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Math 116 Calculus II
Contents
5
6
Exponential and Logarithmic functions
2
5.1
Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
5.1.1
Exponential functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
5.1.2
Logarithmic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
5.1.3
Relation between Exp and Log functions . . . . . . . . . . . . . . . . . . . . .
4
Integration
5
6.1
Antiderivatives and the Rules of integration . . . . . . . . . . . . . . . . . . . . . . . .
5
6.2
Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
6.3
Area and the definite integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
6.4
The fundamental theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
6.5
Evaluating definite integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
6.6
Area between two curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
6.7
Applications of the Definite integral to Business and Economics . . . . . . . . . . . . .
23
1
Chapter 5
Exponential and Logarithmic functions
5.1
5.1.1
Review
Exponential functions
Definition:
• e = 2.7182818.
• f (x) = ex .
Properties:
• domain: (−∞, ∞).
• Range: (0, +∞).
• f (0) = 1.
• Continuous.
• increasing.
Laws of exponents:
• ex ∗ ey = ex+y ;
•
ex
= ex−y ;
ey
2
• (ex )y = exy ;
Differentiation:
•
d x
e = ex ;
dx
•
d f (x)
e
= ef (x) ∗ f 0 (x);
dx
Example 1 Calculate e−10 , e−2 , e0 ,e1 , e10 using calculator.
Example 2 Graph f (x) = e2x ,
[−5, 5] × [0, 100] using calculator.
Example 3 Find the derivative of f (x) = 2xe3x .
Example 4 Determine the intervals where f (x) = e−x
5.1.2
2 /2
is increasing and where it is decreasing.
Logarithmic functions
Definition: Log functions are introduced as inverse functions of exponential functions.
y = ln x ⇔ x = ey .
”ln” denotes the natural log
0 = ln 1 because e0 = 1.
Properties:
• domain (0, +∞).
• Range (−∞, ∞).
• ln 1 = 0, because e0 = 1.
• ln e = 1, because e1 = e.
• ln ex = x, eln x = x. (ln and e can cancel each other. )
• continuous.
• increasing.
Laws of logarithms:
• ln(xy) = ln x + ln y,
(x > 0, y > 0).
3
x
• ln( ) = ln x − ln y(x > 0, y > 0).
y
• ln xy = y ln x.
Differentiation:
•
1
d
ln x =
dx
x
•
d
1 0
ln f (x) =
f (x).
dx
f (x)
Example 5: Calculate ln(0.001), ln(0.01), ln e, ln(1000) by using calculator.
Example 6: Sketch f (x) = ln 2x using calculator.
Example 7: Find f 0 (x) for f (x) = ln(2x + 1).
Example 8: Find f 0 (x) for y = 3x using logarithmic functions.
5.1.3
Relation between Exp and Log functions
• y = ln x ⇔ ey = x.
• eln x = x.
• ln ex = x;
Example 9: Solve 4t−1 = 4;
Example 10: Solve 2e−0.2t − 4 = 6;
Example 11: Solve 3t−1 = 4;
take natural log at both sides, ln 3t−1 = ln 4, t − 1 ln 3 = ln 4, t =
4
ln 4
+1
ln 3
Chapter 6
Integration
6.1
Antiderivatives and the Rules of integration
Example1: the distance that a car travels from its initial position is known to be
d(t) = 4t2
(0 ≤ t ≤ 30)
Find its velocity at any time t, (0 ≤ t ≤ 30).
Soln: v(t) = d0 (t) = 8t.
This process is called differentiation.
Often the velocity of a car can be read recorded from its speedometer. If we know
v(t) = 8t
Can we find the distance that a car travels from its initial position?
This is equivalent to finding d(t) such that
d0 (t) = v(t) = 8t.
Examples of such a function include
d(t) = 4t2 , 4t2 + 10, 4t2 + 10000,
