5.2 Definite Integrals
Greg Kelly, Hanford High School, Richland, Washington
3
1
V  t2 1
8
When we find the area
under a curve by adding
rectangles, the answer is
called a Rieman 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.
partition
n
Area  lim  f  ck  xk
P 0
k 1
if P is a partition
of the interval a, b



n
lim  f  ck  xk
P 0
is called the definite integral of
k 1
f
over
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:
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.

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.

b
a
f  x  dx
We have the notation for integration, but we still need
to learn how to evaluate the integral.

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.
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

3
If the velocity varies:
1
v  t 1
2
Distance:
2
1
1 2
s  t t
4
0
2
x
3
4
1
Area  1  3 4  8
2
(C=0 since s=0 at t=0)
After 4 seconds:
1
1
s  16  4
4
s 8
The distance is still
equal to the area
under the curve!
Notice that the area is a trapezoid.

1 2
What if: v  t  1
8
3
2
1
0
1
2
x
3
4
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.

ds 1 2
v
 t 1
dt 8
3
2
1 3
s
t t
24
1
0
1 3
s
4 4
24
2
s6
3
1
2
x
3
4
2
The area under the curve  6
3
We can use anti-derivatives to
find the area under a curve!

Let’s look at it another way:
Let
Aa  x   area under the
curve from a to x.
(“a” is a constant)
a
x xh
Aa  x  Ax  x  h 
Aa  x  h 
Then:
Aa  x   Ax  x  h   Aa  x  h 
Ax  x  h   Aa  x  h   Aa  x 

min f
max f
xh
x
h
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
h  min f  Aa  x  h   Aa  x   h  max f
min f 
Aa  x  h   Aa  x 
h
 max f

min f 
Aa  x  h   Aa  x 
h
 max f
As h gets smaller, min f and max f get closer together.
lim
h 0
Aa  x  h   Aa  x 
h
 f  x
d
Aa  x   f  x 
dx
initial
value
Take the anti-derivative
of both
sides to find an explicit formula
for area.
AaThis
F definition
x  c
 x  is the
of derivative!
Aa  a   F  a   c
0  F a  c
F  a   c

min f 
Aa  x  h   Aa  x 
h
 max f
As h gets smaller, min f and max f get closer together.
lim
h 0
Aa  x  h   Aa  x 
h
 f  x
d
Aa  x   f  x 
dx
Aa  x   F  x   F  a 
Aa  x   F  x   c
Aa  a   F  a   c
0  F a  c
F  a   c
Area under curve from a to x = antiderivative at x minus
antiderivative at a.

n
Area
 lim  f  ck  xk
P 0
k 1
  f  x  dx
b
a
 F  x  F a

yx
Example:
4
2
Find the area under the curve from
x=1 to x=2.
3

2
2
1
1
2
x dx
2
0
1
2
1 3
x
3 1
1 3 1
 2  1
3
3
8 1
7
 
3 3
3
Area
from
Areax=from
Area
under
thex=0
curve from
1 to x=0
x=2.
to x=2
to x=1

yx
Example:
4
2
Find the area under the curve from
x=1 to x=2.
3
2
To do the same problem on the TI-89:
1
0
1
  x ^ 2, x,1, 2 
2
2nd
ENTER
7

Example:
Find the area between the
x-axis and the curve
y  cos x
3 .
from x  0 to x 
2


2
0
pos.
0
-1
3
2
cos x dx   cos x dx
2
 /2
3 / 2
sin x 0  sin x  / 2

3


 
 sin 
 sin  sin 0    sin
2
2
2

 
1  0   1 1
3
2
1
3

2
neg.
On the TI-89:
  abs  cos  x  , x, 0,3 / 2
3
If you use the absolute
value function, you
don’t need to find the
roots.
