18 A.2 Definite Integrals
3
1
V  t2 1
8
When we find the area
under a curve by adding
rectangles, the answer is
called a Riemann sum.
2
1
The width of a rectangle is
called a subinterval.
0
1
2
subinterval
3
4
The entire interval is
called the partition.
partition
Subintervals do not all have to be the same size.

3
1
V  t2 1
8
If the partition is denoted by P, then
the length of the longest subinterval
is called the norm of P and is
denoted by P .
2
1
0
1
2
3
4
subinterval
As P gets smaller, the
approximation for the area gets
better.
This is the Width
partition
n
Area  lim  f  ck  xk
P 0
k 1
if P is a partition
of the interval a, b


This is the height

n
lim  f  ck  xk
P 0
is called the definite integral of
f over a, b.
k 1
If we use subintervals of equal length, then the length of a
subinterval is:
ba
x 
n
The definite integral is then given by:
n
lim  f  ck  x
n 
k 1

n
lim  f  ck  x
n 
k 1
Leibnitz introduced a simpler notation
for the definite integral:
n
b
k 1
a
lim  f  ck  x   f  x  dx
n 
Note that the very small change
in x becomes dx.


b
a
f  x  dx
We have the notation for integration, but we still need
to learn how to evaluate the integral.

upper limit of integration
Integration
Symbol

f  x  dx
b
a
integrand
lower limit of integration
variable of integration
(dummy variable)
It is called a dummy variable
because the answer does not
depend on the variable chosen.
In Chapter 18 A.1, we considered an object moving at
a constant rate of 3 ft/sec.
Since rate . time = distance:
3t  d
If we draw a graph of the velocity, the distance that the
object travels is equal to the area under the line.
3
After 4 seconds,
the object has
gone 12 feet.
2
velocity
1
0
1
2
3
4
ft
3
 4 sec  12 ft
sec
time

http://mathworld.wolfram.com/RiemannSum.html
Lower Rectangle Sums
Upper Rectangle Sums
Actual Area: 0.67 Units Squared
5
Using the TI 83/84 for  3
1
Find the area under y  3 x on [1,5]
• Graph f(x)
•
•
•
•
•
Press 2nd CALC 7
Enter lower limit 1
Press ENTER
Enter upper limit 5
Press ENTER.
x dx
Set up a Definite Integral for finding the area of the shaded
region. Then use geometry to find the area.
2. f  x   x  1
1. f  x   45
6
4
6
 4 dx
6
1
4
2
 x+1 dx
2
2
5
A  bh
A  (4)(4)  16un
 b1  b2 
A  h

 2 
 3 7 
2
A  4
  20un
 2 
5
2
Use the limit definition to find

3
1
2
3x dx
Graph f(x) y  3x 2
Press 2nd CALC 7
Enter lower limit 1
Press ENTER
Enter upper limit 3
Press ENTER.
Homework
Page 446
Numbers 2 - 3
TITLE PAGE
Chapter 18 B
Anti - differentiation
• A physicist who knows the velocity of a
particle might wish to know its position at a
given time.
• A biologist who knows the rate at which a
bacteria population is increasing might want
to deduce what the size of the population
will be at some future time.
• In each case the rate of change
(the derivative) is known….but what is the
original function?
Think About It
• Suppose this is
the graph of the
derivative of a function
• What do we know about
the original function?
– Cubic (y = ax3)
– Critical numbers
– Where it is increasing, decreasing
• What do we not know?
f '(x)
In this section, we
will learn how to
determine the
original function
when given the
derivative.
We will also learn
21
some applications.
Derivatives
• Let f(x) = 24x
• Some function whose derivative = 24x.
• What is it?
• Let f(x) = 15x2
• Some function whose derivative = 15x2
• What is it?
22
Derivatives
• Let f(x) = 24x3 + 6x.
• Some function whose derivative 24x3 + 6x.
• What is it?
• Let f(x) = e2x
• Some function whose derivative = e2x
• What is it?
23
Derivatives
• Let f(x) = cos(x).
• Some function whose derivative = cos(x).
• What is it?
• Let f(x) = sin(x)
• Some function whose derivative = sin(x)
• What is it?
24
Derivatives
• F(x) is an antiderivative of f(x)
• A function F(x) is an antiderivative of a function f(x)
• Then F '(x) = 24x = f(x)
There is a function, F(x), whose derivative is f(x)
• F(x) = 12x2
(Take the derivative of the anti-derivative)
25
Finding An Antiderivative
• Given f(x) = 12x3
(This is the derivative. Find the original function)
– What is the antiderivative, F(x)?
• Use the power rule backwards
– Recall that for f(x) = xn … f '(x) = n • x n – 1
– Multiply the expression by the exponent
12 4
F ( x)  x  3x 4
– Decrease exponent by 1
4
• Now do opposite (in opposite order)
– Increase exponent by 1
– Divide expression by new exponent
26
Anti-Derivatives
• Derivatives give us the rate of change of a function
• What if we know the rate of change …
– Can we find the original function?
• If F '(x) = f(x)
(This is our Derivative) (Our original function)
• F(x) is an antiderivative of f(x)
The original function is called the
ANTIDERIVATIVE of the rate of change.
27
Antiderivatives
• For instance, let f (x) = x2.
– It is not difficult to discover an
antiderivative of f if we keep the Power
Rule in mind.
21
x
2
F ( x) 
f ( x)  x , Find F( x)
2 1
F ( x)  x
21
x3 1 3
F ( x)   x
3 3
Suppose
f  x   2x
What is its antiderivative?
We can make some guesses
x 2
2
x 3
2
x
2
x  100000000
2
They all fit!
Family of Functions
• All of these could “fit” our f(x) = x3.
– Their graphs are vertical
translates of one another.
– This makes sense, as each
curve must have the same
slope at any given value
of x.
Antiderivatives
• The function G(x) = ⅓ x3 + 100
also satisfies G’(x) = x2.
– Therefore, both F and G are antiderivatives of f.
• Indeed, any function of the form H(x)=⅓ x3 + C,
where C is a constant, is an antiderivative of f.
– The question arises: Are there any others?
Family of Functions
• By assigning “C” to f(x) = x3 + C
– Their graphs are vertical
translates of one another.
– This makes sense, as each
curve must have the same
slope at any given value
of x.
Antiderivatives
Theorem
• If F is an antiderivative of f on an interval I, then
the most general antiderivative of f on I is
F(x) + C
where C is an arbitrary constant.
Antiderivatives
• Going back to the function f (x) = x2, we see that
the general antiderivative of f is ⅓ x3 + C.
Indefinite Integrals of Exponential
Functions
•
e
x
dx  e  C
x
kx
•
e
 e dx  k  C
kx
x
•
a
 a dx  ln a  C
x
k x
•
a
 a dx  k  ln a   C
kx
35
Evaluate the definite integral and check the
result by differentiation.
 12 1 21 


x

x
  2 dx
Rewrite:
1 

  x  2 x dx
1
1
2
Integrate:
 12 1 21 
x
1x


x

x
  2 dx  3  2 1
2
2
3
2
Simplify:
1
1
2
1
 12 1 21 
2x
2


x

x
dx


x
c
  2 
3
3
2
Checking
1
 12 1 21 
2x
2


x

x
dx


x
c
  2 
3
2 3
 x
3 2
3
1
2
1
 x
2

1
1
2
1
2
1
x  x
2
1
2
Homework for 18 B
• Page 449
• 1-3
38