Sec. 4.1 Antiderivatives and Indefinite Integration

By

Dr. Julia Arnold

Antidifferentiation is the art of finding the function F(x) whose derivative would be the given function F /(x).

Antidifferentiation is also called: Indefinite Integration

Just as represents finding a derivative, the

symbol represents finding the antiderivative of the function f(x).The symbol is called the integral symbol and resembles an elongated s which as we will see in the next section is associated with finding sums. The dx shows what variable we are integrating.

dx

d

dxxf )(

Read your text book. I will supplement the information there.

For most of the examples, you will be using the exponent formula

which is like the opposite of the power formula.Example 4 a, b, and c. use this formula.Example 5 shows that you always rewrite radicals and use laws of exponents. You can not integrate numerator and then denominator.

Cn

xx

nn

1

1

In Example 5 some of the missing steps are:

dxxx

xdx

x

x

11

dx

xx

xdx

xx

x

2

1

2

1

11

dxxxdx

xx

x)(

1 2

1

2

11

2

1

2

1

dxxxdxxx )()( 2

1

2

1

2

1

2

11

21

23

)(2

1

2

3

2

1

2

1 xxdxxx

Cxx

2

12

3

23

2

Initial Conditions and Particular Solutions

When we find antiderivatives and add the constant C, we are creating a family of curves for each value ofC. dxx 13 2 Cxx 3

C=0

C=1

C=3

Finding a Particular Solution

Find the general solution of 2

1)('x

xF General Solution means theantiderivative + C.

Cxx

xF

112

12)(

A particular solution means that given some helpfulinformation we will find the value of C.The following information is referred to as the initialcondition: F(1) = 0

01

1)1( CF C=1 1

1)(

xxF

Setting Up A Vertical Motion Problem

When you must set up the motion of a falling object,S is the position function, is the velocity function, and is the acceleration function. Acceleration due to gravity is -32 ft. per second per second. Begin most problems with:

SS

32S

Let’s look at a specific problem.

A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.

A. Find the position function giving the height S as a function of the time t.B. When does the ball hit the ground?

Begin by integrating

The initial velocity is 64 feet per second. This means at t = 0, S’ = 64

CtS 3232S

C

C

64

)0(3264

6432)( ttS

The initial height means at t = 0,

S = 80 Ctt

tS

642

32)(

2

Ctt

tS

642

32)(

2

800642

)0(32)0(

2

CS C = 80

Thus, the position function is: 806416)( 2 tttS

B. When will the ball hit the ground?When S = 0

)54(160

80641602

2

tt

tt

)1)(5(160 tt

1 , 5 t t

Since t must be positive,the ball hit the ground after5 seconds.

Page 243 provides a nice summary of Integration rules.

Sample Homework Problems:8. dxxx 32

dxxx 33Rewrite as

Using the sum and difference formula (4th from the top on page 243), we integrate each term using the power formula.

cxx

2

34

24

Evaluate the definite integral and check the result by differentiation.

18. dx

xx

2

1

Rewrite: dxxx

2

1

2

1

2

1

Integrate:

212

1

232

11

2

11

2

1

2

1

2

1

xxdxxx

Simplify:cx

xdxxx

2

12

3

2

1

2

1

3

2

2

1

cxx

dxxx

2

12

3

2

1

2

1

3

2

2

1Checking

2

1

2

11

2

11

2

3

2

1

2

1

2

3

3

2

xxxx

Find the equation for y, given the derivative andthe indicated point on the curve.

47. )cos(x

dx

dy

Multiply both sides by dx, then integrate both sides. dxxdy )cos(

dxxdy )cos(

cxy )sin(

Of the family of curves the correct one goes through the point (0,4)

Substitute x = 0 and y = 4. Thus c=4 and the solutionis: 4)sin( xy

Solve the differential equation:

52. 3)0(,60,2 ffxxf

Integrate:

cx

xf

xxf

3

3

2

Substitute x=0 and 6 for f’(0) c3

06

3

Thus c = 6 63

3

xxf

Integrate again: cxx

xf 643

1 4

cxx

612

4

Substitute x = 0 and f(0)=3 c 0612

03

4

C=3

Find f(x)

Answer: 3612

4

xx

xf

Vertical Motion: Use a(t)=-32 feet per sec per sec as the acceleration due to gravity. (Neglect air resistance.)

60. A balloon rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant it is 64 feet above the ground.(a) How many seconds after its release will the bag strike the ground?(b) At what velocity will it hit the ground?Solution: Begin with a(t) = -32 and integrate. Remember v’(t)=a(t), thus v(t)= -32t + c.At t = 0 (the moment the sandbag is released, the balloon has a velocity of 16, thus 16 = c.v(t) = -32t + 16 Continued on next page

60 continued

v(t) = -32t + 16 Now integrate v(t) which is s(t) the position function.S(t)= -16t2 + 16t + c

A balloon rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant it is 64 feet above the ground.

At t = 0, s(t) is 64 feet above the ground.Thus c = 64 and s(t) = -16t2 + 16t + 64

(a) How many seconds after its release will the bag strike the ground?

The position of the bag indicates that s = 0 on ground.

s(t) = -16t2 + 16t + 64

0 = -16t2 + 16t + 640 = -16(t2 - t - 4)0 = t2 - t - 4 Since this doesn’t factor we will need to use the quadratic formula to solve. One of these t’s is negative. One is positive.

2

171

12

41411

t

2

171t We want the positive answer.

t = 2.56 approximately Thus the sandbag is on theground in about 2.6 seconds

(b) At what velocity will it hit the ground?

v(t) = -32t + 16 @ t = 2.56 v(2.56) = -32(2.56) + 16 = -65.92The velocity of the sandbag when it hits the groundis approximately -65.92 feet per second.