if necessary.
Z 1
29.
x 4 dx
!2
The integrand is always positive. The integral must therefore be positive, since the signed area has only positive part.
SOLUTION
30.
Z
1
x 3 dx
!2
SOLUTION By symmetry, the positive area from the interval Œ0; 1! is cancelled by the negative area from Œ!1; 0!. With the interval
Œ!2; !1! contributing more negative area, the definite integral must be negative.
Z 2!
31.
x sin x dx
0
As you can see from the graph below, the area below the axis is greater than the area above the axis. Thus, the definite
integral is negative.
SOLUTION
y
0.2
−0.2
−0.4
−0.6
32.
Z
2!
0
sin x
dx
x
+
1
2
x
3
4
5
−
6
7
580
THE INTEGRAL
CHAPTER 5
SOLUTION
From the plot below, you can see that the area above the axis is bigger than the area below the axis, hence the integral
is positive.
y
1
0.8
0.6
0.4
0.2
+
1
2
x
4 − 5
3
6
In Exercises 33–42, use properties of the integral and the formulas in the summary to calculate the integrals.
Z 4
33.
.6t ! 3/ dt
0
SOLUTION
34.
Z
2
!3
Z
4
0
.6t ! 3/ dt D 6
Z
4
t dt ! 3
0
Z
0
4
1
.4/2 ! 3.4 ! 0/ D 36.
2
1 dt D 6 "
.4x C 7/ dx
SOLUTION
Z
2
!3
.4x C 7/ dx D 4
Z
9
SOLUTION
Z
SOLUTION
37.
0
1
!3
D4
D4
Z
!
0
x dx C
!3
2
x dx !
0
Z
2
dx
!3
Z
0
Z
!
2
x dx C 7.2 ! .!3//
!3
0
!
x dx C 35
"
1 2 1
2 ! .!3/2 C 35 D 25:
2
2
Z
9
1 3
.9/ D 243:
3
x 2 dx D
x 2 dx
2
Z
By formula (5),
0
5
x dx C 7
x 2 dx
0
36.
2
Z
D4
35.
Z
Z
5
2
x 2 dx D
Z
5
0
x 2 dx !
Z
2
x 2 dx D
0
1
1
.5/3 ! .2/3 D 39.
3
3
.u2 ! 2u/ du
SOLUTION
Z
1
0
38.
Z
0
1=2
.u2 ! 2u/ du D
Z
1
0
u2 du ! 2
Z
1
0
u du D
! "
1 3
1
1
2
.1/ ! 2
.1/2 D ! 1 D ! :
3
2
3
3
.12y 2 C 6y/ dy
SOLUTION
Z
1=2
0
.12y 2 C 6y/ dy D 12
Z
D 12 "
D
1=2
0
1
3
y 2 dy C 6
Z
1=2
y dy
0
! "3
! "
1
1 1 2
C6"
2
2 2
1
3
5
C D :
2
4
4
The Definite Integral
S E C T I O N 5.2
39.
Z
1
!3
.7t 2 C t C 1/ dt
SOLUTION
First, write
Z
1
!3
Z
2
.7t C t C 1/ dt D
0
!3
Z
D!
Z
2
.7t C t C 1/ dt C
!3
0
1
0
.7t 2 C t C 1/ dt
Z
.7t 2 C t C 1/ dt C
1
0
.7t 2 C t C 1/ dt
Then,
!
" !
"
1
1
1
1
.7t 2 C t C 1/ dt D ! 7 " .!3/3 C .!3/2 ! 3 C 7 " 13 C 12 C 1
3
2
3
2
!3
!
" !
"
9
7
1
196
D ! !63 C ! 3 C
C C1 D
:
2
3
2
3
Z
40.
Z
3
!3
1
.9x ! 4x 2 / dx
SOLUTION
First write
Z
3
!3
.9x ! 4x 2 / dx D
Z
0
.9x ! 4x 2 / dx C
!3
D!
Z
!3
0
Z
3
0
.9x ! 4x 2 / dx C
.9x ! 4x 2 / dx
Z
3
0
.9x ! 4x 2 / dx:
Then,
!
" !
