Transcript Document

Homework questions thus far???



Section 4.10?
5.1?
5.2?
sin 2x
 sin x dx
The Definite Integral
Chapters 7.7, 5.2 & 5.3
January 30, 2007
Estimating Area
vs
Exact Area
Pictures
Riemann sum rectangles, ∆t = 4 and n = 1:
Better Approximations

Trapezoid Rule uses straight lines
Trapezoidal Rule
Better Approximations


The Trapezoid Rule uses small lines
Next highest degree would be
parabolas…
Simpson’s Rule
Mmmm…
parabolas…
Put a parabola across each
pair of subintervals:
Simpson’s Rule
Mmmm…
parabolas…
Put a parabola across each
pair of subintervals:
So n must be even!
Simpson’s Rule Formula
Like trapezoidal
rule
Simpson’s Rule Formula
Divide by 3
instead of 2
Simpson’s Rule Formula
Interior
coefficients
alternate:
4,2,4,2,…,4
Simpson’s Rule Formula
Second from
start and end
are both 4
Simpson’s Rule

Uses Parabolas to fit the curve

b
a
x
f (x)dx 
[ f (x0 )  4 f (x1 )  2 f (x2 )  4 f (x3 )  ...
3
4 f (xn 1 )  f (xn )]
Where
n is even and
∆x = (b - a)/n
S2n=(Tn+
2Mn)/3
Use Simpson’s Rule to Approximate the
definite integral with n = 4
g(x) = ln[x]/x on the interval [3,11]
Use T4.
Runners:
A radar gun was used
to record the speed of
a runner during the first
5 seconds of a race
(see table) Use
Simpsons rule to
estimate the distance
the runner covered
during those 5
seconds.
t(s)
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
v(m/s)
0
4.67
7.34
8.86
9.73
10.22
10.51
10.67
10.76
10.81
10.81
Definition of Definite Integral:
If f is a continuous function defined for a≤x≤b, we
divide the interval [a,b] into n subintervals of equal
width ∆x=(b-a)/n. We let x0(=a),x1,x2,…,xn(=b) be the
endpoints of these subintervals and we let x1*, x2*, …
xn* be any sample points in these subintervals so
xi*lies in the ith subinterval [xi-1,xi]. Then the Definite
Integral of f from a to b is:
b
n
 f (x)dx  lim  f (x )x
a
n
*
i
i 1
Express the limit as a Definite Integral
 1 4ni  
n
 e
4
lim  
n
n
4i
i 1 2 


n
7 
7i  
7i 
lim 
2 3     3  
n

n 
n
i 1 n
n
2
Express the Definite Integral as a limit
2
 (2  x
2
)dx
0
5
 tan 2x dx
1
Properties of the Definite Integral
Properties of the Definite Integral
Properties of the Definite Integral
Properties of the Integral
b
a
a
b
 f (x)dx    f (x)dx
1)
a
 f (x)dx
2)
=0
a
b
3)
b
 cf (x)dx  c  f (x)dx
a
a
for “c” a constant
Properties of the Definite Integral

Given that:
1
 2 f (x)dx  8
2
 Evaluate the following:
1
 f (x)dx  ?
4
4
 f (x)dx  3
1
2
 g(x)dx  5
2
4
 g(x)dx   7
2
1
 f (x)dx  ?
2
1
 3dx  ?
1
Properties of the Definite Integral

Given that:
 Evaluate the following:
1
 2 f (x)dx  8
2
4
 f (x)dx  3
1
2
 g(x)dx  5
2
4
 g(x)dx   7
2
4
 [3 f (x)  2g(x)]dx  ?
2
2
 3g(x)dx  ?
2
4
 f (x)dx
1
Given the graph of f, find:
Evaluate:

3
1
f (x)dx
 1  x2

f (x)   1
2  x

1  x  0
0  x 1
1 x  3
Integral Defined Functions
Let f be continuous. Pick a constant a.
Define:
x
F(x) 
 f (t)dt
a
Integral Defined Functions
Let f be continuous. Pick a constant a.
Define:
x
F(x) 
 f (t)dt
a
Notes:
• lower limit a is a constant.
Integral Defined Functions
Let f be continuous. Pick a constant a.
x
Define:
F(x) 
 f (t)dt
a
Notes:
• lower limit a is a constant.
• Variable is x: describes how far to integrate.
Integral Defined Functions
Let f be continuous. Pick a constant a.
x
Define:
F(x) 
 f (t)dt
a
Notes:
• lower limit a is a constant.
• Variable is x: describes how far to integrate.
• t is called a dummy variable; it’s a placeholder
Integral Defined Functions
Let f be continuous. Pick a constant a.
x
Define:
F(x) 
 f (t)dt
a
Notes:
• lower limit a is a constant.
• Variable is x: describes how far to integrate.
• t is called a dummy variable; it’s a placeholder
• F describes how much area is under the curve up to x.
Example
x
Let f (x)  2  x . Let a = 1, and F(x) 
 f (t)dt .
Estimate F(2) and F(3).
a
x
F(x) 

2  tdt
1
2
F(2) 

1
2  tdt 
1/ 2
 f (1)  4 f (1.5)  f (2) 
3
1.8692
Example
x
Let f (x)  2  x . Let a = 1, and F(x) 
 f (t)dt .
Estimate F(2) and F(3).
a
x
F(x) 

2  tdt
1
3
F(3) 

2  tdt
1
1/ 2

 f (1)  4 f (1.5)  2 f (2)  4 f (2.5)  f (3)
3
 1.8692
x
Where is F(x) 
 f (t)dt
increasing and decreasing?
a
f (t )
is given by the graph below:
F is increasing.
(adding area)
F is decreasing.
(Subtracting area)
Fundamental Theorem I
Derivatives of integrals:
Fundamental Theorem of Calculus, Version I:
If f is continuous on an interval, and a a number on that
interval, then the function F(x) defined by
x
F(x) 
 f (t)dt
a
has derivative f(x); that is, F'(x) = f(x).
Example
x
Suppose we define F(x) 
2
cos(t
)dt .

2.5
Example
x
Suppose we define F(x) 
2
cos(t
)dt .

2.5
Then F'(x) = cos(x2).
Example
x
2
F(x)

(t
Suppose we define
  2t  1)dt .
7
Then F'(x) =
Example
x
2
F(x)

(t
Suppose we define
  2t  1)dt .
7
Then F'(x) = x2 + 2x + 1.
Examples:
x


d
sin(t)dt 


dx   

2
y


d


2
d
2
  5x dx      5x dx 
dy  y

dy  2

d
dr
r


cost dt
Examples:
d 2r
3
tant dt

dr  
d  2 
  x dx 
d  0

3
d
F[g(x)]  F '[g(x)]g'(x)
dx
Fundamental Theorem of Calculus (Part 1)

If f is continuous on [a, b], then the function
defined by
x
F(x) 
 f (t)dt
axb
a

is continuous on [a, b] and differentiable on (a, b)
and
F '(x)  f (x)
Fundamental Theorem of Calculus (Part 1)
(Chain Rule)

If f is continuous on [a, b], then the function
defined by
u( x )
F(x) 

f (t)dt a  x  b
a

is continuous on [a, b] and differentiable on (a, b)
and
F '(x)  f (u(x))u '(x)
In-class Assignment
1.
Find:

d  cos x
ln t dt 
1


dx  2

2.
a.
b.
c.
Estimate (by counting the squares) the total area between f(x) and the xaxis.
8
Using the given graph, estimate  0 f (x)dx
Why are your answers in parts (a) and (b) different?
Consider the function f(x) = x+1 on the interval [0,3]

First let the bottom bound = 1, if x >1, we calculate
the area using the formula for trapezoids:
1
b1  b2 h 
2
Consider the function f(x) = x+1 on the interval [0,3]

Now calculate with bottom bound = 1, and x < 1, :
Consider the function f(x) = x+1 on the interval [0,3]

So, on [0,3], we have that


1 2
F(x)  x  2x  3
2

And F’(x) = x + 1 = f(x)
as the theorem claimed!
Very Powerful!
Every continuous function is the derivative of some
x
other function! Namely:
 f (t)dt
a