Asia Pacific University of Technology and Innovation

Introductory calculus

Learning Objectives

At the end of this module, YOU should be able to:

Understand the terms function, variable and constant. Understand the idea of derived functions. Differentiate the polynomial functions [Integer powers of x, inverse

functions] together with sums, difference, product and quotient functions using Basic rules.

Find the derivatives of functions involving radicals using Basic rules. Apply differentiation to find minimum and maximum points of the

functions [curve].

Find higher-order derivatives using the notations dydx

, d2 ydx2 etc.,

Find the profit function, cost function, marginal cost, average cost, and break-even point of the functions by means of applications of differentiation.

Understand integration as the reverse process of differentiation. Integrate sums of terms in powers of x. Evaluate definite integrals. Apply integration to the evaluation of areas under the curve.

(i) Areas between a curve and the x axis(ii) Areas between a curve and the y axis

Introductory Calculus

Introduction This Mathematics measures changes in one quantity in relation to another quantity. Calculus is used for functions involving curves. Differentiation and integration can help us solve many types of real-world problems. We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.). Derivatives are met in many engineering and science problems, especially when modeling the behavior of moving objects. We can use integration to find the area when the sides are curved.

Let us see some important definitions which are fundamental in differential calculus.]

(i) Variable: A quantity that takes on various numerical values is known as variable. Usually it is represented by x, y, z and θ.

(ii) Constant: A quantity that does not change is called Constant. Usually it is represented by a, b, c, k and so on.

(iii) Function: A relationship between two things in which the value of one thing depends on the value of the other. Functions can be represented by graph.

The concept of derivative:

The derivative of a function expresses its rate of change with respect to an independent variable. The derivative is also the slope of the tangent line to the curve. The Limit Method utilizes the definition of a derivative of a function. The Limit Method, as seen below, is imperative because some problems may not work or may take longer with the differentiation rules.Let f(x) be a function defined in the interval (a, b). letcx be any point of this interval. Let (c + h) be any neighbouring point, (h may be positive or negative). So f(c) and f(c + h) may be obtained. As x changes from c to c + h, the rate of change of f(x) is   f(c + h) - f(c)                          h           when h = o, the above is not defined. However,         lim     f(c + h) - f(c) exists, we call it as differential coefficient h-->0            h of f(x) at x = c.

f’(c) = lim     f(c + h) - f(c)                         h-->0            h

[or] dydx =

limh .→0

f ( x+h )− f ( x )h

Basic Differentiation rules

1. Differentiation of a constantWhen y = c

dydx = 0

2. Differentiation of cf(x) where c is a constantWhen y = cf(x)dydx = c

ddx {f(x)}

3. Differentiation of xn

When y = xn

dydx = nxn-1

Derivatives of polynomial functions

Examples

Differentiate the following functions with respect to x.1. 1. y = x3

dydx

=3 x2

[using rule 2]2. 2. y = -5x4

dydx

=−5[ 4 x3 ]=−20 x3

[ using rule 2 and 3]3. y = 5

dydx

=0

4. y = √ x Function Rewrite Differentiate Simplify

y = √ x y = x1/2

dydx

=12

x−

12

dydx

= 12√ x

Note: If the function involving the radical, rewrite the question to differentiate using basic rules as given in (d).Review questions

1. y = 16

x6

2. y = 5x

Answers

1.dydx

= 16

6 x5=x5

2.dydx

=5(1)=5

Rules of Differentiation

1. Differentiation of sums and differences

Given y =p(x) + q(x) – r(x)

dydx =

ddx {p(x)} +

ddx {q(x)} -

ddx {r(x)}

Where p(x), q(x) and r(x) are functions of x

2. Differentiation of products of a function

Given that y = uv, where u and v are functions of xdydx = u

dvdx + v

dudx

3. Differentiation of quotients

Given that y =uv , where u and v are functions of x

dydx

=

v dudx

−u dvdx

v2

Examples

Differentiate the following functions with respect to x.