general forms: d(t) = 4t2 + C, C is a constant.
A particular function, d(t) = 4t2 or d(t) = 4t2 + 10 is called an antiderivative. The general form of of
antiderivatives is called the indefinite integral.
That is
5
• Antiderivatives of 8t: 4t2 , 4t2 + 10, 4t2 + 10000, ...
• Indefinite integral of 8t: 4t2 + C.
The process of finding the distance function from velocity is called integration, i.e.,
dif f erentiation (unique)
−−
distance )
−−
−−
−−
−−
−−
−−
−−
−−
−−
−−
−−
−−
−−
−*
− velocity
integration (notunique)
Question: How can we determine which antiderivative is the distance we want?
Answer: An extra condition — initial condition d(t) = 4t2 + C.
initially::the distance from initial position is 0. I.e.,
d(0) = 0
0 = d(0) = 4 ∗ 02 + C ⇒ C = 0.
d(t) = 4t2 .
Summary:
Definition: An antiderivative of a function f (x) is a function, g(x), such that
g 0 (x) = f (x).
Definition: The indefinite integral of a function f (x) is the general form or the family of antiderivatives
of f (x), denoted by
Z
.
≡
f (x) d
x
g(x)
+
C
|{z}
|{z}
|{z}
|{z}
anantiderivative
arbitrary constant
General integration rules:
Z
•
Z
[f (x) ± g(x)]dx =
Z
•
Z
f (x)dx ±
g(x)dx.
Z
cf (x)dx = c
f (x)dx.
Integrations rules:
Z
•
1dx = x + C.
6
integrant
integral variable
Z
•
xn dx =
Z
1
dx = ln |x| + C.
x
Z
ex dx = ex + C.
•
•
xn+1
+ C,
n+1
n 6= −1.
Example 2: Find f (x) such that f 0 (x) = 2x − 1.
Z
Example 3: (2 + x + 2x2 + ex )dx.
Z
1 4
(x − 2x2 + 1)dx
x2
Z 4 √
t + 3t
dt.
Example 5:
t2
Z
Example 6: (2x + 1)(3x − 1)dx.
Example 4:
Example 6: find the function f given that the slope of the tangent line to the graph of the function at (0, 3)
(initial condition) is f 0 (x) = ex + x.
1
Answers: f (x) = ex + x2 + 3.
2
Example 7 Find the solution of the initial value problems:
f 0 (x) = 3x2 − 4x + 8 and f (1) = 9
| {z }
IC
Answers: f (x) = x3 − 2x2 + 8x + 2.
6.2
Integration by Substitution
Z
Example 1: Find
2x(x2 + 3)4 dx.
Solution:
1. define the substitution function:
u = g(x) = x2 + 3;
7
(6.1)
2. find the total differential:
du = g 0 (x)dx = 2xdx;
3. Rewrite the integral:
Z
2
Z
4
2x(x + 3) dx =
(x2 + 3)4 (2xdx
)
| {z } | {z }
u
4. Make the substitution:
Z
=
du
1
u4 du = u5 + C
5
5. Replace u by g(x):
1
= (x2 + 3)5 + C;
5
6. Verification:
Z
Example 2:
Z
Example 3:
0
= 2x(x2 + 3)4 .
e−3x dx. Let u = −3x.
x
dx. Let u = 2x2 + 1.
+1
2x2
Z
Example 4:
1 2
(x + 3)5
5
1
dx. Let u = ln x.
x(ln x)2
3x2
Example 5: Find f (x) given the slope of the tangent line f 0 (x) = √
at (1, 1).
2 x3 − 1
Example 6 (Application Problem):
The rate of change of the unit price p of Apex Ladies’ boots is given by
p0 (x) =
−250x
(16 + x2 )3/2
where x is the quantity demanded daily in units of a hundred. Find the demand function for these boots if
the quantity demanded daily is 300 pairs (x = 3) where the unit price is 50/pair.
Z
If F (x) =
Z
f (ax + b)dx =
proof:
8
f (x)dx
1
F (ax + b)
a
1. u = ax + b
1
du
a
Z
Z
Z
1
1
1
3.
f (ax + b)dx = f (u) · du =
f (u)du = F (u)
a
a
a
2. du = adx, or dx =
4.
1
F (ax + b)
a
Examples:
Z
e2x+3 dx.
Z
1
ex dx = ex ⇒ Soln = e2x+3
2
•
Z
•
6.3
2
√
dx.
3x + 3
Z
Z
1
1
x1/2
2
√ dx = 2x− 2 dx = 2 ·
= 4x 2
1/2
x
1
1
1
4
Soln = [4(3x + 3) 2 ] = (3x + 3) 2
3
3
Area and the definite integral
We know how to calculate areas of the following shapes:
a
h
b
A = a2
1
A = h(a + b)
2
A = πr2
How about
9
2
1
1
0
2
Example 1 Compute the area of the region bounded by y = 0.5x2 , x = 1, x = 2 and x-axis.
