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Average Value of a Function and the Second Fundamental Theorem of Calculus. Day 2 – Section 5.4. Nancy Powell 2008. Average Value of a Function. The Mean Value Theorem for Integrals focuses on the fact that there is a “c” such that - PowerPoint PPT Presentation
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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