Integration

AntidfferentiationThe Fundamental Theorem of Calculus (FTC)

and the Algebra of IntegrationSecond FTC

Theorem 4.5 The Definite Integral as the Area of a Region

Fundamental Theorem of Calculus

4.3

The Fundamental Theorem of Calculus

• If a function f is continuous on the closed [a,b] and F is an antiderivative of f on the interval [a,b], then:

( ) ( ) ( )b

af x dx F b F a

FTC and the Constant of Integration

It is not necessary to include a constant of integration C in the antiderivative because:

FTC Example

( ) ( ) ( ) ( )b b

aaf x dx F x F b F a

33 13 3

11

3 1

xx dx

34 4 4

1

3 1 81 120

4 4 4 4 4

x

Using FTC to Find Area

Example 3 in the text: Find the area of the region bounded by the graph of

y= 2x2-3x+2, the x-axis and the vertical lines x=0 and x=2 Note that all of the function is above the x-axis in the intervalStart by integrating the function over the closed interval [0,2]

Then find the antiderivative and apply the FTC

Simplify.

2 2

0(2 3 2)Area x x dx

2

3 2 3 2 3 2

0

2 3 2 3 2 32 2 2 2 2 0 0 2 0

3 2 3 2 3 2x x x

10

3

Theorem 4.4 Continuity Implies Integrability

Theorem 4.6 Additive Interval Property

Absolute Value Integration Example:

• Example 2 in the text: keep in mind the definition of absolute value and where your function is positive and where it would be negative:

• From this, you can rewrite the integral in two parts:

(2 1), 1/ 22 1

2 1, 1/ 2

x xx

x x

2 1/ 2 2

0 0 1/ 22 1 (2 1) 2 1x dx x dx x dx

21/ 22 2

0 1/ 2( )

1 1 1 1 5(0 0) (4 2)

4 2 4 2 2

x x x x

Theorem 4.7 Properties of Definite Integrals

Definitions of Two Special Definite Integrals

Average Value of a FunctionJulia S. Lucas

Definition of the Average Value of a Function on an Interval and Figure 4.32

Let’s get started:

• You already know about functions and how to take the average of some finite set.

• Today we’re going to take the average over infinitely many values (those that the function takes on over some interval)…which means CALCULUS!

• Before I show how to do this…let’s talk about WHY we might want to do this?

Consider the following picture:

• How high would the water level be if the waves all settled?

Okay! So, now that you have seen that this an interesting

question...

Let’s forget about real life, and...

Do Some Math

Suppose we have a “nice”function and we need to find its average value over the interval [a,b].

Let’s apply our knowledge of how to find the average over a finite set of values to this problem:

First, we partition the interval [a,b] into n subintervals ofequal length to get back to the finite situation:

In the above graph, we have n=8

nabx /)(

Let us set up our notation:

ix comes from the i-th interval

Now we can get an estimate for theaverage value:

n

xfxfxff

naverage

)(...)()( 21

Let’s try to clean this up a little:

n

xfxfxff

naverage

)(...)()( 21

)](...)()([ 21 nxfxfxfab

x

Since nabx /)(

In a more condensed form, we now get:

n

i

iaverage xxfab

f1

)(1

But we want to get out of the finite,and into the infinite!

How do we do this?

Take Limits!!!

b

a

n

i

inaverage dxxfab

xfab

f )(1

)(1lim1

In this way, we get the average value of f(x) over the interval [a,b]:

So, if f is a “nice” function (i.e. wecan compute its integral) then we havea precise solution to our problem.

Let’s look back at our graph:

So we’ve solved our problem!

If I give you the equation 13)( 2 xxf

and ask you to find it’s average valueover the interval [0,2], you’ll all say

NO PROBLEM!!!! the answer is…

b

a

average dxxfab

f )(1

dxx 1302

1 2

0

2

2

0

3 ][2

1xx

5)010(2

1

Now we can answer our fish tank question!(That is, if the waves were described by an integrable function)

Theorem 4.10 Mean Value Theorem for Integrals and Figure 4.30

Finding the x values where we GET the average value of the function.