"
1
1
1
1
.9x ! 4x 2 / dx D ! 9 " .!3/2 ! 4 " .!3/3 C 9 " .3/2 ! 4 " .3/3
2
3
2
3
!3
!
" !
"
81
81
D!
C 36 C
! 36 D !72:
2
2
Z
41.
Z
1
!a
3
.x 2 C x/ dx
SOLUTION
First,
Z
1
Rb
!a
0
.x 2 C x/ dx D
D
42.
Z
a2
Rb
.x 2 C x/ dx D
Z
0
!a
!
x 2 dx C
0
Rb
0
x dx D 13 b 3 C 12 b 2 . Therefore
.x 2 C x/ dx C
Z
1
0
.x 2 C x/ dx D
Z
0
1
.x 2 C x/ dx !
Z
0
!a
.x 2 C x/ dx
" !
"
1 3 1 2
1
1
1
1
5
"1 C "1 !
.!a/3 C .!a/2 D a3 ! a2 C :
3
2
3
2
3
2
6
x 2 dx
a
SOLUTION
Z
a2
a
x 2 dx D
Z
a2
0
x 2 dx !
Z
a
0
x 2 dx D
In Exercises 43–47, calculate the integral, assuming that
Z 5
f .x/ dx D 5;
0
43.
Z
0
Z
0
5
0
g.x/ dx D 12
5
.f .x/ C g.x// dx
SOLUTION
44.
Z
1 $ 2 %3 1
1
1
a
! .a/3 D a6 ! a3 :
3
3
3
3
5!
Z
5
0
.f .x/ C g.x// dx D
"
1
2f .x/ ! g.x/ dx
3
Z
0
5
f .x/ dx C
Z
5
0
g.x/ dx D 5 C 12 D 17.
581
582
Z
SOLUTION
45.
Z
0
Z
SOLUTION
Z
5!
0
"
Z 5
Z
1
1 5
1
2f .x/ ! g.x/ dx D 2
f .x/ dx !
g.x/ dx D 2.5/ ! .12/ D 6.
3
3 0
3
0
g.x/ dx
5
46.
THE INTEGRAL
CHAPTER 5
5
0
g.x/ dx D !
5
Z
5
g.x/ dx D !12.
0
.f .x/ ! x/ dx
0
Z
SOLUTION
5
.f .x/ ! x/ dx D
0
Z
47. Is it possible to calculate
5
Z
5
0
f .x/ dx !
5
0
1
15
x dx D 5 ! .5/2 D ! .
2
2
g.x/f .x/ dx from the information given?
0
It is not possible to calculate
SOLUTION
Z
R5
0
g.x/f .x/ dx from the information given.
48. Prove by computing the limit of right-endpoint approximations:
Z
b
0
x 3 dx D
b4
4
Let f .x/ D x 3 , a D 0 and #x D .b ! a/=N D b=N . Then
0
1
!
N
N
N
3
X
b X
b4 @ X 3A
b4
3 b
D #x
f .xk / D
k " 3 D 4
k
D 4
N
N
N
N
9
SOLUTION
RN
kD1
Hence
Z
b
0
kD1
kD1
b4
b4
b4
C
C
4
2N
4N 2
3
x dx D lim RN D lim
N !1
N !1
!
D
N4
N3
N2
C
C
4
2
4
!
D
b4
b4
b4
C
C
:
4
2N
4N 2
b4
.
4
In Exercises 49–54, evaluate the integral using the formulas in the summary and Eq. (9).
Z 3
49.
x 3 dx
0
SOLUTION
50.
Z
3
0
1
34
81
D
.
4
4
Z
3
1
x 3 dx D
Z
3
x 3 dx !
0
Z
1
0
x 3 dx D
1 4 1 4
.3/ ! .1/ D 20.
4
4
3
SOLUTION
52.
x 3 dx D
.x ! x / dx
0
Z
3
x 3 dx
SOLUTION
51.
Z
0
3
1
Z
By Eq. (9),
Z
3
0
.x ! x 3 / dx D
Z
3
0
x dx !
Z
3
0
x 3 dx D
1 2 1 4
63
3 ! 3 D! .
