Average Value of a Average Value of a FunctionFunctionand theand the

Second Fundamental Second Fundamental Theorem of CalculusTheorem of Calculus

Day 2 – Section 5.4

Nancy Powell 2008

P1

2 4 6

9

8

7

6

5

4

3

2

1

j y=f(1.6510)

Area P1 = 5.69 cm2

Area WZYX = 5.68 cm2

f x = 9

x3

YZ

W X

Average ValueAverage Value of a Function of a Function• The Mean Value

Theorem for Integrals focuses on the fact that there is a “c” such that

• The value of f(c)f(c) given in the MVT is called the AVERAGE AVERAGE ValueValue of f on the interval [a,b].

C

( ) ( )( )b

af x dx f c b a

Average ValueAverage Value of a Function of a FunctionThe Average Value of a FunctionAverage Value of a Function on an interval

states:

If If f f is integrable on the closed interval [a,b], then is integrable on the closed interval [a,b], then thethe average valueaverage value of of ff on the interval ison the interval is

1( )

b

af x dx

b a

P1

2 4 6

9

8

7

6

5

4

3

2

1

j y=f(1.6510)

Area P1 = 5.69 cm2

Area WZYX = 5.68 cm2

f x = 9

x3

YZ

W X

So let’s revisit a problem from yesterday…So let’s revisit a problem from yesterday…

We found that c=1.6510 and that the height of our rectangle was f(c) = 2.

Let’s calculate the average valueaverage value of of ff on [1,3]

C

3

31

33 -3

211

1 9

3 1 x

9 9 1x

2

1(

2 2

9 1 12

2 18 2

) b

a

dx

dxx

f x dxb a

The Second Fundamental Theorem The Second Fundamental Theorem of Calculusof Calculus

The definite integral as a NUMBER

( ) b

af x dx

The definite integral as a Function

( ) ( ) x

aF x f t dt

Constant

Constant Constantf is a function of x

F is a Function of x

f is a function of t

Getting ready for

Integrals:Integrals as Accumulated Area Functions

You are given the graph of a function y = f (x) on a grid. Assuming that the grid lines are spaced 1 unit apart both horizontally and vertically,

sketch the graph of the area function

over the interval [1,5] for the function. Then sketch the graph of the derivative of each area function and compare it with the original function’s graph.

1( ) ( )

xA x f t dt

Accumulated Area under a curve You are given the graph of a function f - sketch the graph of the area function over the interval [1,5] for the function.

IntervalApprox. Area

Accumulated Area

1 to 1

1 to 2

2 to 3

3 to 4

4 to 5

1( ) ( )

xA x f t dt

6

4

2

-2

-4

-6

5

- 1.0- 1.0

0.50.5

1.01.0

- 0.5- 0.5

- 1.0- 1.0

00

2.52.5

1.51.5

00

3.53.5

4

2

-2

5

This is where the area = 0 (our initial condition!)

The Definite Integral as a FunctionThe Definite Integral as a FunctionEvaluate the function below at x = 0, /6, /4, /3,

and /2

0( ) cos( )

xF x t dt

You could evaluate 5 different integrals or better yet, temporarily fix x as a constant and use the Fundamental Theorem as shown below

00( ) sin sin( ) sin(0)cos( ) sin( )

xxF x t xt dt x So, what do we know?

The Definite Integral as a FunctionThe Definite Integral as a Function

x sin(x)

0 0

/6 .5

/4 .7071

/3 .8660

/2 1

00( ) sin sin( ) sin(0)cos( ) sin( )

xxF x t xt dt x

So, what

do we know?

1.5

1

0.5

-0.5

-1

1 2

1.5

1

0.5

-0.5

-1

1 2

f x = sin x

What if we changed the integral to

1c ( ) os( )

xtF dtx

The Definite Integral as a FunctionThe Definite Integral as a Function00

( ) sin sin( ) sin(0)cos( ) sin( )xx

F x t xt dt x

2

c ( ) os( )x

tG dtx What if we changed the integral to

-1 1 2 3

1.5

1

0.5

-0.5

-1

-1.5

F(x)

G(x)

2

cos( )( ) x

t dtH x

What if we changed the integral to

The Second Fundamental Theorem The Second Fundamental Theorem

( ) ( )x

a

df t dt f x

dx

If If ff is continuous on an open is continuous on an open interval interval II containing a, then, containing a, then, for every for every x x in the interval,in the interval,

The Second Fundamental Theorem The Second Fundamental Theorem

( ) ( )x

a

df t dt f x

dx

2

01

xdt

dx

Remember

Find:

2 1x

The Second Fundamental Theorem The Second Fundamental Theorem 3

2

( ) cos( ) x

F x t dt

3 '( )dF du

Let u x F xdu dx

( )d du

F xdu dx

Find the derivative of

Because x 3 is our upper limit of integration, we need to look at this integral with what we know about the chain rule.

Chain Rule

Definition of dFdu

Substitute3

2

cos( ) xd du

du dxt dt

The Second Fundamental Theorem The Second Fundamental Theorem 3

2

( ) cos( ) x

F x t dt3 '( )

dF duLet u x F x

du dx ( )

d duF x

du dx

Find the derivative of

Substitute u for x3

3

2

cos( ) xd du

du dxt dt

2

cos( ) u

t dtd

u x

d

d

u

d

23cos( )u x Apply the 2nd Fundamental Theorem of Calculus

3 23cos( )x x Rewrite as a function of x