First take the equation 13)( 2 xxf

and find it’s average valueover the interval [0,2],

NO PROBLEM!!!! the answer is…

b

a

average dxxfab

f )(1

dxx 1302

1 2

0

2

2

0

3 ][2

1xx

5)010(2

1

Finding the x values where we GET the average value of the function.

NOW take the equation 13)( 2 xxf

and it’s average valueover the interval [0,2], average f = 5and set that equal to the function… And then solve for x…

NO PROBLEM!!!! the answer is…

513)( 2 xxf

Mean Value Theorem (find the value of x that gives you the

average value of your function)

513)( 2 xxf

547.13

4

3

4

43

513)(

2

2

2

x

x

x

xxf

Average Value of a Function

Definition:

If f is integrable on the closed interval [a,b], then the average value of f on the interval is:

1( )

b

af x dx

b a

Average Value Example

• Find the average value of

2( ) 3 2 ,[1,4]f x x x

4 2

1

1 1( ) 3 2

3

b

af x dx x x dx

b a

43 2

1

1

3x x

3 2 3 21 14 4 1 1

3 3

164 16 (1 1) 16

3

Mean Value Theorem

• If f is continuous on the closed interval [a,b], then there exists a number c in the closed interval [a,b] such that:

( ) ( )( )

1: ( ) ( )( )

b

a

b

a

f x dx f c b a

OR f x dx f cb a

Mean Value Theorem Example:

• Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval:

3

9( ) ,[1,3]f x

x

3 3 331 1

1 9 19

2 2dx x dx

x

3 3 33 1 2

211 1

1 9 9 9

2 3 1 4 4

x x

x

2 2

9 9 1 92

4 44 3 4 1

3 33

9 9 92 ( ) 2

2 2f c x x

x

Now we need to find the x coordinate where we get this average y value:

MVT-I

• The mean value theorem does not specify how to find c, just that there exists at least one number c in the interval that will give you the average value of the function.

Theorem 4.8 Preservation of Inequality

The Second Fundamental Theorem of Calculus

4.4

Second Fundamental Theorem of Calculus

• Earlier you saw that the definite integral of f on the interval [a,b] was defined using the constant b as the upper limit of integration and x as the variable of integration. However, a slightly different situation may arise in which the variable x is used as the upper limit of integration. To avoid the confusion of using x in two different ways, t is temporarily used as the variable of integration

Theorem 4.11 The Second Fundamental Theorem of Calculus

Definite Integral diagrams

The Definite Integral as a Function

0( ) cos , 0, , , ,

6 4 3 2

xF x tdt x and

00cos sin sin sin 0 sin

x xtdt t x x

(0) sin(0) 0

1sin

6 6 2

2sin

4 4 2

3sin

3 3 2

sin 12 2

F

F

F

F

F

You can think of the function F(x) as accumulating the area under the curve f(t)=cost from t=0 to t=x. For x=0, the area is 0 and F(x)=0. for x=pi/2, F(pi/2)=1 gives the accumulated area under the cosine curve on the entire interval [0,pi/2]. This interpretation of an integral as an accumulation function is used often in applications of integration. See p. 288 for a graphical example of this.

Figure 4.35

The Second Fundamental Theorem of Calculus

• If f is continuous on an open interval I containing a, then, for every x in the interval,

• The proof of this is on p. 289 in the text.

( ) ( )x

a

df t dt f x

dx

Using the SFTC

• Note that is continuous on the entire real line. So, using the SFTC you can write

• The differentiation shown in tthis example is a straightforward application of the SFTC. The next example shows an application of this combined with the chain rule to find the derivative of a function

2 1t

2 2

01 1

xdt dt x

dx

Using the SFTC

• Chain Rule• Definition of dF/du• Substitute the integral

for F(x)• Substitute u for x3

• Apply the SFTC• Rewrite as a function

of x.