2
4
4
.2x 3 ! x C 4/ dx
SOLUTION Applying the linearity of the definite integral, Eq. (9), the formula from Example 4 and the formula for the definite
integral of a constant:
Z 1
Z 1
Z 1
Z 1
1
1
.2x 3 ! x C 4/ dx D 2
x 3 dx !
x dx C
4 dx D 2 " .1/4 ! .1/2 C 4 D 4:
4
2
0
0
0
0
Z 1
53.
.12x 3 C 24x 2 ! 8x/ dx
0
SOLUTION
Z
1
0
.12x 3 C 24x 2 ! 8x/ dx D 12
Z
1
0
x 3 dx C 24
Z
1
0
x2 ! 8
1
1
1
D 12 " 14 C 24 " 13 ! 8 " 12
4
3
2
D 3C8!4D 7
Z
0
1
x dx
The Definite Integral
S E C T I O N 5.2
54.
Z
2
!2
.2x 3 ! 3x 2 / dx
SOLUTION
Z
2
!2
3
2
.2x ! 3x / dx D
D
Z
0
3
.2x ! 3x / dx C
!2
Z
2
0
D2
Z
2
.2x 3 ! 3x 2 / dx !
2
0
x 3 dx ! 3
Z
2
0
Z
Z
2
0
.2x 3 ! 3x 2 / dx
!2
0
.2x 3 ! 3x 2 / dx
x 2 dx ! 2
Z
!2
0
x 3 dx C 3
Z
!2
x 2 dx
0
1
1
1
1
D 2 " .2/4 ! 3 " .2/3 ! 2 " .!2/4 C 3 " .!2/3
4
3
4
3
D 8 ! 8 ! 8 ! 8 D !16:
In Exercises 55–58, calculate the integral, assuming that
Z 1
Z
f .x/ dx D 1;
0
55.
Z
2
SOLUTION
Z
1
SOLUTION
Z
Z
f .x/ dx D
Z
f .x/ dx D
Z
4
0
f .x/ dx C
Z
f .x/ dx !
Z
1
0
4
f .x/ dx D 1 C 7 D 8.
1
Z
2
1
2
0
1
f .x/ dx D 4 ! 1 D 3.
0
f .x/ dx
4
58.
f .x/ dx D 7
1
f .x/ dx
1
57.
4
f .x/ dx
SOLUTION
Z
0
f .x/ dx D 4;
4
0
56.
Z
2
4
Z
Z
1
4
f .x/ dx D !
4
f .x/ dx D !7.
1
f .x/ dx
2
SOLUTION
R4
From Exercise 55,
0
f .x/ dx D 8. Accordingly,
Z
4
2
f .x/ dx D
Z
4
0
f .x/ dx !
Z
2
0
f .x/ dx D 8 ! 4 D 4:
In Exercises 59–62, express each integral as a single integral.
Z 3
Z 7
59.
f .x/ dx C
f .x/ dx
0
SOLUTION
60.
Z
9
SOLUTION
61.
2
3
3
0
f .x/ dx C
f .x/ dx !
2
Z
Z
Z
2
Z
3
f .x/ dx D
Z
7
f .x/ dx.
0
f .x/ dx
Z
f .x/ dx D
Z
f .x/ dx D
Z
9
4
f .x/ dx C
Z
f .x/ dx C
Z
4
2
9
4
!
Z
!
Z
f .x/ dx !
f .x/ dx D
Z
f .x/ dx D
Z
9
4
4
f .x/ dx:
2
5
f .x/ dx
2
9
2
7
4
f .x/ dx !
f .x/ dx !
SOLUTION
9
9
9
Z
Z
Z
f .x/ dx !
Z
5
2
5
2
9
5
f .x/ dx !
5
2
9
5
f .x/ dx:
583
584
62.
THE INTEGRAL
CHAPTER 5
Z
3
f .x/ dx C
7
SOLUTION
Z
Z
9
f .x/ dx
3
3
f .x/ dx C
7
Z
Z
9
f .x/ dx D !
3
7
f .x/ dx C
3
Z
7
3
f .x/ dx C
In Exercises 63–66, calculate the integral, assuming that f is integrable and
63.
