Antiderivative/ Indefinite Integral

Find all possible functions F(x) whose derivative is

f(x) = 2x+1

F(x) = x2

+ x

F(x) = x2 + x + 5F(x) = x2 + x - 1000F(x) = x2 + x + 1/8

F(x) = x2 + x - π

Definition

A function F is called an antiderivative (also an indefinite integral) of a function f in the interval I if

for every value x in the interval I.

The process of finding the antiderivative of a given function is called antidifferentiation or integration.

'( ) ( )F x f x

Find all antiderivatives F(x) of f(x) = 2x+1

F(x) = x2

+ x

F(x) = x2 + x + 5F(x) = x2 + x - 1000F(x) = x2 + x + 1/8

F(x) = x2 + x - π

In fact, any function of the form F(x) = x2 + x + c where c is a constant is an

antiderivative of 2x + 1

Theorem

If F is a particular antiderivative of f on an interval I, then every antiderivative of f on I is given by

where c is an arbitrary constant, and all the antiderivatives of f on I can be obtained by assigning particular values for c. .

( )F x c

Notation

The symbol denotes the operation of

antidifferentiation, and we write

where F’(x)=f(x), and c is an arbitrary constant.

This is read “The indefinite integral of f(x)

with respect to x is F(x) + c".

( ) ( )f x dx F x c

In this notation,

is the integral sign;

f(x) is the integrand;

dx is the differential of x which denotes the variable of integration; and

c is called the constant of integration.

If the antiderivative of the function on interval I exists, we say that the function is integrable over the interval I.

( ) ( )f x dx F x c

Integration Rules

1. Constant Rule. If k is any real number, then the indefinite integral of k with respect to x is

2. Coefficient Rule. Given any real number coefficient a and integrable function f,

kdx kx C

( ) ( )af x dx a f x dx

Integration Rules

3. Sum and Difference Rule. For integrable functions f and g,

4. Power Rule. For any real number n,

where n ≠ -1, the indefinite integral xn of is,

1 2 1 2[ ( ) ( )] ( ) ( )f x f x dx f x dx f x dx

1

1

nn x

x dx Cn

Example 1.

211 22

252

(5 7) 5 7

5 7

5( ) 7

7

x dx xdx dx

xdx dx

x C x C

x x C

Example 2.

1

2

12

32

4 2

4 2

4 2

5 36 9 25 3 3

6 9

6 9

6 9

x x x dx

x dx x dx x dx

x dx x dx x dx

x x x C

Example 3.

543 3

53

543 3

7 23 3

3

15 37 2

5 25 2

5 2

xdx x x dx

x

x dx x dx

x x C

Integration Formulas for Trigonometric Functions

Cxdxxx

Cxdxxx

Cxdxx

Cxdxx

Cxdxx

Cxdxx

csccotcsc

sectansec

cotcsc

tansec

sincos

cossin

2

2

Example 4.

Cxx

CxCx

dxxxdxx

dxxxx

tan7csc3

tan7csc3

sec7cotcsc3

)sec7cotcsc3(

21

2

2

Example 5.

C

dd

dd

d

sin4sec3

cos4tansec3

cos4tancos

13

cos

cos4tan3 2

Exercises:

d

dxx

x

dyy

yy

dxxx

dyyy

)tan3cot2(.5

cos

sin.4

12.3

)1(.2

)32(.1

22

2

24

23

Integration by Chain Rule/Substitution

For integrable functions f and g

where is an F antiderivative of f and C is an arbitrary constant.

( ( ))[ '( ) ] ( ( ))f g x g x dx F g x C

Example 6.

14

54

54

2 34

3 24

3 2

3

54

385

36 6 5

2 6 5 (18 )

2 (6 5) (18 )

(6 5)2

(6 5)

x x dx

x x dx

x x dx

xC

x C

Let g(x) = 6x3+5

g’(x)=18x2

Example 6. Take 2!

14

54

54

54

2 3 3 24 4

4

54

85

385

36 6 5 2 6 5 (18 )

2

2 ( )

2

(6 5)

x x dx x x dx

u du

u du

uC

u C

x C

Let u = 6x3 + 5

du = 18x2 dx

Example 7.

3 3

7 74 4

74 3

64

64

2 1 1 2(2 1)

22 2

12 2(2 1)

2

21

2 61

12 2

t t dxdx

t t t t

t t t dx

t tC

Ct t

Let g(t) = t4 + 2t

g’(t) = 4t3 + 2

= 2(2t3 + 1)

Example 8.

5 2 12

22 2 12

2 12

14 13 12

15 14 131 2 115 14 13

2 15 2 14 2 131 1 115 7 13

( 1) 2

( 1) 2

( 1)

( 2 )

( 1) ( 1) ( 1)

x x dx

x x xdx

u u du

u u u du

u u u C

x x x C

Let u = x2 -1

du = 2x dx

x2 = u+1

Example 9.