Differentiation of sums and differences

1. y = 4x2 +1x+ 3

x3

dydx

=8 x− 1x2

− 9x4

Note:

Function Rewrite Differentiate Simplifyy = 1

x+ 3

x3 y = x−1+3 x−3 y ' =−x−2+3(−3 ) x−4

y ' =− 1x2

− 9x4

Differentiation of products of a function

2. y = (x – 2)(3x + 1)

Solution: Separate the terms of product of the function and make up the table as follows. Differentiate each using basic rules. Substitute the terms as shown below using product rule.

u = x – 2 v = 3x + 1dudx

= 1 dvdx

= 3

Then, we get

dydx

=( x−2 )(3)+(3 x+1) .(1)=3 x−6+3 x+1=6 x−5

Differentiation of quotients

3. y =

x2+4x

Solution: Separate the terms of quotient of the function and make up the table as follows. Differentiate each using basic rules.Substitute the terms as shown below using Quotient rule.

u = x2 + 4 v = x dudx

= 2 x dvdx

= 1

Then, we get

dydx

=x (2 x )−( x2+4 )(1)

x2 =2 x2−x 2−4x2 = x2−4

x2

Review questions

Differentiate the following functions with respect to x

1. y = (x2 – 4x + 6)(1- 3x2)2. y = (3x-2)(4x+1)

3. y = (4x -x)(x -

1x2 )

4. y =

x3

1+ x2

5. y =

x2

2 x2−3

6. y = x−1x+1

Answers

1.dydx

=( x2−4 x+6 ) (−6 x )+(1−3 x2 ) (2 x−4 )

2.dydx

=(3 x−2 ) (4 )+ (4 x+1 ) (3 )=24 x−5

3.

dydx =( 4

x −x)(1+ 2x3 )+(x− 1

x2 )(− 4x2−1)

4.

dydx

=( 1+x2 ) (2 x )−( x2 ) (2 x )

(1+x2)2

5.

dydx

=( 2 x2−3 ) (2 x )− ( x2) (4 x−3 )

(2 x2−3 )2

6.

dydx

=(x+1 ) (1 )−( x−1 ) (1 )

( x+1 )2= x+1−x+1

( x+1 )2= 2

( x+1 )2

Differential of composite function

Chain Rule

If y = f(u) and u =f(x), function within a function is called composite function.Therefore

dydx

= dydu

× dudx

Examples:

Find dydx

:

1. y = (2 x−3 )5

Solution:dydx

=5(2 x−3)4 ddx

(2 x−3 )=5(2 x−3)4(2 )=10(2 x−3 )4

2. y = (5+x2 )10

Solution:dydx

=10(5+x2 )9 ddx

(5+x2 )=10(5+x2 )9 (2 x )=20x (5+x2 )9

Review questions

1. Given that h(x )= 1

(2 x−3 )2, evaluate h’ (1)

2.If y = (1 – 3x)2, find y’.

Answers

1.h( x )= 1

(2 x−3)2= (2 x−3)−2

h' ( x )=−2(2 x−3)−2−1 d

dx(2 x−3)

=−2(2 x−3)−3(2)

=−4(2 x−3 )−3

h''(1 )=−4(2(1 )−3 )−3=4

2 . y '=2(1−3 x )(−3 )=−6+18 x

Higher order derivatives

We know that the derivative of a function y = f(x) is itself a function, f '( x ).

If we differentiate the function f ‘(x), the resulting function is called the second derivative of f at x. Continuing this way, we get higher-order derivatives.

Some notations for Higher-order derivatives:

When y = f(x)

First order derivative: y ' f '( x )

dydx

ddx

[ f (x )]

Second order derivative: y ' ' f ' ' ( x )

d2 ydx2

d2

dx2 [ f ( x ) ]

Third order derivative: y ' ' ' f ' ' ' ( x )

d3 ydx3

d3

dx3 [ f ( x )]

Finding Higher-order derivatives

If f(x) = 6x3 – 12x2 + 6x – 2, find third order derivative.

Solution:

Differentiating the function using the rules of differential calculus givesf '( x )=18 x2−24 x+6

Differentiating further the above function[1st derivative] yieldsf ' ' ( x )=36 x−24

Similarly,f ' ' ' ( x )=36

Review question

1. Given y = 500 + 20x + 4x2 – 2x3, find the value of x for which y’’ = 0

Answer1. y=500+20 x+4 x2−2 x3

y '=20+8 x−6 x2

y” = -12x + 4-12x + 4 = 0 when y” = 0

x =

412 =

13

Maximum and Minimum Values on the curve

A turning point on a curve is defined as a point on the curve where the gradient is zero.

If y = f(x) is a curve and P is a point on the curve, such that dydx

= 0 at point P, then P is a

turning point.