Z
5
SOLUTION
65.
6
SOLUTION
66.
Z
9
f .x/ dx:
7
f .x/ dx D 1 ! b !1 for all b > 0.
Z
5
4
.
5
f .x/ dx D 1 ! 5!1 D
1
Z
5
f .x/ dx D
3
Z
5
1
f .x/ dx !
Z
3
1
!
" !
"
1
1
2
f .x/ dx D 1 !
! 1!
D
.
5
3
15
.3f .x/ ! 4/ dx
1
Z
7
D
f .x/ dx
3
Z
1
f .x/ dx
!
f .x/ dx
SOLUTION
Z
b
9
5
1
64.
Z
Z
1
Z
6
1
.3f .x/ ! 4/ dx D 3
Z
6
1
f .x/ dx ! 4
Z
6
1
1 dx D 3.1 ! 6!1 / ! 4.6 ! 1/ D !
35
.
2
f .x/ dx
1=2
! "!1 !
1
SOLUTION
f .x/ dx D !
f .x/ dx D ! 1 !
D 1.
2
1=2
1
Z b
Z
67.
Explain the difference in graphical interpretation between
f .x/ dx and
Z
1
Z
1=2
a
b
a
jf .x/j dx.
Rb
When f .x/ takes on both positive and negative values on Œa; b!, a f .x/ dx represents the signed area between f .x/
Rb
and the x-axis, whereas a jf .x/j dx represents the total (unsigned) area between f .x/ and the x-axis. Any negatively signed
Rb
Rb
areas that were part of a f .x/ dx are regarded as positive areas in a jf .x/j dx. Here is a graphical example of this phenomenon.
SOLUTION
Graph of | f (x)|
Graph of f (x)
10
−4
−2
−10
30
2
4
x
20
10
−20
−30
68.
−4
−2
2
4
x
Use the graphical interpretation of the definite integral to explain the inequality
ˇZ
ˇ Z
ˇ b
ˇ
b
ˇ
ˇ
f .x/ dx ˇ $
jf .x/j dx
ˇ
ˇ a
ˇ
a
where f .x/ is continuous. Explain also why equality holds if and only if either f .x/ # 0 for all x or f .x/ $ 0 for all x.
SOLUTION Let AC denote the unsigned area under the graph of y D f .x/ over the interval Œa; b! where f .x/ # 0. Similarly, let
A! denote the unsigned area when f .x/ < 0. Then
Z
b
a
f .x/ dx D AC ! A! :
Moreover,
ˇZ
ˇ
Z b
ˇ b
ˇ
ˇ
ˇ
f .x/ dx ˇ $ AC C A! D
jf .x/j dx:
ˇ
ˇ a
ˇ
a
Equality holds if and only if one of the unsigned areas is equal to zero; in other words, if and only if either f .x/ # 0 for all x or
f .x/ $ 0 for all x.
S E C T I O N 5.2
69.
SOLUTION
Let f .x/ D x. Find an interval Œa; b! such that
ˇZ
ˇ
ˇ b
ˇ
1
ˇ
ˇ
f .x/ dx ˇ D
ˇ
ˇ a
ˇ
2
Z
and
b
a
jf .x/j dx D
The Definite Integral
585
3
2
If a > 0, then f .x/ # 0 for all x 2 Œa; b!, so
ˇZ
ˇ Z
ˇ b
ˇ
b
ˇ
ˇ
f .x/ dx ˇ D
jf .x/j dx
ˇ
ˇ a
ˇ
a
by the previous exercise. We find a similar result if b < 0. Thus, we must have a < 0 and b > 0. Now,
Z
b
jf .x/j dx D
a
1 2 1 2
a C b :
2
2
Because
Z
b
a
f .x/ dx D
1 2 1 2
b ! a ;
2
2
then
ˇZ
ˇ
ˇ b
ˇ
1
ˇ
ˇ
f .x/ dx ˇ D jb 2 ! a2 j:
ˇ
ˇ a
ˇ
2
If b 2 > a2 , then
1 2 1 2
3
a C b D
2
2
2
yield a D !1 and b D
1 2
1
.b ! a2 / D
2
2
and
1 2
1
.a ! b 2 / D
2
2
p
2. On the other hand, if b 2 < a2 , then
1 2 1 2
3
a C b D
2
2
2
p
yield a D ! 2 and b D 1.
