XEQ 201: Calculus II

Contents
Course description
References

iv iv Chapter 1. Applications of Differentiation
1.1. Mean value theorems of differential calculus
1.2. Using differentials and derivatives
1.3. Extreme Values

iii

1
1
5
7

Course description
Application of differentiation. Taylor theorem. Mean Value theorem of differential calculus. Methods of integration. Applications of integration.
References
1. Calculus: A complete course by Robert A. Adams and Christopher
Essex.
2. Fundamental methods of mathematical economics by Alpha C. Chiang.
3. Schaum’s outline series: Introduction to mathematical economics by Edward T. Dowling

iv

CHAPTER 1

Applications of Differentiation
1.1. Mean value theorems of differential calculus
Theorem 1.1.1 (Mean Value Theorem).
Suppose that the function f is continuous on the closed finite interval [a, b] and that it is differentiable on the interval (a, b). Then ∃ a point c ∈ (a, b) such that f (b) − f (a)
= f (c) . b−a It means that the slope of the chord joining the points (a, f (a)) and (b, f (b)) is equal to the slope of the tangent line to te curve y = f (x) at the point
(c, f (c)) so that the two lines are parallel.

Fig 2.28
Example 1.1.1.
√
Verify the conclusion of the mean value theorem for f (x) = x on the interval [a, b], where a ≤ x ≤ b.

Solution. We are to show that ∃ c ∈ (a, b) such that f (b) − f (a)
= f (c) b−a 1

XEQ 201 so long as f is continuous on [a, b] and is differentiable on (a, b). Now
√
√
1
f (x) = 2√x , f (a) = a, f (b) = b.
√
√
√
√
1
b− a b− a
1
∴ √ =
= √
=√
√ .
√
√
√
b−a
2 c b+ a b− a b+ a
The above equality implies that a < b, we have
√
a=

a+
2

√

a

√

√

c=

√

2

<

b+
2

√ b+ a
2

√

a

√

so that c =
√

2

<

b+
2

√

b

√
2
b+ a
.
2

Since

2

=b

which implies that c ∈ (a, b).
Example 1.1.2.
Show that sin x < x for all x > 0.
Solution. If x > 2π, then sin x ≤ 1 < 2π < x. If 0 < x ≤ 2π, then by
MVT, ∃ c ∈ (0, 2π) such that sin x sin x − sin 0 d =
= [MVT on [0, x]] = sin x x x−0 dx = cos c < 1 x=c which implies that sin x < x in this case too.

increasing decreasing functions

Definition 1.1.1 (Increasing and decreasing functions). Suppose that the function f is defined on an interval I and that x1 and x2 are two points in I.
(a) If f (x2 ) > f (x1 ) whenever x2 > x1 , we say that f is increasing on
I.
(b) If f (x2 ) < f (x1 ) whenever x2 > x1 , we say that f is decreasing on
I.
(c) If f (x2 ) ≥ f (x1 ) whenever x2 > x1 , we say that f is non-decreasing on I.
(d) If f (x2 ) ≤ f (x1 ) whenever x2 > x1 , we say that f is non-increasing on I.
Diagram Fig 2.31
2

XEQ 201
Theorem 1.1.2.
Let J be an open interval and let I be an interval consisting of all points in J and possibly one or both of the end points of J. Suppose that f is continuous on I and differentiable on J.
(a)
(b)
(c)
(d)

If
If
If
If

f f f f (x) > 0
(x) < 0
(x) ≥ 0
(x) ≤ 0

for for for for all all all all x ∈ J, x ∈ J, x ∈ J, x ∈ J,

then then then then f f f f is is is is increasing on I. decreasing on I. non-decreasing on I. non-increasing on I.

derivative of increasing and decreasing functions

Example 1.1.3.
On what intervals is the function f (x) = x3 − 12x + 1 increasing? On what intervals is it decreasing?
Solution. f (x) = 3x2 − 12 = 3 (x − 2) (x + 2). It follows that f (x) >
0 when x < −2 or x > 2 and f (x) < 0 when −2 < x < 2. Therefore f is increasing on the intervals (−∞, −2) and (2, ∞) and is decreasing on the interval (−2, 2).
Diagram Fig 2.32

Example 1.1.4.
Show that f (x) = x3 is increasing on any interval.
Solution. Let x1 , x2 be any two real numbers satisfying x1 < x2 . Since f (x) = 3x2 > 0 for all x = 0, we have that f (x1 ) < f (x2 ) if either x1 < x2 ≤ 0 or 0 ≤ x1 < x2 . If x1 < 0 < x2 , then f (x1 ) < 0 < f (x2 ). Thus f is increasing on every interval.
Theorem 1.1.3.
If f is continuous on an interval I and f (x) = 0 at every interior point of
I, then f (x) = C, a constant on I.
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derivetive of constant function