The steps to determine if a turning point is a maximum or a minimum point on a curve of y = f(x) are as follows:

(a) Find dydx

(b) Find the value of x whendydx

=0

(c) Find the value of y by substituting the value of x into the function y = f(x)(d) Determine if the turning point is a maximum or minimum by differentiating further of

dydx .(i) When the second derivative is positive, it signifies a minimum.(ii) When the second derivative is negative, it signifies a maximum.

Examples

1. A certain manufacturing concern has total cost function c = 15 + 9x – 6x2 + x3. Find x value when the total cost is minimized.

Solution

At turning points, dydx

=0;

dcdx

=9−12 x+3 x2

. dcdx

=3 x2−12 x+9=0, solving this equation gives x=1 and x=3

At x = 1,

d2cdx2 =6 x−12=6 (1)−12=−6<0

Hence, total cost is maximized.

At x = 3,

d2cdx2 =6 x−12=6 (3)−12=6

> 0Hence, total cost is minimized.

2. The total revenue of a firm is given by R = 83x – 4x2 – 21. Find the output when the revenue is maximized.

Solution

Revenue R= 83x – 4x2 – 21Differentiating with respect to x,

dRdx

=83−8 x ; d2 Rdx 2 =−8

Revenue is maximum when dRdx

=0and

d2 Rdx 2 <0

dRdx

=0→ 83−8 x=0 . Therefore, x =

838

d2 Rdx 2 =−8<0.

Therefore, R is maximum.

When the output x = 838 units, revenue is maximized.

Review questions

1. Find the turning points of the curve and determine whether each of them is a maximum or a minimum point.y = 2x3-12x2+18x+3

2. Find the turning points for each of the following curves and determine if each of them is a maximum or a minimum.(a) y= 2x3+3x2-12x+4(b) y= (x-3)2

(c) y= x(x-2)2

(d) y = x+1x

3. The average cost function and the revenue function of a firm are given as

AC=2x−60+360000x and R(x) = -2x2 + 916x respectively, where x is the level of

production. Find(a) the level of production when cost is minimized(b) The price per unit when cost is minimized.

Answers

1.dydx

=6 x2−24 x+18=0

=x2−4 x+3=0 = (x−1 ) ( x−3 )=0 x = 1 or x = 3Substituting the values of x into y = 2x3-12x2+18x+3 we get, y =11, when x = 1 and y = 3, when x = 3∴ The turning points are (1, 11) and (3, 3)

To check whether the points are a maximum or minimum, we need to findd2 ydx2 =12 x−24

Substituting the value of x =1 and x =3, we get, -12 and 12, respectively.

∴ The maximum point is (1, 11) and the minimum point is (3, 3)

2. (a)dydx

=6 x2+6 x−12=0

= x2 + x – 2 = 0 = (x – 1)(x + 2) x =1 and x = -2 Turning points (1,-3) and (-2, 24)d2 ydx2 =12x+6

Substituting x = 1 and x = -2 into the above equation we get that (1, -2) is the

minimum and (-2, 24) is the maximum.

(b)dydx

=2 x−6=0

Turning points (3, 0)d2 ydx2 =2

(c)

dydx

=x (2 x )+( x−2)2=2 x2+ x2−4 x+4=3 x2−4 x+4=0

(d)dydx

=1− 1x2

=0

x – x2 = 0

x(x -1) = 0

x = 0 and x =1

(0, infinity) and (1, 2)

Point (1,2) is the minimum.

3. (a) To find minimum costdTCdx

=4 x−60=0

4x = 60

x = 15

Therefore the minimum Total cost is achieved at a production level of 15

units, becaused2 TC

dx=4 which is greater than 0

(b) Price per unit is equal to the following.R ( x )

x=−2 x+916

Substituting x = 15 into the equation we get, 886 as the price per unit at the

minimized cost.

Applications of differentiation in Economics and Commerce

Profit function In any commercial environment, the profit function can be considered as a simple function of the difference between revenue from the sale of a number of products [total revenue] and the costs involved in their production[total cost].

Profit = Revenue – Cost

Or P(x) = R(x) – C(x)

Cost function

The cost involved in standard processes can be normally categorized as follows: Fixed cost Variable cost Special / optional cost

Variable cost: It is single valued function of output.Fixed cost: It is independent of the level of output.

Total cost = Fixed cost + variable cost + Special cost

Definition of Average cost

Average cost =

Marginal cost

Total costtotal quantity [ output ]

It is defined as the increase in cost per unit increase in sales ie., the rate of change of total cost. In calculus, if cost function is given, the marginal cost can be found using above formula.ddx

=C( x )=C' ( x )

Cost maximization or minimization

They occur when dydx

=0

The second differential coefficient is used to determine whether there is a maximum or minimum.