Method:Approximation.
Procedure:
1. divide [1, 2] into n = 1 subinterval.
Let x1 = 1, x2 = 2,
f
2−1
∆x =
= 1 (length of subintervals)
n
A ' ∆xf (x1 ) = ∆x · 0.5x21 = 0.5.
2
f (x2 )
1
f (x1 )
1
0
x1
10
2
x2
2. divide [1, 2] into n = 2 subintervals:
1
2−1
= = 0.5
n
2
x1 = 1, x2 = x1 + ∆x = 1.5, x3 = x2 + ∆x = 2
∆x =
A = ∆xf (x1 ) + ∆xf (x2 )
f
= ∆x(f (x1 ) + f (x2 ))
1
= (0.5x21 + 0.5x22 )
2
= 0.8125
f (x3 )
2
f (x2 )
1
f (x1 )
1
0
x1
3. divide [1, 2] into n = 4 subintervals:
1
2−1
= = 0.25
n
4
x1 = 1, x2 = 1.25, x3 = 1.5, x4 = 1.75, x5 = 2
∆x =
A = ∆xf (x1 ) + ∆xf (x2 ) + ∆xf (x3 ) + ∆xf (x4 )
= ∆x(f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ))
1
= (0.5x21 + 0.5x22 + 0.5x23 + 0.5x24 )
2
= 0.9845
11
2
x2
x3
2
1
1
0
4. n = 8, ∆x =
#
1
2
3
4
5
6
7
8
xi
1
1.125
1.25
1.375
1.5
1.625
1.75
1.875
2
2−1
1
= f = 0.125
8
8
f (x) = 0.5x2
0.5
0.633
0.7815
0.9455
1.125
1.3205
1.5315
1.758
sum= 8.595
2
1
0
Approximation = ∆x · Sum =
1
2
1
× 8.595 = 1.0745.
8
5. When n → ∞, the approximations go to the actual area as shown in the following graph.
12
2
1
1
0
2
General Procedure: Approximate the area between y = f (x) = 0.5x2 and the x-axis, on [a, b] = [1, 2]
1. Given n > 0, divide [a, b] into n subintervals. Length of subintervals : ∆x =
b−a
.
n
2. Calculate points:
x1 = a, x2 = a + ∆x, xi = a + (i − 1)∆x, ..., xn+1 = b
3. Compute approximation (using the left end-point)
A = ∆xf (x1 ) + ∆xf (x2 ) + ... + ∆xf (xn ) .
{z
}
|
n terms
= ∆x[f (x1 ) + ... + f (xn )]
In summary,
n
Approximations
1
2
4
8
→
0.5
0.8125
0.9845
1.0745
→
∞
1.1666 =
This shows
∆x[f (x1 ) + ... + f (xn )] −−−→
n→∞
7
6
7
6
Z
2
This limit is called the definite integral of f (x) = 0.5x from x = 1 to x = 2, denoted by
1
Definition: (definite integral)
Z
a
b
f (x)dx = lim ∆x[f (x1 ) + ... + f (xn )] = lim
n→∞ |
{z
} n→∞
Riemann Sum
13
n
X
”width”
z}|{
f (xi ) ∆x .
| {z }
i=1 ”height”
2
0.5x2 dx.
Here
n
X
∆x = b − a, or ∆x =
i=1
b−a
n
Z
b
Existence Theorem: If f is continuous on [a, b], then the definite integral
f (x)dx exists, or f is
a
integrable on [a, b].
Geometric interpretation:
Z b
f (x)dx is not the area of the region bounded by y = f (x), x = a, x = b and
Generally speaking:
a
x-axis. It is the area above the x-axis minus the area below the x-axis.
Z 1
(−x + 1)dx = 0
Example The area of the region below = 1. However
0
1
y = f (x) = −x + 1
1
0
2
−1
Z
b
Actually, if f (x) ≥ 0 on [a, b], then
f (x)dx is the area of the region bounded by y = f (x), x = a,x =
Z b
f (x)dx is negative of the area of the region bounded by
b and the x-axis. if f (x) ≤ 0 on [a, b], then
a
a
y = f (x), x = a,x = b and the x-axis.
Z
0
Z
1
1
1
1
(−x + 1)dx = b · h = · 1 · 1 =
2
2
2
2
1
1
1
(−x + 1)dx = − b · h = − · 1 · 1 = −
2
2
2
1
Z 2
Z 1
Z 2
(−x + 1)dx =
(−x + 1)dx +
(−x + 1)dx = 0
0
0
1
14
6.4
The fundamental theorem of Calculus
Theorem Let F (x) be any antiderivative of f (x): F 0 (x) = f (x). Then
Z b
f (x)dx = F (x)|ba ≡ F (b) − F (a). (definite integral, which has unique value)
a
Example 1 : Find the area of the region bounded by y = x2 , x = 1, x = 2 and the x-axis.
2
Z 2
x3 7
23 13
2
A=
x dx =
−
=
=
3
3
3
3
1
1
Example 2 Find the area of the region bounded by y = −x2 + 4, x = 1, x = 3 and the x-axis.
−x2 + 4 = 0,
Z
Area 6=
x2 = 4,
x = ±2
3
(−x2 + 4)dx
1
Z
2
Z
2
(−x + 4)dx −
Area =
1
3
(−x2 + 4)dx
2
