calculus

1. 5.2 Definite Integrals
Greg Kelly, Hanford High School, Richland, Washington

2. When we find the area
under a curve by adding
rectangles, the answer is
called a Rieman sum.
0
1
2
3
1 2 3 4
2
1
1
8
V t
 
subinterval
partition
The width of a rectangle is
called a subinterval.
The entire interval is
called the partition.
Subintervals do not all have to be the same size.


3. 0
1
2
3
1 2 3 4
2
1
1
8
V t
 
subinterval
partition
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
As gets smaller, the
approximation for the area gets
better.
P
 
0
1
Area lim
n
k k
P
k
f c x


 
 if P is a partition
of the interval  
,
a b


4.  
0
1
lim
n
k k
P
k
f c x



 is called the definite integral of
over .
f  
,
a b
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:
 
1
lim
n
k
n
k
f c x






5.  
1
lim
n
k
n
k
f c x



 Leibnitz introduced a simpler notation
for the definite integral:
   
1
lim
n b
k a
n
k
f c x f x dx


 
 
Note that the very small change
in x becomes dx.


6.  
b
a
f x dx

Integration
Symbol
lower limit of integration
upper limit of integration
integrand
variable of integration
(dummy variable)
It is called a dummy variable
because the answer does not
depend on the variable chosen.

7.  
b
a
f x dx

We have the notation for integration, but we still need
to learn how to evaluate the integral.


8. 0
1
2
3
1 2 3 4
time
velocity
After 4 seconds,
the object has
gone 12 feet.
In section 5.1, 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.
ft
3 4 sec 12 ft
sec
 


9. 0
1
2
3
1 2 3 4
x
If the velocity varies:
1
1
2
v t
 
Distance:
2
1
4
s t t
 
(C=0 since s=0 at t=0)
After 4 seconds:
1
16 4
4
s   
8
s 
 
1
Area 1 3 4 8
2
  
The distance is still
equal to the area
under the curve!
Notice that the area is a trapezoid.


10. 2
1
1
8
v t
 
What if:
We could split the area under the curve into a lot of thin
trapezoids, and each trapezoid would behave like the large
one in the previous example.
It seems reasonable that the distance will equal the area
under the curve.

0
1
2
3
1 2 3 4
x

11. 2
1
1
8
ds
v t
dt
  
3
1
24
s t t
 
3
1
4 4
24
s  
2
6
3
s 
The area under the curve
2
6
3

We can use anti-derivatives to
find the area under a curve!

0
1
2
3
1 2 3 4
x

12. Let’s look at it another way:
a x
Let area under the
curve from a to x.
(“a” is a constant)
 
a
A x 
x h

 
a
A x
Then:
     
a x a
A x A x h A x h
   
     
x a a
A x h A x h A x
   
 
x
A x h

 
a
A x h



13. x x h

min f max f
The area of a rectangle drawn
under the curve would be less
than the actual area under the
curve.
The area of a rectangle drawn
above the curve would be
more than the actual area
under the curve.
short rectangle area under curve tall rectangle
 
   
min max
a a
h f A x h A x h f
     
h
   
min max
a a
A x h A x
f f
h
 
 


14.    
min max
a a
A x h A x
f f
h
 
 
As h gets smaller, min f and max f get closer together.
   
 
0
lim a a
h
A x h A x
f x
h

 
 This is the definition
of derivative!
   
a
d
A x f x
dx

Take the anti-derivative of both
sides to find an explicit formula
for area.
   
a
A x F x c
 
   
a
A a F a c
 
 
0 F a c
 
 
F a c
 
initial value


15.    
min max
a a
A x h A x
f f
h
 
 
As h gets smaller, min f and max f get closer together.
   
 
0
lim a a
h
A x h A x
f x
h

 

   
a
d
A x f x
dx

   
a
A x F x c
 
   
a
A a F a c
 
 
0 F a c
 
 
F a c
 
     
a
A x F x F a
 
Area under curve from a to x = antiderivative at x minus
antiderivative at a. 

16.  
0
1
lim
n
k k
P
k
f c x


 

 
b
a
f x dx
 
   
F x F a
 
Area


17. Area from x=0
to x=1
0
1
2
3
4
1 2
Example: 2
y x

Find the area under the curve from
x=1 to x=2.
2
2
1
x dx

2
3
1
1
3
x
3
1 1
2 1
3 3
  
8 1
3 3

7
3

Area from x=0
to x=2
Area under the curve from x=1 to x=2.


18. 0
1
2
3
4
1 2
Example: 2
y x

Find the area under the curve from
x=1 to x=2.
To do the same problem on the TI-89:
 
^ 2, ,1,2
x x


ENTER
7
2nd

19. -1
0
1
Example:
Find the area between the
x-axis and the curve
from to .
cos
y x

0
x 
3
2
x

 2

3
2

3
2 2
0
2
cos cos
x dx x dx
 


 
/2 3 /2
0 /2
sin sin
x x
 


3
sin sin 0 sin sin
2 2 2
  
   
  
   
   
   
1 0 1 1
    3

On the TI-89:
 
 
 
abs cos , ,0,3 / 2
x x 

3

If you use the absolute
value function, you
don’t need to find the
roots.
pos.
neg.
