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5.2 Definite Integrals
When we find the area under a curve by adding rectangles, the answer is called a Riemann sum.
211
8V 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.
211
8V 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 limn
k kP
k
f c x
if P is a partition of the interval ,a b
0
1
limn
k kP
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
xn
The definite integral is then given by:
1
limn
kn
k
f c x
1
limn
kn
k
f c x
Leibniz introduced a simpler notation for the definite integral:
1
limn b
k ank
f c x f x dx
Note that the very small change in x becomes dx.
b
af x dx
IntegrationSymbol
lower limit of integration
upper limit of integration
integrandvariable of integration
(dummy variable)
It is called a dummy variable because the answer does not depend on the variable chosen.
b
af x dx
We have the notation for integration, but we still need to learn how to evaluate the integral.
Note: all continuous functions are integrable. That is, if a Function is continuous on [a,b] then its definite integral over[a,b] exists.
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.
ft3 4 sec 12 ft
sec
If the velocity varies:
11
2v t
Distance:21
4s t t
(C=0 since s=0 at t=0)
After 4 seconds:1
16 44
s
8s
1Area 1 3 4 8
2
The distance is still equal to the area under the curve!
Notice that the area is a trapezoidThat can be solved graphically.
211
8v 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. We would use the sum expressions to solve.
It seems reasonable that the distance travelled will again equal the area under the curve.
211
8
dsv t
dt
31
24s t t
314 4
24s
26
3s
The area under the curve2
63
We can use anti-derivatives to find the area under a curve!
Note – for some problems, using algebraic integration is either very difficult or even impossible!
– drawing a diagram and solving pictorially is still an excellent method!
0
5
2x25
Use the graph of the integrand and areas to evaluate the integral:
Since this is really asking for the area of a quartercircle of radius 5, we canuse the area formula!
x
y
Area
4254)5(
4r
2
2
Let’s look at it another way:
a x
Let area under the
curve from a to x.
(“a” is a constant)
aA x
x h
aA x
Then:
a x aA x A x h A x h
x a aA x h A x h A x
xA x h
aA x h
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 ah f A x h A x h f
h
min max a aA x h A x
f fh
min max a aA x h A x
f fh
As h gets smaller, min f and max f get closer together.
0
lim a a
h
A x h A xf x
h
This is the definition
of derivative!
a
dA x f x
dx
Take the anti-derivative of both sides to find an explicit formula for area.
aA x F x c
aA a F a c
0 F a c
F a c initial value
min min a aA x h A x
f fh
As h gets smaller, min f and max f get closer together.
0
lim a a
h
A x h A xf x
h
a
dA x f x
dx
aA x F x c
aA a F a c
0 F a c
F a c aA x F x F a
Area under curve from a to x = antiderivative at x minus
antiderivative at a.
0
1
limn
k kP
k
f c x
b
af x dx
Area
= F(b) – F(a)
Area from x=0to x=1
Example: 2y x
Find the area under the curve from x=1 to x=2.
2 2
1x dx
23
1
1
3x
31 12 1
3 3
8 1
3 3
7
3
Area from x=0to x=2
Area under the curve from x=1 to x=2.
=
=
=
Example: 2y x
Find the area under the curve from x=1 to x=2.
To do the same problem on the TI-83:
ENTERfnInt(x^2,x,1,2)
Math 9
Example:
Find the area between the
x-axis and the curve
from to .
cosy x
0x 3
2x
2
3
2
3
2 2
02
cos cos x dx x dx
/ 2 3 / 2
0 / 2sin sinx x
3sin sin 0 sin sin
2 2 2
1 0 1 1 3
On the TI-83:
3
If you use the absolute value function, you don’t need to find the roots.
pos.
neg.
fnInt(abs(cos(x)),x,0,3 /2)
In general, if a function f takes on both positive and negativevalues, then we can interpret the integral as the difference of areas or the net area of the region.
21)( AAdxxfb
a
A1
A2
Where A1 is the regionabove the x-axis and below the graph of f
Where A2 is the regionbelow the x-axis and above the graph of f
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