Example 3 : Find the area of the region bounded by y = x2 + 1 from x = −1 to x = 2.
2
Z 2
x3 23 (−1)3
2
2
A=
(x + 1)dx =
−
+ (2 − (−1)).
+
x|
=
−1
3 −1
3
3
−1
Z 3
Example 4 Calculate
(3x2 + ex )dx.
1
Z
1
3
3
3x3 (3x + e )dx =
+ ex |31 = 33 − 13 + e3 − e1 = 26 + e3 − e.
3 1
2
x
Example 5: Find the area of region bounded by the graph of y = −x2 , x = −1, x = 2 and the x-axis.
y=0
−2
−1
0
1
−2
y = −x2
−4
15
2
3
Soln:
2
x3 23 (−1)3
A=
[0 − (−x )]dx =
=
−
= 3.
3 −1
3
3
−1
Z
2
2
Note that area is always a positive number. But the integral of −x2 over [−1, 2] is negative.
Z 2
−x2 dx = −3.
−1
Ex 3: Find the area of the region bounded by y = −x + 1, x = 0, x = 2 and x-axis.
1
1
0
2
−1
Z
2
(−x + 1)dx is positive in the black region and negative in the red region. So the
Note that the integral
0
area is the sum absolute value of these two parts, which leads to
Z 2
Z 1
(−x + 1)dx + −
(−x + 1)dx = 2
A=
1
0
or
Z 1
Z 2
A = (−x + 1)dx + (−x + 1)dx = 2
0
1
Ex 4: Find the area enclosed by the graph of f (x) and x-axis on the interval [a, b].
(a) f (x) = (x − 1)(x − 3),
[0, 4]
• Find the roots of f (x) = 0: (x − 1)(x − 3) = 0,
x = 1, 3
• x = 1, 3 divides [0, 4] into [0, 1], [1, 3] and [3, 4].
• Find the area of each interval and add them up:
Z 1
Z 3
Z 4
f (x)dx
Area=
f (x)dx + f (x)dx + 0
(b) f (x) = (x − 1)(x − 3),
1
3
[2, 4]
• Find the roots of f (x) = 0: x = 1, 3.
• x = 1, 3 divides [2, 4] into [2, 3] and [3, 4].
16
3
Z
• Area= 2
f (x)dx + 43 f (x)dx.
(c) f (x) = (x − 1)(x − 3),
[4, 5]
• Find the roots of f (x) = 0: x = 1, 3
• x = 1, 3 divides [4, 5] into [4, 5].
Z 5
f (x)dx
• Area=
4
6.5
Evaluating definite integrals
Integration Rules:
Z
b
b
Z
•
cf (x)dx = c
f (x)dx.
a
Z
a
b
•
Z
[f (x) ± g(x)]dx =
a
Z
b
Z
b
f (x)dx +
a
g(x)dx.
a
a
•
f (x)dx = 0.
a
Z
b
•
a
Z
f (x)dx = −
f (x)dx.
a
Z
•
b
b
Z
c
f (x)dx =
a
f (x)dx.
a
Substitution ZMethod
2
c
2 +1
xex
Example 1 :
b
Z
f (x)dx +
dx. Let u = x2 + 1, du = xdx. then
0
Z
2
xe
0
x2 +1
Z
dx =
0
2
u
z }|
{
Z 22 +1
2
(x
+
1)
e
xdx =
eu du = eu |51 = e5 − e.
|{z}
2
0 +1
du
Since we are performing the integration with respect to a new variable u now, the range of integration has
to be changed to reflect the fact that the integration is being performed with respect to the new variable u.
That is saying, the upper limit and lower limit has to be changed correspondingly. In this example, since
u = x2 + 1, the upper limit is changed to 22 + 1 = 5, and the lower limit is changed to 02 + 1 = 1.
Z b
Z u(b)
General Case: if x → u = u(x),
−→
a
u(a)
17
Z
1
x2 (x3 + 1)4 dx.
Example 2:
−1
Let u = x3 + 1, du = 3x2 dx.
Z 1
Z
x2 (x3 + 1)4 dx =
1
2
(x3 + 1)4 x
|
{zdx} =
|
{z
}
−1
−1
1
du
3
u
Z
0
2
1 4
u du
3
2
32
1 u5 1 25 05
= .
=
=
−
3 5 0 3 5
5
15
Average of a function:
f (x1 ) + f (x2 ) + ... + f (xn )
n
1
=
∆x[f (x1 ) + ... + f (xn )],
{z
}
b−a|
average of f (x) on [a, b] '
∆x =
b−a
n
Riemann Sum
⇒
1
b−a
b
Z
f (x)dx.
a
average of f (x) on [a, b] =
1
b−a
Z
b
f (x)dx.
a
Example 1: A car is traveling at a velocity v(t) from time a to b. Find the average velocity of this car
during this time period.
Soln:
Z
• Total Distance=
b
v(t)dt
a
Total Distance
• average velocity =
=
Total time
Rb
v(t)dt
b−a
a
Examples
Z
2
2
xe2x dx
•
0
Z
1
•
0
Z
•
x2
dx
+1
x3
4
x
p
9 + x2 dx
0
18
6.6
Z
Area between two curves
b
f (x)dx is the area of the region bounded by y = f (x), x = a, x = b and x-axis. If f (x) ≥ 0, for all