Break –even point

It is the level of production where profit is zero or when the total revenue for a period will equal the total costs for the revenue period R = C. Hence, Pq = F + Vq

Examples

1. A manufacturing company has determined that the cost function of manufacturing a product is given by C = a + bx, where a is the fixed cost, b is the cost of producing every item, and x is the number of items produced and sold.(a) When 20 items are produced, the cost is $2,500 and when 25 items are

produced, the cost is $3,000. Determine the values of a and b, and derive the cost function in terms of x.

(b) If the average revenue function is 145, determine the revenue function in terms of x.

(c) Determine the profit function in terms of x. Find the profit when 15 units are produced and sold, and interpret the answer.

Solution:(a) C = a + bx

a + 20b = 2,500 ----------(1)a + 25b = 3,000 ----------(2)(2) – (1)5b = 500

b = 100Substitute b = 100 into (1)a = 2,500 – 20(100) = 500C = 500 + 100x

(b) R = AR × x = 145x(c) P(x) = R(x) – C(x)

P(x) = 145x – (500 + 100x) = -500 + 45xP(15) = -500 + 45(15) = 175The company makes a loss of $175 when 15 units are produced.

2. On a national tour of a rock band, the demand for T-shirts is given byPr ( x )=4 x+15

where x is the number of T-shirts (in hundred) that can be sold during a single concert at a price$ Pr . If the shirts cost the band RM5 each, (a) derive the total cost function, C(x).(b) derive the total revenue function R(x),(c) derive the total profit function,(d) calculate the level of activity that minimizes profit. Please provide evidence

that the value you obtained is minimizing the profit.Solution:

(a) C ( x )=5 x

(b) R( x )=xPr ( x )=4 x2+15 x

(c) P = (4x2+ 15x) – (5x)P = 4x2 + 15x – 5x

P = 4x2 + 10x

(d) P' ( x )=8x+10 . For minimum value, P' ( x )=0 ; 8 x+10=0

x=−54 P''( x )=8>0 At

x=−54 , profit is at minimum.

Integration

Integration is the reverse process of Differentiation.

Integration is a process of obtaining y from dydx , where y is a function of x.

The basic rules of integration

1.

2.∫ axn dx=axn+1

n+1+c

where a and c are constants, n is an integer and n ≠1

3. ∫ ( f ( x )∓g ( x )) dx=∫ f ( x ) dx∓∫ g ( x ) dx

Examples:

Integrate the following functions.

1) ∫¿¿+x)dx2) ∫ (1−2x2 ) dx

3) ∫5 ( x−2 )(x+1)dx =∫5 (x2−x−2 )=∫5 x2−5 x−10

4) ∫(12

x2− 4x2 ¿+1)dx¿

Solution:

1) x3

3+ x2

2+c

2) x−2 x3

3+c

3 ) 5 x3

3−5 x2

2−10 x+c

4 ) 12

x3

3−4 x−1

−1+x+c= x3

6+ 4

x+x+c

Note: In general, the constant of integration must be included in the result. If you omit it, your work will be completely wrong.

Review questions

1.∫(2 x3+6 x2+

10x2 )dx

2. ∫ (12 x3+8 x3+4 x2−10 ) dxAnswers

1.∫(2 x3+6 x2+10

x2 )dx= x4

2+2 x3−10

x+C

2.∫ (12 x3+8 x3+4 x2−10 ) dx=5 x4+ 4 x3

3−10 x+C

Definite Integrals

Definite integrals have lower and upper limits of integration

∫b

a

f ( x )dx is known as the definite integrals of f(x) for the values of x from a to b

If ddx g(x) = f(x) then ∫a

b

f ( x )dx = g(b) – g(a)

Basic laws of definite Integrals

1. ∫a

b

kf (x ) dx = k∫a

b

f ( x )dx

2. ∫a

b

{f 1 ( x )+ f 2 ( x )}dx =∫a

b

f 1 (x ) dx + ∫a

b

f 2 (x ) dx

3. ∫a

b

f ( x )dx= -∫b

a

f ( x )dx

4. ∫a

c

f ( x )dx =∫a

b

f ( x )dx + ∫b

c

f ( x )dx

Example

Evaluate the integral function:∫0

110 t−3 t2 dt

Solution:

∫0

110 t−3 t2 dt= [5 t 2−t3 ]0

1=5(1)2−13=4

Review questions

Evaluate the following integral functions:

1. ∫1

3

( 4 x−1 ) dx

2. ∫−1

1

( x+1 ) ( x−3 ) dx

3. ∫−1

2 x2+2 x+1x4 dx

Answers

1.∫1

3

( 4 x−1 ) dx =

[ 4 x2

2−x ]

1

3

=[2 (3 )2−3 ]− [2 (1 )2−1 ]=14

2. ∫−1

1( x2−2 x−3 ) dx

=[ x3

3−2 x2

2−3 x]

−1

1

−[ (1 )3

3−(1 )2−3 (1 )]−[ (−1 )3−(−1 )2−3 (−1 ) ]=−14

3

3.∫−1

2 ( 1x2+

2x3 +

1x4 )dx

=[−1x −

1x2−

3x3 ]

−1

2

−[− (1 )2 −

122−

323 ]−[− 1

(−1 )−

1(−1 )2

−3 1(−1 )3 ]=−

338

Area under a curve

Areas between a curve and the x axis

The area of a region bounded by a curve y = f(x), the x-axis and the lines of x =a and x = b is

A = ∫a

b

y dx

The value of A is positive if the area is above the x-axis and is negative if the area is below the x-axis.

Examples

1. Find the area underneath the curve y = 6 – x - x2 from x = -3 to x = 2.

Solution:

1. Area = ∫−3

2(6−x−x2) dx

= [6 x− x2

2− x3

3 ]−3

2

= [(6 (2)−

(2 )2

2−

(2 )3

3 )−(6 (−3 )−(−3 )2

2−

(−3)3

3 )] =

20 56 units2

2. Find the area of the shaded region in the diagram.

Solution:

2. Area =|∫

0

2

( x3−x2−2 x) dx|

=|[ x 4

4− x3

3−x2]

0

2

|

=|4−8

3−4−0|=8

3

Areas between a curve and the y axis

The area of a region bounded by a curve y = f(x), the y-axis and the lines of y =a and y = b is

A = ∫a

b

f−1( y )dy

The value of A is positive if the area is to the right of the y-axis and is negative if the area is to the left of the y-axis.

Example

Find the area of the region bounded by y = 2x + 4, y = 1, y = 3 and y-axis.

Solution:

Area of the region by A = ∫1

3(−x ) dy=∫1

3−( y−4

2 )dy=∫1

3 12

(4− y ) dy

=

12 [ 4 y− y2

2 ]1

3

Y

X0 2

y = x(x+1)(x–2)

=

12 [(4 (3)−

(3)2

2 )−(4 (1)−(1 )2

2 )]=2 unit2

Review questions

1. Calculate the area bounded by the graph:

∫0

12 3

2 x dx

2. Calculate the area bounded by the graph:

∫0

4

( 8√x−x2 ) dx

Answers

1.3

16

2.643

Points to remember:

Formula/ConceptFunction: A relationship between two things in which the value of one thing depends on the value of the other. Functions can be represented by graph.The derivative is used to determine the maximum and minimum values of particular functions. [eg., cost, profit etc.,] Integration is used to find the area when the sides are curved.

Four Basic rules: FunctionDifferential coefficient

[ dydx ]

Sum rule u ± vdudx

± dvdx

Product rule uvu dv

dx+v du

dx

Quotient ruleuv

v dudx

−u dvdx

v2

Chain rule Unnun−1 du

dx

Constant 0

Maximum MinimumNecessary condition

Sufficient

dydx

=0 dydx

=0

conditiondydx

=0 . d2 ydx 2 <0 dy

dx=0 . d2 y

dx 2 >0

Profit function P(x) = R(x) – C(x)

Cost function Total cost = Fixed cost + variable cost + Special costAverage cost

Marginal cost

ddx

=C( x )=C' ( x )

Break–even point Pq = F + VBasic integral formula ∫ axndx=axn+1

n+1+c

Definite integral∫a

b

f ( x )dx = f(

b) – f(a)

Area A = ∫

a

b

y dx [or] A = ∫a

b

x dy

The value of A is positive if the area is above the x-axis and is negative if the area is below the x-axis.The value of A is positive if the area is to the right of the y-axis and is negative if the area is to the left of the y-axis.

Total costtotal quantity [ output ]