Z
70.
Evaluate I D
that I C J D 2".
and
2!
0
sin2 x dx and J D
Z
2!
0
cos2 x dx as follows. First show with a graph that I D J . Then prove
The graphs of f .x/ D sin2 x and g.x/ D cos2 x are shown below at the left and right, respectively. It is clear that
the shaded areas in the two graphs are equal, thus
Z 2!
Z 2!
I D
sin2 x dx D
cos2 x dx D J:
SOLUTION
0
0
Now, using the fundamental trigonometric identity, we find
Z 2!
Z
2
2
I CJ D
.sin x C cos x/ dx D
0
2!
0
1 " dx D 2":
Combining this last result with I D J yields I D J D ".
y
y
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
x
1
In Exercises 71–74, calculate the integral.
Z 6
71.
j3 ! xj dx
0
2
3
4
5
6
x
1
2
3
4
5
6
586
THE INTEGRAL
CHAPTER 5
SOLUTION Over the interval, the region between the curve and the interval Œ0; 6! consists of two triangles above the x axis, each
of which has height 3 and width 3, and so area 92 . The total area, hence the definite integral, is 9.
y
3
2
1
x
1
2
3
4
5
6
Alternately,
Z
6
0
j3 ! xj dx D
Z
72.
Z
.3 ! x/ dx C
0
D3
Z
3
Z
Z
3
dx !
0
6
.x ! 3/ dx
3
Z
3
0
x dx C
6
x dx !
0
Z
3
0
!
x dx ! 3
Z
6
dx
3
1
1
1
D 9 ! 32 C 62 ! 32 ! 9 D 9:
2
2
2
3
j2x ! 4j dx
1
SOLUTION The area between j2x ! 4j and the x axis consists of two triangles above the x-axis, each with width 1 and height 2,
and hence with area 1. The total area, and hence the definite integral, is 2.
y
2
1.5
1
0.5
x
0.5
1
1.5
2
2.5
3
Alternately,
Z
3
j2x ! 4j dx D
1
Z
73.
.4 ! 2x/ dx C
1
D4
Z
2
Z
1
D4!2
1
!1
Z
2
dx ! 2
!
Z
3
2
.2x ! 4/ dx
2
0
x dx !
Z
1
!
Z
x dx C 2
0
3
0
x dx !
"
!
"
1 2 1 2
1 2 1 2
2 ! 1 C2
3 ! 2 ! 4 D 2:
2
2
2
2
Z
2
0
!
x dx ! 4
Z
3
dx
2
jx 3 j dx
SOLUTION
3
(
x 3 dx D
Z
jx j D
x3
!x 3
x#0
x < 0:
Therefore,
74.
Z
0
Z
1
!1
2
jx 3 j dx D
Z
0
!1
!x 3 dx C
Z
1
0
!1
0
x 3 dx C
Z
0
1
x 3 dx D
2
jx ! 1j dx
SOLUTION
2
jx ! 1j D
(
x2 ! 1
!.x 2 ! 1/
1$x$2
0 $ x < 1:
Therefore,
Z
0
2
jx 2 ! 1j dx D
Z
1
0
.1 ! x 2 / dx C
Z
2
1
.x 2 ! 1/ dx
1
1
1
.!1/4 C .1/4 D :
4
4
2
The Definite Integral
S E C T I O N 5.2
D
Z
1
0
dx !
Z
1
75. Use the Comparison Theorem to show that
Z 1
Z
x 5 dx $
0
SOLUTION
x dx C
0
1
D 1 ! .1/ C
3
1
Z
2
!
x dx !
0
Z
x 4 dx;
2
1
On the interval Œ0; 1!, x 5 $ x 4 , so, by Theorem 5,
1
0
2
x 5 dx $
Z
1
x 4 dx $
Z
1
0
"
1
1
.8/ ! .1/ ! 1 D 2:
3
3
0
Z
2
Z
2
2
!
x dx !