a
x ∈ [a, b].
Finding the area between two curves: If f (x) ≥ g(x) on [a, b], then the area of the region bounded
above by y = f (x) and below by y = g(x) on [a, b] is given by
b
Z
[f (x) − g(x)]dx
a
5
y = f (x)
4
3
2
y = g(x)
1
−1
0
1
2
3
4
5
6
−1
Approximation: Height=f (x) − g(x), width=∆x =
b−a
.
n
n
X
b−a
Area '
[f (xi ) − g(xi )] ∗
n
i=1
Z
a
b
”width”
n
z}|{
X
f (xi ) − g(xi )∆x
f (x) − g(x) dx = lim
n→∞
|
{z
}
i=1
”Height”
n=1
19
y = f (x)
4
3
2
1
−1
y = g(x)
0
1
2
3
4
5
6
4
5
6
4
5
6
−1
n=4
y = f (x)
4
3
2
1
−1
y = g(x)
0
1
2
3
−1
n = 20
y = f (x)
4
3
2
1
−1
y = g(x)
0
1
2
3
−1
20
Example 1: Find the area of the region that is completely enclosed by the graphs of y = 2x − 1 and
g(x) = x2 − 4.
6
A2
4
y = 2x − 1
2
−4 −3 −2 −1
0
−2
1
2
3
4
5
y = x2 − 4
A1
−4
Soln: We first find out the intersection point of these two curves by solving
2x − 1 = x2 − 4,
x2 − 2x − 3 = 0,
(x − 3)(x + 1) = 0,
x = −1, 3.
Z
3
Then A =
[(2x − 1) − (x2 − 4)]dx.
−1
Example 2: Find the area of the region completely enclosed by the graphs of the functions f (x) =
x3 − 3x + 3, and g(x) = x + 3
21
6
f (x) = x3 − 3x + 3
g(x) = x + 3
A3
4
A2
2
A1
−3
−2
−1
1
0
2
3
4
5
We get the x-values of the points of intersection of f (x) and g(x) by solving
x3 − 3x + 3 = x + 3
x3 − 4x = 0 ⇒ (x2 − 4)x = 0 ⇒ (x − 2)(x + 2)x = 0
x = −2, 0, 2.
Z
In the red region, since f (x) ≥ g(x), its area is
Z 2
its area is
g(x) − f (x)dx. So
0
f (x)−g(x)dx. In the black region, since g(x) ≥ f (x),
−2
0
Z
0
Z
f (x) − g(x)dx +
A=
−2
2
g(x) − f (x)dx = 8
0
Examples
√
(a) f (x) = x x + 2, g(x) = x2
√
√
• f (x) = g(x), x x + 2 = x2 , x( x + 2 − x) = 0
√
⇒ x = 0, or x + 2 − x = 0
√
x + 2 = x, x + 2 = x2 , x2 − x − 2 = 0, (x − 2)(x + 1) = 0,
√
However, x = −1 is not a solution because −1 + 2 6= −1.
x = 0, 2
22
x = −1, 2
Z
• A = 0
2
√
2
x x + 2 − x dx
(b) f (x) = x3 − 3x, g(x) = 2x2
• f (x) = g(x), x3 − 2x2 − 3x = 0, x(x − 3)(x + 1) = 0,
Z 3
Z 0
f (x) − g(x)dx
f (x) − g(x)dx + • Area= −1
6.7
x = −1, 0, 3
0
Applications of the Definite integral to Business and Economics
Consumers’ and Producers’ Surplus:
Consumers’ Surplus:
Demand
function — relates unit price p of a commodity to the quantity x demanded of it.
if p is high, x is small
if p is low, x is large
p = D(x) indicates how much people is willing to pay for the unit price at a level quantity x. Suppose
that a fixed unit market price p̄ has been established for the commodity and the corresponding quantity
demanded is x̄. Thus
the actual payment by consumers = p̄x̄
Z x̄
D(x)dx.
the amount consumers are willing to pay =
0
consumers’ Surplus = amount consumers are willing to pay
− amount consumers actually pay
Z x̄
D(x)dx − p̄x̄.
=
0
30
20
Demand function p = D(x)
(willing to pay)
10
p̄
0
10
20
23
30
x̄
20
20
consumers
surplus
10
consumers
actually pay
10
p̄
p̄
0
10
20
x̄
consumers
willing to pay
10
p̄
0
30
20
10
20
0
30
x̄
10
20
30
x̄
Producers’ Surplus
Supply function— relates the unit price of a commodity p and the quantity x produced by the supplier.
Let p = S(x), if p is high, then x is large; if p is low, then x is small.
p = S(x) indicates how much the producers are willing to receive for the unit price at the level of quantity
x.
20
p = S(x)
10
p̄
0
10
20
30
x̄
40
−10
Suppose that a fixed unit market price p̄ has been established for the commodity and the corresponding