XEQ 201

derivative at interior extreme point Theorem 1.1.4.
If f is defined on an open interval (a, b) and achieves a maximum (or a minimum) at the point c ∈ (a, b), and if f (c) exists, then f (c) = 0.
Values of x where f (x) = 0 are called critical points of the function f .
Theorem 1.1.5 (Rolle’s Theorem).
Suppose that the function g is continuous on the closed finite interval [a, b] and if it is differentiable on the open interval (a, b). If g (a) = g (b), ∃ a point c ∈ (a, b) such that g (c) = 0.
Theorem 1.1.6 (The Generalized Mean Value Theorem).
If functions f and g are both continuous on [a, b] and differentiable on (a, b), and if g (x) = 0 for every x ∈ (a, b), then ∃ a number c ∈ (a, b) such that f (c) f (b) − f (a)
=
. g (b) − g (a) g (c)
Exercise 1.1.

1. Illustrate the MVT by finding any points in the open interval (a, b) where the tangent line is parallel to the chord joining (a, f (a)) and
(b, f (b)).
(a) f (x) = x2 on [a, b]; Ans: c = b+a .
2
2
(b) f (x) = x3 − 3x + 1 on [−2, 2]; Ans: c = ± √3 .
2. Show that tan x > x for 0 < x < π .
2
3. Find intervals of increase and decrease of the following functions
2
(a) f (x) = x3 − 4x + 1. Ans: Increasing on −∞, − √3 and
2
√ ,∞
3

2
2
; decreasing on − √3 , √3 .
2

(b) f (x) = x2 − 4 . Ans: Increasing on (−2, 0) and (2, ∞); decreasing on (−∞, −2) and (0, 2).
(c) f (x) = x3 (5 − x)2 . Ans: Increasing on (−∞, 3) and (5, ∞); decreasing on (3, 5).
(d) f (x) = x + sin x. Ans: Increasing on (−∞, ∞).
4

XEQ 201
1.2. Using differentials and derivatives
Suppose dx is regarded as a new independent variable called the differential of x we can define a new dependent variable dy, called the differentail of y as a function of x and dx by dy dx = f (x) dx. dx For example if y = x2 , then dy = 2xdx means the same thing as dy/dx = 2x.
If f (x) = 1/x, then df (x) = − 1/x2 dx.
If y is a function of x, y = f (x), then denoting a small change in x by dx instead of ∆x, the corresponding small change in y, ∆y is approximated by the differential dy, i.e. dy =

∆y ≈ dy = f (x) dx.
Diagram Fig 2.25

Example 1.2.1.
Without using a scientific calculator, determine by a pproximately how much the value of sin x increases as x increases from π/3 to (π/3) + 0.006. To 3 decimal places, what is the value of sin ((π/3) + 0.006)?
Solution. If y = sin x, then dx = 0.006. Therefore

dy dx = cos x. Now x =

π
3

≈ 1.0472 and

dy π 1 dx = cos xdx = cos
· 0.006 = (0.006) = 0.003. dx 3
2
This means that the change in the value of sin x is approximately 0.003. Now dy =

sin

π π + 0.006 ≈ sin
+ 0.003 = · · · = 0.869(3 d. p.).
3
3
5

XEQ 201
Suppose changes in x are measured with respect to the size of x, then

relative change in x =

dx dx and percentage change in x = 100 . x x

Differentials and point elasticity1
For a demand function Q = f (P ), the elasticity is defined as relative change in Q
∆Q/Q
=
.
∆P/P relative change in P point elasticity

If the change in P is infinitesimal, then the expressions ∆Q and ∆P reduce to the differentials dP and dQ. In that case the elasticity measure assumes the sense of point elasticity of the demand function which is denoted by εd .
Thus
dQ/Q dQ/dP εd ≡
=
. dP/P Q/P
The numerator in the right hand is the derivative (or marginal2) function of the demand function while the denominator is the average function of the demand function. Thus the point elasticity is a ratio of the two functions.
In general, for any given total function y = f (x), the point elasticity of y
w.r.t. x is dy/dx marginal function εyx =
=
. y/x average function
The absolute value of the point elasticity measure is used in deciding whether the function is elastic at a particular point. In the case of a demand function, 

elastic
|εd | > 1




The demand is of unit elasticity if |εd | = 1





 inelastic |εd | < 1. at a given point.

1Elasticity is the measure of how an economic variable responds to change in another

variable. An elastic variable is one which responds more than proportionally to changes in other variables. An inelastic variable is one which changes less than proportionally in response to changes in other variables.
2Marginal denotes the rate of change of a quantity with respect to a variable on which it depends. 6

XEQ 201
Example 1.2.2.
Find εd for the demand function Q = 100 − 2P . Determine the point elasticity at P = 25.
Solution.

dQ dP = −2 and εd =

Q
P

=

100−2P
.
P

Therefore

−2
P
=
.
(100 − P ) /P
P − 50

Thus
= · · · = −1.

