174 Chapter 6. Integration
Area between Graphs of Functions Let f andg be functions that are continuous on a closed and bounded interval [a,b]. Sup- pose that f(x) ≤g(x) for allx ∈[a,b] (this means that the graph of f lies below that of g). Then the area Aof the region that is bounded by the graphs of fandgand the vertical linesx=aand x=bis given by
A= Z b
a
hg(x)− f(x)i
dx. a
b A
y= f(x) y=g(x)
Figure 6.11
Proof For the case where f is nonnegative (hence the graphs of both f andgare above the x-axis), we have A = Ag−Af, where Ag(respectivelyAf) is the area of the region that lies below the graph ofg(respectively the graph of f) and above thex-axis fromx = ato x = b(see Figure 6.12a). Hence, using rules for definite integrals (Int1) and (Int2), we have
A= Z b
a
g(x) dx− Z b
a
f(x) dx= Z b
a
hg(x)− f(x)i dx.
a b
Ag
Af
Figure 6.12a
y=g(x) y=g1(x)
y= f(x) y= f1(x)
Figure 6.12b
For the general case, we can move the region upward suitably so that the graph of f is above thex-axis and then apply the result for the case where f is nonnegative (see Figure 6.12b). Indeed, since f is continuous on [a,b], there exists a constantksuch that f(x)+k≥0 for allx∈[a,b]. Let f1andg1be the functions from [a,b]
intoRgiven by
f1(x)= f(x)+k and g1(x)=g(x)+k fora≤ x≤b.
Since area is translation invariant, the required areaAis equal to the area of the region that is bounded by the graphs of f1andg1and the vertical linesx=aandx=b. Hence by what we obtain for the special case (since
f1is nonnegative), we have A=
Z b
a
hg1(x)− f1(x)i dx=
Z b
a
hg(x)− f(x)i dx.
Example Find the area of the region bounded by the parabolay=x2and the liney= x+2.
6.4. Application of Integration 175
ExplanationThe parabola and the line divide the plane into five regions—
four of them are unbounded (R1,R2,R3andR5) and one of them is bounded (R4, the required one).
Solution Solving for the x-coordinates of the intersection points of the parabola and the line:
x2 = x+2 x2−x−2 = 0 (x−2)(x+1) = 0, we getx1 =−1 andx2=2.
R1
R2
R3
R4 R5
Figure 6.13a
-1 2
A
y=x2 y=x+2
Figure 6.13b
The region under consideration lies below the graph ofy = x+2, above that ofy = x2 (and between the vertical linesx=−1 andx=2). The areaAof the region is
A = Z 2
−1
h(x+2)−x2i dx
=
"
x2
2 +2x− x3 3
#2
−1
= 2+4− 8 3
!
− 1
2−2+ 1 3
!
= 9
2 (square units).
Example Find the area of the combined region bounded by the curvey=x3−5x2+6xand thex-axis.
ExplanationThe curve and the x-axis divide the plane into six regions—
four of them are unbounded (R1,R3,R4 and R5) and two of them are bounded (R2 andR6). The two bounded regions intersect at one point and their union forms a combined region. The question is to find the area of R2∪R6.
R1
R2 R3
R4 R5
R6
Figure 6.14a
Solution Solving for the x-coordinates of the intersection points of the curve and thex-axis:
x3−5x2+6x = 0 x(x2−5x−6) = 0 x(x−2)(x−3) = 0, we getx1 =0,x2=2 andx3 =3.
The required areaAisA=A1+A2(see Figure 6.14b).
Note that for 0≤ x≤2, the curve is above thex-axis, for 2≤ x≤3, thex-axis is above the curve.
2 3
y=x3−5x2+6x
A1
A2 Figure 6.14b
176 Chapter 6. Integration
Therefore, we have
A = Z 2
0
(x3−5x2+6x)−0 dx+
Z 3
2
0−(x3−5x2+6x)] dx
=
"
x4 4 − 5x3
3 +3x2
#2
0
−
"
x4 4 − 5x3
3 +3x2
#3
2
= 8 3 −0
!
− 9 4− 8
3
!
= 37 12.
Next we will give some examples that can be done using definite integrals as well as indefinite integrals.
Before that, we give a result that is also known as the Fundamental Theorem of Calculus.
Fundamental Theorem of Calculus, Version 3 Let f be a function such that f0 is continuous on an open interval (a,b). Then for everyx0∈(a,b), we have
f(x)= Z x
x0
f0(t) dt+ f(x0) for allx∈(a,b).
Proof Letgbe the function from (a,b) intoRdefined by g(x)=
Z x
x0
f0(t) dt fora<x<b.
From the Fundamental Theorem of Calculus (Version 1), we see thatgis an antiderivative for f0on (a,b). Since f is also an antiderivative of f0on (a,b), it follows from Theorem 6.3.1 that there exists a constantksuch that
f(x)−g(x)=k for allx∈(a,b).
Putting x = x0, we get f(x0)−g(x0) = kwhich yieldsk = f(x0) sinceg(x0) = Rx0
x0 f(t) dt = 0. Therefore we have
f(x)=g(x)+ f(x0) for all x∈(a,b)
and the required result follows.
Example Find an equation for the curve that passes through the point (1,0) and has slope function given by x3−2x+1.
Solution Let the curve be given byy = f(x). Since the curve passes through the point (1,0), it follows that f(1) = 0. From the given slope function, we have f0(x) = x3−2x+1. Taking x0 = 1 in the Fundamental Theorem of Calculus (Version 3), we have
f(x) = Z x
1
f0(t) dt+ f(1) = Z x
1
(t3−2t+1) dt+0
=
"
t4
4 −t2+t
#x
1
= x4
4 −x2+x
!