quantity supplied is x̄.
The actual amount received by producers = p̄x̄ (at the current market)
Z x̄
The amount producers are willing to receive =
S(x)dx
0
Producers’ surplus = amount actually received − amount that producers are willing to received
Z x̄
= p̄x̄ −
S(x)dx.
0
24
Actual received
10
Willing to receive
10
p̄
p̄
0
10
20
0
30
x̄
10
20
30
x̄
Producers’ surplus
10
p̄
0
10
20
30
x̄
Example 1 The demand function for a commodity is
p = D(x) = −0.001x2 + 250
where p is the unit price in dollars and x is the quantity demanded in units of a thousand. The supply
function for the commodity is
p = S(x) = 0.0006x2 + 0.02 + 100
Determine the consumers’ and producers’ surplus if the market price of the commodity is set at the equilibrium price.
25
S(x)
D(x)
p̄
x1
x̄
x2
• x1 : supply < demand, x ↑
• x2 : supply > demand, x ↓
• x̄: supply = demand, market price, stable.
Soln: x̄ is the solution of S(x) = D(x). Solve the equation, we get x =
negative, x̄ = 300. And p̄ = D(x̄) = 160.
Z
−625
, 300. Since x̄ can’t be
2
300
D(x)dx − p̄x̄ = 18, 000.
Consumers’ surplus =
0
Z
Producers’ surplus = p̄x̄ −
300
S(x)dx = 11, 700.
0
The future and present value of an income stream
• a firm generates a stream of income over a period of time.
R(t) — rate of income generation at time t. E.g., R(t) = 3000 dollar/hour.
• The realized income is reinvested (in the bank) and earns interest at a fixed rate:
r — interest rate compounded continuously
• The period of time = T
26
• If put in the bank, and interest is compounded continuously,
Total interest + Principle = Principal · ert
Principal: the initial money you borrowed from or deposited to the bank.
t is the total amount of time.
Future value & present value
R(t)=rate of income reinvested
and earns interest rate r
P
F
0 (Now)
T
What is the total amount
F ? (future value)
Z
Future value f =
T
R(t)er(T −t) dt = erT
Z
T
R(t)e−rt dt.
0
0
Y = R(t)
R(t)
t
t + ∆t
• R(t)∆t: Profit generated from time period (t, t + ∆t).
• R(t)∆t ·er(T −t) : future value of profit generated from time period (t, t + ∆).
| {z }
Principal
• lim
∆t→0
X
R(t)∆t · e
T −t
Z
=
T
R(t)er(T −t) dt: total income
0
27
Present Value
P
R(t)=rate of income reinvested
and earns interest rate r
F
T
0 (Now)
Present value P = ?
future value F
The present value P that will yield the same accumulated value as the income stream itself when P is
invested for the same period of time at the same rate of interest.
rT
Pe
rT
Z
=F =e
T
R(t)e−rt dt
0
Z
⇒ P =
T
R(t)e−rt dt .
0
Example 2: The owner of a local cinema is studying a plan for renovating and improving the theater. the
plan calls for an immediate outlay of $ 250,000. It has been estimated that the plan would result in a net
income stream generated at the rate of
R(t) = $630, 000 per year
If the prevailing interest rate for the next 5 years is 10% annually, determine the net income in the present
value and the futher value at the end of 5 years.
Soln:
R(t) = $630, 000, r = 0.1, T = 5
0.1×5
Z
Future value of the plan F = e
5
630, 000e−0.1t dt
0
= 1, 632, 845 × e0.1×5
= 2, 692, 106
future value of the cost 250, 000 if put in the bank B = 250000e0.1∗5 = 412, 180.
A − B = 2, 279, 926 (future value of profit)
Z 5
present value P =
630, 000e−0.1t dt = 1, 632, 845
0
P − cost = 1, 632, 845 − 250, 000 = 1, 382, 845
28
Review: Chapter 5 & 6
• e ' 2.7182818
 x+y
e
= ex ey