Z
587
2
1 dx
1
x 5 dx
1
x 4 dx:
0
On the other hand, x 4 $ x 5 for x 2 Œ1; 2!, so, by the same Theorem,
Z 2
Z 2
x 4 dx $
x 5 dx:
76. Prove that
SOLUTION
1
$
3
Z
1
6
1
1
1
dx $ .
x
2
4
On the interval Œ4; 6!,
1
6
$
1
x,
so, by Theorem 5,
Z
1
D
3
On the other hand,
1
x
1
4
$
1
dx $
6
4
Z
6
4
1
dx:
x
on the interval Œ4; 6!, so
Z
6
4
R6
1
dx $
x
Z
6
4
1
1
1
dx D .6 ! 4/ D :
4
4
2
dx $ 12 , as desired.
R 0:3
77. Prove that 0:0198 $ 0:2
sin x dx $ 0:0296. Hint: Show that 0:198 $ sin x $ 0:296 for x in Œ0:2; 0:3!.
Therefore
1
3
6
$
1
4 x
SOLUTION For 0 $ x $
0:2 $ x $ 0:3, we have
!
6
% 0:52, we have
d
dx
.sin x/ D cos x > 0. Hence sin x is increasing on Œ0:2; 0:3!. Accordingly, for
m D 0:198 $ 0:19867 % sin 0:2 $ sin x $ sin 0:3 % 0:29552 $ 0:296 D M
Therefore, by the Comparison Theorem, we have
Z 0:3
Z
0:0198 D m.0:3 ! 0:2/ D
m dx $
0:2
78. Prove that 0:277 $
SOLUTION
Z
0:3
0:2
sin x dx $
!=4
!=8
Z
0:3
0:2
M dx D M.0:3 ! 0:2/ D 0:0296:
cos x dx $ 0:363.
cos x is decreasing on the interval Œ"=8; "=4!. Hence, for "=8 $ x $ "=4,
Since cos."=4/ D
p
cos."=4/ $ cos x $ cos."=8/:
2=2,
p
Z !=4 p
Z !=4
"
2
2
0:277 $ "
D
dx $
cos x dx:
8 2
2
!=8
!=8
Since cos."=8/ $ 0:924,
Z
!=4
!=8
Therefore 0:277 $
R !=4
!=8
cos x $ 0:363:
cos x dx $
Z
!=4
!=8
0:924 dx D
"
.0:924/ $ 0:363:
8
588
THE INTEGRAL
CHAPTER 5
79. Prove that 0 $
SOLUTION
Z
!=2
!=4
p
sin x
2
dx $
.
x
2
Let
f .x/ D
sin x
:
x
As we can seepin the sketch below, f .x/ is decreasing on the interval Œ"=4; "=2!. Therefore f .x/ $ f ."=4/ for all x in Œ"=4; "=2!.
2 2
! ,
f ."=4/ D
so:
Z
!=2
!=4
sin x
dx $
x
Z
!=2
!=4
p
p
p
2 2
"2 2
2
dx D
D
:
"
4 "
2
y
y = sin x
2 2/
x
2/
x
/4
80. Find upper and lower bounds for
SOLUTION
Z
1
0
Let
/2
dx
p
.
5x 3 C 4
f .x/ D p
1
5x 3 C 4
:
f .x/ is decreasing for x on the interval Œ0; 1!, so f .1/ $ f .x/ $ f .0/ for all x in Œ0; 1!. f .0/ D
Z
1
0
1
dx $
3
1
$
3
Z
Z
1
0
1
0
f .x/ dx $
Z
1
0
1
2
and f .1/ D 13 , so
1
dx
2
1
f .x/ dx $ :
2
81.
Suppose that f .x/ $ g.x/ on Œa; b!. By the Comparison Theorem,
f 0 .x/ $ g 0 .x/ for x 2 Œa; b!? If not, give a counterexample.
Rb
a
f .x/ dx $
Rb
a
g.x/ dx. Is it also true that
The assertion f 0 .x/ $ g 0 .x/ is false. Consider a D 0, b D 1, f .x/ D x, g.x/ D 2. f .x/ $ g.x/ for all x in the
interval Œ0; 1!, but f 0 .x/ D 1 while g 0 .x/ D 0 for all x.