− 1
4−1+1
!
= x4
4 −x2+x− 1 4.
6.4. Application of Integration 177
Therefore, an equation for the curve is: y= x4
4 −x2+x− 1
4.
Alternative solutionThe function f can also be found using indefinite integral:
f(x)= Z
(x3−2x+1) dx= x4
4 −x2+x+c
where the first inequality means that f is an antiderivative for (x3−2x+1) and the second equality means that f is the function given by x4
4 −x2+x+candcis a specific constant (which is determined by f).
Puttingx=1, we get
0= f(1)= 1
4 −1+1+c, that is,c=−1
4. Therefore, we have f(x)= x4
4 −x2+x− 1
4.
Example Find the cost function if the marginal cost is 3+40x−5x2and the fixed cost is 45.
ExplanationFixed cost is the cost whenx=0.
Solution
(Method 1) Let the cost function beC. By the Fundamental Theorem of Calculus (Version 3), we have C(x) =
Z x
0
C0(t) dt+C(0)
= Z x
0
(3+40t−5t2) dt+45
=
"
3t+20t2−5 3t3
#x
0
+45
= 3x+20x2− 5
3x3+45.
(Method 2) Let the cost function beC. Then we have C(x)=
Z
(3+40x−5x2) dx=3x+20x2− 5 3x3+c, for some constantc. Puttingx=0, we get 45=C(0)=cand so
C(x)=3x+20x2− 5
3x3+45.
The following result is a simple consequence of the Fundamental Theorem of Calculus (Version 3). It is known as theNet Change Theoremsince f(x1)− f(x0) is the net change of the values of f asxchanges from x0tox1.
Theorem 6.4.1 Let fbe a function such that f0is continuous on an open interval (a,b). Then for every pair of numbersx0,x1in (a,b), we have Z x1
x0
f0(t) dt= f(x1)− f(x0).
178 Chapter 6. Integration
Example A particle moves along a line so that its velocity at timetisv(t) = t2−t(measured in meters per second). Find the displacement of the particle during the time period 1≤t≤2.
ExplanationThe question is to finds(2)−s(1), wheres(t) is the position of the particle at timet. Note that the derivative ofsisv.
Solution
(Method 1) By Theorem 6.4.1, the required displacement is s(2)−s(1) =
Z 2
1
s0(t) dt
= Z 2
1
(t2−t) dt
=
"
t3 3 −t2
2
#2
1
= 8 3 −2
!
− 1 3− 1
2
!
= 5
6 (meter).
(Method 2) The displacement function of the particle is given by s(t)=
Z
(t2−t) dt= t3 3 − t2
2 +c, wherecis the initial position of the particle. The required displacement is
s(2)−s(1)= 8
3−2+c
!
− 1 3 − 1
2+c
!
= 5
6 (meter).
Exercise 6.4
1. For each of the following, find the area of the region bounded by the given curve, the x-axis, and the given vertical line(s).
(a) y= x3, x=3
(b) y= x2−4x, x=1, x=2 (c) y=|x+1|+2, x=−2, x=3 (d) y=√
x+3, x=1 Hint: move the region appropriately.
2. For each of the following, find the area of the (combined) region bounded by the given curves (or lines).
(a) y=√
x and y= x
(b) y= x2−4x−8 and y=2x−x2 (c) y= x and y= x(x−2)2
(d) y= x2−4x+4, y=10−x2 and y=16
3. Suppose f is a function such that f0(x)= x2+1 and f(1)=2. Find f(x).
4. Suppose f is a function such that f00(x)=(x+1)(x−2), f(0)=1 and f(1)=0. Find f(x).
5. Water flows from the bottom of a storage tank at a rate of r(t) = 150− 5t liters per minute, where 0≤t≤30. Find the amount of water that flows from the tank during the first 15 minutes.
Chapter 7
Trigonometric Functions
7.1 Angles
Idea of DefinitionAnangleis formed by rotating a ray about its endpoint.
• The initial position of the ray is called theinitial side.
• The endpoint of the ray is called thevertex.
• The final position is called theterminal side.
An angle is said to be instandard positionif its vertex is at the origin and its initial side is along the positive x-axis.
Note An angle in standard position is uniquely determined by the direction and magnitude of rotation. So we can use numbers to represent angles.
• The direction of rotation may be counterclockwise or clockwise which will be considered to be positive or negative respectively.
• Magnitudes of rotation are traditionally measured indegreeswhere one revolution is defined to be 360 degrees, written 360◦.
Figures 7.1(a), (b) and (c) show three angles in standard position: Although the angles have the same terminal sides, their measures are different.
60◦
Figure 7.1(a)
−300◦
Figure 7.1(b)
420◦
Figure 7.1(c)
Another unit for measuring angles is the radian. To define radian, we consider unit circles.
180 Chapter 7. Trigonometric Functions
Terminology A circle with radius 1 is called aunit circle. The circle with radius 1 and center at the origin is calledthe unit circle.
Definition The angle determined by an arc of length 1 along the circum- ference of a unit circle is said to be of measureone radian.
1 1
1 rad.
length=1
Figure 7.2
Since the circumference of a unit circle has length 2π, there are 2πradians in one revolution. Therefore, we have 360◦ =2πradians. The conversion between degrees and radians is given by
d◦=d× π
180 radians Thus, we have 90◦ = π
2 (radian) and 60◦= π
3 (radian) for example.
RemarkIn calculus, it is more convenient to consider angles in radians and the unitradianis usually omitted.
Exercise 7.1
1. Convert the following degree measures to radians:
(a) 270◦ (b) 210◦
(c) 315◦ (d) 750◦
2. Convert the following radian measures to degrees:
(a) π
6 (b) 3π
4
(c) 5π
2 (d) 7π