x


 ex−y = e
eyxy
• ex
x y
(e
)
=
e




d
d f (x)


= ex ,
e
= ef (x) f 0 (x)
dx
dx

ln(xy)

= ln x + ln y



x


= ln x − ln y
ln


y


ln xy = y ln x
• ln x

d
1


ln x =


dx
x



d
1 0


ln f (x) =
f (x)
dx
f (x)
• Relation:ln ex = x, eln x = x.
• Chapter 6 Integration: see page 543–544
• Problems: (Review of chapter 6, Page 544–547) 14, 32, 34, 38, 50.
Review: Integration
Concepts:
• differentiation ↔ integration.
• antiderivatives F 0 (x) = f (x).
Z
• indefinite integral f (x)dx = F (x) + C.
Z
• definite integral
b
f (x)dx = F (b) − F (a).
a
• Riemann Sum
• relation between area and definite integral.
Z x̄
• Consumers’ surplus: CS =
D(x)dx − p̄x̄.
0
29
x̄
Z
• Producers’ surplus: P S = p̄x̄ −
S(x)dx.
0
Z
T
• Present Value: P =
R(t)e−rt dt.
0
• Future Value: F = e
rT
Z
T
R(t)e−rt dt.
0
Integration Techniques:
Z
•
Z
f (x) ± g(x)dx =
Z
•
Z
f (x)dx +
g(x)dx.
Z
cf (x)dx = c
f (x)dx.
Z
•
1dx = x + C.
xn+1
+ C, n 6= −1.
n+1
Z
xn dx =
Z
1
dx = ln(x) + C.
x
Z
ex dx = ex + C.
Z
eax dx =
•
•
•
•
Z
1 ax
e + C, a 6= 0.
a
a
•
f (x)dx = 0.
a
Z
b
•
a
Z
f (x)dx = −
f (x)dx.
a
Z
•
b
b
Z
f (x)dx =
a
c
Z
f (x)dx +
a
b
f (x)dx.
c
• Method of substitution.
Z b
• fundamental theorem of calculus:
f (x)dx = F (x)|ba = F (b) − F (a).
a
Applications:
• Compute areas between curves.
30
• compute CS, PS, FV,PV
Z b
1
f (x)dx.
• average=
b−a a
31