3

/ 2cos

xdtdt

dx

( )dF du

F xdu dx

( ) ( )d du

F x F xdu dx

3

/ 2cos

xd dutdt

du dx

/ 2cos

ud dutdt

du dx

2cos 3u x

3 2cos 3x x

Find if

What is u and du?

U-Substitution

Antidifferentiation of a Composite Function

• Let g be a function whose range is an interval I and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then

• If u = g(x), then du = g’(x)dx and

( ( )) '( ) ( ( ))f g x g x dx F g x c

( ) ( )f u du F u c

Pattern Recognition

In this section you will study techniques for integrating composite functions.

This is split into two parts, pattern recognition and change of variables.

u-substitution is similar to the techniques used for the chain rule in differentiation.

Recognizing Patterns

2 4 22 ( 1) 1 2x x dx u x du x

2 3 3 23 1 1 3x x dx u x du x

2 2sec tan 3 tan 3 secx x dx u x du x

Pattern Recognition for Finding the Antiderivative

• Find• Let g(x)=5x and we have g’(x)=5 So we have f(g(x))=f(5x)=cos5xFrom this, you can recognize that the integrand

follows the f(g(x))g’(x) pattern. Using the trig integration rule, we get

You can check this by differentiating the answer to obtain the original integrand.

5cos5xdx

cos5 5 sin 5x dx x c

Multiplying and Dividing by a Constant

• Example 3 p. 297

22 1x x dx2_ 1 2let u x du xdx

_ _ _ _ , 2 ... _ _ _ lgbut we have only x not x so we do a ebra

1_ :

2du xdx now substitute

2 1

2u du

2 31 1

2 6u du u c

321_ _ : 1

6now back substitute x c

Change of VariablesExample 4 p. 298

• P. 298 2 1x dx

1

2

1 311 32 2

2 21 3

12 2

3

2

_ 2 1 2

_ _ _ _ , 2 ... _ _ _ lg

1_ :

2

1

2

1 1 1 1

2 2 2 3

1_ _ : 2 1

3

let u x du dx

but we have only dx not dx so we do a ebra

du dx now substitute

u du

u uu du c c u c

now back substitute x c

Change of Variables

• Example 5 p. 298 2 1x x dx_ 2 1 2let u x du dx

_ _ : ( 1) / 2solve for x u 1

21

: 2 12 2

u duSubstitute x x dx u

1 3 1

5/ 2 3/ 22 2 21 1 1 2 2

14 4 4 5 3

u u du u u du u u c

5 3

2 21 1

_ : 2 1 2 110 6

Back Substitute x x c

Change of Variables Example

• Example 6 p. 299

Guidelines for Making a Change of Variables

1. Choose a substitution u=g(x). Usually it is best to choose the INNER part of a composite function, such as a quantity raised to a power.

2. Compute du=g’(x)dx

3. Rewrite the integral in terms of the variable u

4. Find the resulting integral in terms of u

5. Replace u by g(x) to obtain an antiderivative in terms of x.

6. Check your answer by differentiating.

General Power Rule for Integration

• Theorem: The General Power Rule for Integration

If g is a differentiable function of x, then

Equivalently, if u=g(x), then

1

,1

nn g x

g x g x dx c nn

1

, 11

nn u

u du c nn

Theorem 4.14 Change of Variables for Definite Integrals

Change of Variables for Definite Integrals

• Theorem 4.14 p 301

Definite Integrals and Change of Variables

• Example 8 p. 301

Definite Integrals and Change of Variables

• Example 9 p. 302

Theorem 4.15 Integraion of Even and Odd Functions and Figure 4.39

Integration of Even and Odd Functions

• P. 303

Integration of an Odd Function

• Example 10 p. 303

Figure 4.41

Theorem 4.16 The Trapezoidal Rule

Theorem 4.17 Integral of p(x) =Ax2 + Bx + C

Theorem 4.18 Simpson's Rule (n is even)

Theorem 4.19 Errors in the Trapezoidal Rule and Simpson's Rule