εd
P =25

Therefore the demand is of unit elasticity when P = 25.
Exercise 1.2.
1. Use differentials to determine approximate change in the values of the given function as its argument changes from the given value to the given amount. What is the approximate value of the function after the change?
(a) y = 1/x as x increases from 2 to 2.01.
(b) h (t) = cos (πt/4) as t increases from 2 to 2 + (1/10π).
2. Find the approximate percentage changes in the given function that will result from an increase of 2% in the value of x.
(a) y = x2

(b) y = 1/x2

3. Given the consumption function C = a + bY (with a > 0; 0 < b <
1);
(a) Find its marginal function and its average function.
(b) Find the income elasticity of consumption εCY , and determine its sign, assuming Y > 0.
(c) Show that the consumption is inelastic at all positive income levels. 1.3. Extreme Values
Maximum and Minimum Values
Definition 1.3.1. Function f has an absolute maximum value f (x0 ) at x0 in its domain if f (x) ≤ f (x0 ) holds for every x in the domain of f .
7

absolute extreme values

XEQ 201
Similarly, f has an absolute minimum value f (x0 ) at x0 in its domain if f (x) ≥ f (x0 ) holds for every x in the domain of f .
Remark 1. extreme value is unique, can occur at several points existence of extreme value not guaranteed

existence of extreme values for closed finite intervals local extreme values 1. A function will have only one absolute maximum (or minimum) value if it exists. However the value can occur at many points. For example, f (x) = sin x has absolute maximum of 1 but it occurs at every point π + 2nπ, n ∈ Z.
2
2. A function need not have any extreme value. The function f (x) =
1
x becomes arbitrarily large as x approaches 0 from the right, and so has no finite absolute maximum value.
Theorem 1.3.1.
If the domain of the function f is a closed, finite interval or a union of finitely many such intervals, and if f is continuous on that domain, then f must have an absolute maximum value and absolute minimum value.
Definition 1.3.2. Function f has a local maximum value (loc. max.) f (x0 ) at the point x0 in its domain provided ∃ a number h > 0 such that f (x) ≤ f (x0 ) whenever x is in the domain of f and |x − x0 | < h.
Similarly, f has a local minimum value (loc. min.) f (x1 ) at the point x1 in its domain provided ∃ a number h > 0 such that f (x) ≥ f (x1 ) whenever x is in the domain of f and |x − x1 | < h.
Diagram Fig 4.17

From the above figure we see that local extreme values can occur at any of the following points.
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XEQ 201 critical, singular or end points

(i) critical points of f ; points x ∈ D (f ) where f (x) = 0.
(ii) singular points of f ; points x ∈ D (f ) where f (x) is not defined.
(iii) endpoints of the domain of f ; points that do belong to D (f ) but are not interior points of D (f ).

In the figure above, x1 , x3 , x4 are critical points, x2 and x5 are singular points and a and b are endpoints.
Theorem 1.3.2.
If the function f is defined on an interval I and has a local maximum (or local minimum) value at the point x = x0 in I, then x0 must be either a critical point of f , a singular point of f , or and endpoint of I.
Example 1.3.1.
Find the maximum and minimum values of the function g (x) = x3 − 3x2 −
9x + 2 on the interval −2 ≤ x ≤ 2.

Solution. Since g is a polynomial it can never have a singular point.
For critical points we calculate g (x) = 3x2 − 6x − 9 = 3 (x + 1) (x − 3) = 0.
Thus x = −1 or x = 3. But x = 3 is not in the domain of g and so we ignore it. We then investigate the endpoints x = −2 and x = 2 and critical point x = −1. g (−2) = 0, g (−1) = 7, g (2) = −20
The maximum value of g on −2 ≤ x ≤ 2 is at the critical point x = −1, and the minimum value is at the endpoint x = 2.
Diagram Fig 4.19
9

exteme values occur at critical, singular or end points

XEQ 201
Example 1.3.2.
Find the maximum and minimum values of h (x) = 3x2/3 −2x on the interval
[−1, 1].
Solution. The derivative of h is h (x) = 2x−1/3 − 2.
Note that x−1/3 is not defined at x = 0 in D (h), so x = 0 is a singular point of h. Also h (x) = 0 at x−1/3 = 1, that is at x = 1, which also happens to be an endpoint of the domain of h. We therefore examine the values of h at endpoints x = −1 and x = 1 and at the singular point x = 0. h (−1) = 5, h (0) = 0, h (1) = 1
The function h has a maximum value 5 at the endpoint x = −1 and a minimum value 0 at the singular point x = 0.
Diagram Fig 4.20

10