SOLUTION
82.
State whether true or false. If false, sketch the graph of a counterexample.
Z b
(a) If f .x/ > 0, then
f .x/ dx > 0.
(b) If
Z
a
b
f .x/ dx > 0, then f .x/ > 0.
a
SOLUTION
Rb
(a) It is true that if f .x/ > 0 for x 2 Œa; b!, then a f .x/ dx > 0.
Rb
R1
(b) It is false that if a f .x/ dx > 0, then f .x/ > 0 for x 2 Œa; b!. Indeed, in Exercise 3, we saw that !2 .3x C 4/ dx D 7:5 > 0,
yet f .!2/ D !2 < 0. Here is the graph from that exercise.
y
6
4
2
−2
−1
x
−2
1
The Definite Integral
S E C T I O N 5.2
589
Further Insights and Challenges
83. ZExplain graphically: If f .x/ is an odd function, then
a
!a
f .x/ dx D 0.
SOLUTION If f is an odd function, then f .!x/ D !f .x/ for all x. Accordingly, for every positively signed area in the right
half-plane where f is above the x-axis, there is a corresponding negatively signed area in the left half-plane where f is below
the x-axis. Similarly, for every negatively signed area in the right half-plane where f is below the x-axis, there is a corresponding
positively signed area in the left half-plane where f is above the x-axis. We conclude that the net area between the graph of f and
the x-axis over Œ!a; a! is 0, since the positively signed areas and negatively signed areas cancel each other out exactly.
y
4
−1
−2
2
1
−2
2
x
−4
84. Compute
Z
1
!1
SOLUTION
sin.sin.x//.sin2 .x/ C 1/ dx.
Let f .x/ D sin.sin.x//.sin2 .x/ C 1//. sin x is an odd function, while sin2 x is an even function, so:
f .!x/ D sin.sin.!x//.sin2 .!x/ C 1/ D sin.! sin.x//.sin2 .x/ C 1/
D ! sin.sin.x//.sin2 .x/ C 1/ D !f .x/:
Therefore, f .x/ is an odd function. The function is odd and the interval is symmetric around the origin so, by the previous exercise,
the integral must be zero.
85. Let k and b be positive. Show, by comparing the right-endpoint approximations, that
Z
b
0
k
x dx D b
kC1
Z
1
x k dx
0
Rb
Let k and b be any positive numbers. Let f .x/ D x k on Œ0; b!. Since f is continuous, both 0 f .x/ dx and
0 f .x/ dx exist. Let N be a positive integer and set #x D .b ! 0/ =N D b=N . Let xj D a C j#x D bj=N , j D 1; 2; : : : ; N
Rb
Rb
be the right endpoints of the N subintervals of Œ0; b!. Then the right-endpoint approximation to 0 f .x/ dx D 0 x k dx is
0
1
"
N
N !
N
X
b X bj k
1 X kA
kC1 @
RN D #x
f .xj / D
Db
j
:
N
N
N kC1 j D1
j D1
j D1
SOLUTION
R1
In particular, if b D 1 above, then the right-endpoint approximation to
SN D #x
N
X
f .xj / D
j D1
R1
0
f .x/ dx D
R1
0
x k dx is
"
N !
N
X
1 X j k
1
1
D
jk D
R
kC1
kC1 N
N
N
N
b
j D1
j D1
In other words, RN D b kC1 SN . Therefore,
Z
b
0
x k dx D lim RN D lim b kC1 SN D b kC1 lim SN D b kC1
N !1
N !1
86. Verify for 0 $ b $ 1 by interpreting in terms of area:
Z
0
b
N !1
Z
1
x k dx:
0
p
1 p
1
1 ! x 2 dx D b 1 ! b 2 C sin!1 b
2
2
p
The function f .x/ D 1 ! x 2 is the quarter circle of radius 1 in the first quadrant. For 0 $ b $ 1, the area
p
Rbp
represented by the integral 0 1 ! x 2 dx can be divided into two parts. The area of the triangular part is 12 .b/ 1 ! b 2 using the
Pythagorean Theorem. The area of the sector with angle $ where sin $ D b, is given by 12 .1/2 .$/. Thus
SOLUTION
Z
b
0
p
1 p
1
1 p
1
1 ! x 2 dx D b 1 ! b 2 C $ D b 1 ! b 2 C sin!1 b:
2
2
2
2
590
CHAPTER 5
THE INTEGRAL
y
1
θ
x
1
b
87.