Let u = 2 – cos2x

du = 0 – (-sin2x)(2dx)

=2sin2xdx

Cx

Cu

duu

duu

xdxx

dxxx

2/331

2/332

21

2/121

21

21

)2cos2(

)(

)(

)2sin2(2cos2

2cos22sin

Example 10.

xdxx

dxxx

dxxx

xx

dxx

x

x

x

dxxx

2csc2sec

2sin2cos

1

2sin2cos

2cos2sin

2sin

2cos

2cos

2sin

)2cot2(tan

22

22

222

2

2

Example 10.

Cxx

Cxx

xdxdxxx

xdxdxxx

xdxxdxx

dxxx

xdxx

2tan2cot

2tan)2(tan

2sec22sec2)2(tan

2sec2sec)2(tan

2sec2cot2sec

)12(cot2sec

2csc2sec

21

21

21

21

22122

21

222

222

22

22

1

Exercises:

drr

r

xx

dxxx

7

3 22

62

)1(

2.3

)49(5.2

)12(7.1

424

2

22

)123(

)13(.6

3sin21

3cos.5

3cot3csc.4

xx

dxxx

dxx

x

dyyyy

Applications of Indefinite Integrals

1. Graphing

Given the sketch of the graph of the function, together with some function values, we can sketch the graph of its antiderivative as long as the antiderivative is continuous.

Example 11. Given the sketch of the function f =F’(x) below, sketch the possible graph of F if it is continuous, F(-1) = 0 and F(-3) = 4.

-1

4

5

-3 -2 -1 0 1 2 3 4 5

1

-4-5

3

2

-2

-3

-4

-5

F(x) F’(x) F’’(x) Conclusion

X<-3 + - Increasing, Concave down

X=-3 4 0 - Relative maximum

-3<x<-2 - - Decreasing, Concave down

X=-2 - 0 Decreasing, Point of inflection

-2<x<-1 - + DecreasingConcave up

X=-1 0 0 + Relative minimum

X>-1 + + Increasing,Concave up

The graph of F(x)

-1

4

5

-3 -2 -1 0 1 2 3 4 5

1

-4-5

3

2

-2

-3

-4

-5

1. Boundary/Initial Valued Problems

There are many applications of indefinite integrals in different fields such as physics, business, economics, biology, etc.

These applications usually desire to find particular antiderivatives that satisfies certain conditions called initial or boundary conditions, depending on whether they occur on one or more than one point.

Applications of Indefinite Integrals

Example 11.

6 1dy

xdx

Suppose we wish to find a particular antiderivative satisfying the equation

and the initial condition y=7 when x =2.

Sol’n of Example 11

2

2

(6 1)

(6 1)

3

2 7,

7 3(2) 3 7

dy x dx

dy x dx

y x x C

but x when y then

C C

23 7y x x Thus the particular antiderivative desired,

Example 12.

The volume of water in a tank is V cubic meters when the depth of water is h meters. The rate of change of V with respect to h is π(4h2 +12h + 9), find the volume of water in the tank when the depth is 3m.

Sol’n of Example 12

2

2

32

32

32

32 3

4h 12h 9

4h 12h 9

46h 9h

3

4(0 )0 6(0 ) 9(0)

3

0

46h 9h

3

4(3 )6(3 ) 9(3) 207

3

dV

dh

dV dh

hV C

C

C

hThus V

V m

Volume V=0 if depth h =0

The Differential Equations

Equation containing a function and its derivative or just its derivative is called differential equations.

Applications occur in many diverse fields such as physics, chemistry, biology, psychology, sociology, business, economics etc.

The order of a differential equation is the order of the derivative of highest order that appears in the equation.

The function f defined by y= f(x) is a solution of a differential equation if y and its derivatives satisfy the equation.

2

2

6 1

(6 1)

(6 1)

3

2 7,

7 3(2) 3 7

dyx

dxdy x dx

dy x dx

y x x C

but x when y then

C C

23 7y x x

Thus find the particular solution

If each side of the differential equations involves only one variable or can be reduced in this form, then, we say that these are separable differential equations.

Complete solution (or general solution)

y = F(x) + C

Particular solution – an initial condition is given

Example 13. Find the complete solution of the differential equation

2

2

21

4 3

(4 3)

(4 3)

2 3

d y dyx

dx dxdy x dx

dy x dx

y x x C

21

21

3 2321 23 2

2 3

(2 3 )

dyx x C

dx

dy x x C dx

y x x C x C

2

2

d y dyddx dxdx

dydxlet y

342

2

xdx

yd

Example 14. Find the particular solution of the differential equation in Ex. 13 for which y=2 and y’=-3 when x=1.

21

21

1

2 3

3 2(1) 3

8

y x x C

x C

C

3 2321 23 2

3 23223 2

472 6

2 (1) (1) 8(1)

y x x C x C

C

C

3 23 4723 2 68y x x x

Example 16.

A stone is thrown vertically upward from the ground with an initial velocity of 20ft/sec.

(a) How long will the ball be going up?Ans. 0.625 sec

(b) How high will the ball go?Ans. 6.25 ft

(c) With what velocity will the ball strike the ground?

Ans. 20 ft/sec

That’s all Folks!

Thank you!!!