Suppose that f and g are continuous functions such that, for all a,
Z
a
!a
f .x/ dx D
Z
a
g.x/ dx
!a
Give an intuitive argument showing that f .0/ D g.0/. Explain your idea with a graph.
Ra
Ra
SOLUTION Let !a f .x/ dx D !a g .x/ dx. Consider what happens as a decreases in size, becoming very close to zero.
Intuitively, the areas of the functions become .a ! .!a//.f .0// D 2a.f .0// and .a ! .!a//.g.0// D 2a.g.0//. Because we know
these areas must be the same, we have 2a.f .0// D 2a.g.0// and therefore f .0/ D g.0/.
88. Theorem 4 remains true without the assumption a $ b $ c. Verify this for the cases b < a < c and c < a < b.
Rc
Rb
Rc
SOLUTION The additivity property of definite integrals states for a $ b $ c, we have a f .x/ dx D a f .x/ dx C b f .x/ dx.
Rc
Ra
Rc
# Suppose that we have b < a < c. By the additivity property, we have
b f .x/ dx D b f .x/ dx C a f .x/ dx. Therefore,
Rc
Rc
Ra
Rb
Rc
a f .x/ dx D b f .x/ dx ! b f .x/ dx D a f .x/ dx C b f .x/ dx.
Rb
Ra
Rb
# Now suppose that we have c < a < b. By the additivity property, we have
c f .x/ dx D c f .x/ dx C a f .x/ dx.
Rc
Ra
Rb
Rb
Rb
Rc
Therefore, a f .x/ dx D ! c f .x/ dx D a f .x/ dx ! c f .x/ dx D a f .x/ dx C b f .x/ dx.
# Hence the additivity property holds for all real numbers a, b, and c, regardless of their relationship amongst each other.
5.3 The Fundamental Theorem of Calculus, Part I
Preliminary Questions
1. Suppose that F 0 .x/ D f .x/ and F .0/ D 3, F .2/ D 7.
(a) What is the area under y D f .x/ over Œ0; 2! if f .x/ # 0?
(b) What is the graphical interpretation of F .2/ ! F .0/ if f .x/ takes on both positive and negative values?
SOLUTION
(a) If f .x/ # 0 over Œ0; 2!, then the area under y D f .x/ is F .2/ ! F .0/ D 7 ! 3 D 4.
(b) If f .x/ takes on both positive and negative values, then F .2/ ! F .0/ gives the signed area between y D f .x/ and the x-axis.
2. Suppose that f .x/ is a negative function with antiderivative F such that F .1/ D 7 and F .3/ D 4. What is the area (a positive
number) between the x-axis and the graph of f .x/ over Œ1; 3!?
Z 3
SOLUTION
f .x/ dx represents the signed area bounded by the curve and the interval Œ1; 3!. Since f .x/ is negative on Œ1; 3!,
1
Z 3
f .x/ dx is the negative of the area. Therefore, if A is the area between the x-axis and the graph of f .x/, we have:
1
AD!
3.
(a)
(b)
(c)
Z
3
1
f .x/ dx D ! .F .3/ ! F .1// D !.4 ! 7/ D !.!3/ D 3:
Are the following statements true or false? Explain.
FTC I is valid only for positive functions.
To use FTC I, you have to choose the right antiderivative.
If you cannot find an antiderivative of f .x/, then the definite integral does not exist.
SOLUTION
(a) False. The FTC I is valid for continuous functions.
(b) False. The FTC I works for any antiderivative of the integrand.
(c) False. If you cannot find an antiderivative of the integrand, you cannot use the FTC I to evaluate the definite integral, but the
definite integral may still exist.