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5.4
Indefinite Integrals and
the Net Change Theorem
Wednesday, February 10, 2010
In this section, we will learn about:Indefinite integrals and their applications.
INTEGRALS
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INDEFINITE INTEGRALS
Both parts of the FTC establish
connections between antiderivatives
and definite integrals.
Part 1 says that if, fis continuous, thenis an antiderivative of f.
Part 2 says that can be found by evaluatingF(b) F(a), where Fis an antiderivative of f.
( )x
a f t dt
( )
b
a f x dx
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Due to the relation given by the FTC between
antiderivatives and integrals, the notation
f(x) dxis traditionally used for an
antiderivative of fand is called an indefiniteintegral.
Thus, f(x) dx= F(x) means F(x) = f(x)
INDEFINITE INTEGRAL
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INDEFINITE INTEGRALS
For example, we can write
Thus, we can regard an indefinite integralas representing an entire family of functions(one antiderivative for each value of the constant C).
3 32 2
because3 3
x d x x dx C C x
dx
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INDEFINITE VS. DEFINITE INTEGRALS
You should distinguish carefully between
definite and indefinite integrals.
A definite integral is a number.
An indefinite integral f(x) dxis a function(or family of functions).
( )
b
a f x dx
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INDEFINITE INTEGRALS
Any formula can be verified by differentiating
the function on the right side and obtaining
the integrand.
For instance, 2
2
sec tan
because
(tan ) sec
x dx x C
d x C x
dx
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TABLE OF INDEFINITE INTEGRALS
1
2 2
( ) ( ) [ ( ) ( )]
( ) ( )
( 1)1
sin cos cos sin
sec tan csc cot
sec tan sec csc cot csc
nn
cf x dx c f x dx f x g x dx
f x dx g x dx
x
k dx kx C x dx C nn
x dx x C x dx x C
x dx x C x dx x C
x x dx x C x x dx x C
Table 1
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INDEFINITE INTEGRALS
Thus, we write
with the understanding that it is valid on
the interval (0, ) or on the interval (-, 0).
21 1dx Cxx
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INDEFINITE INTEGRALS
This is true despite the fact that the general
antiderivative of the function f(x) = 1/x2,
x 0, is:
1
2
1 if 0
( )1
if 0
C xx
F x
C xx
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INDEFINITE INTEGRALS
Find the general indefinite integral
(10x4 2 sec2x) dx
Using our convention and Table 1, we have:
(10x4 2 sec2x) dx= 10 x4 dx 2 sec2x dx= 10(x5/5) 2 tan x+ C= 2x5 2 tan x+ C
You should check this answer by differentiating it.
Example 1
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INDEFINITE INTEGRALS
Evaluate
This indefinite integral isnt immediately apparentin Table 1.
So, we use trigonometric identities to rewritethe function before integrating:
Example 2
2
cos 1 cos
sin sinsin
csc cot csc
d d
d C
2
cos
sin d
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INDEFINITE INTEGRALS
Evaluate
Using FTC2 and Table 1, we have:
Compare this with Example 2 b in Section 5.2
Example 33
3
0
( 6 ) x x dx
34 23
3
00
4 2 4 21 1
4 4
81
4
( 6 ) 64 2
3 3 3 0 3 0
27 0 0 6.75
x x x x dx
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INDEFINITE INTEGRALS
Find
Example 4
12
0( 12sin ) x x dx
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The FTC gives:
This is the exact value of the integral.
INDEFINITE INTEGRALS Example 4
2
12 12
00
212
12sin |12( cos )2
(12) 12(cos12 cos0)
72 12cos12 12
60 12cos12
x
x x dx x
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INDEFINITE INTEGRALS
If a decimal approximation is desired, we can
use a calculator to approximate cos 12.
Doing so, we get:
12
012sin 70.1262
x x dx
Example 4
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The figure shows the graph of the integrand
in the example.
We know from Section 5.2
that the value of theintegral can beinterpreted as the sumof the areas labeled
with a plus sign minusthe area labeled witha minus sign.
INDEFINITE INTEGRALS
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INDEFINITE INTEGRALS
Evaluate
First, we need to write the integrand in a simplerform by carrying out the division:
Example 5
2 29
21
2 1t t tdtt
2 2
9 9 1 2 221 1
2 1 (2 )t t t dt t t dt t
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Then,9
1 2 2
1
93 2 1
32
19
3 2
1
3 2 3 22 1 2 1
3 9 3 1
1 2 4
9 3 9
(2 )
21
2 12
3
(2 9 9 ) (2 1 1 )
18 18 2 1 32
t t dt
t tt
t tt
INDEFINITE INTEGRALS Example 5
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APPLICATIONS
The FTC2 says that, if fis continuous on
[a, b], then
where Fis any antiderivative of f.
This means that F= f.
So, the equation can be rewritten as:
( ) ( ) ( )b
af x dx F b F a
'( ) ( ) ( )b
aF x dx F b F a
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We know F(x) represents the rate of change
of y= F(x) with respect to xand F(b) F(a) is
the change in ywhen xchanges from ato b.
Note that ycould, for instance, increase,then decrease, then increase again.
Although ymight change in both directions,F(b) F(a) represents the net change in y.
APPLICATIONS
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NET CHANGE THEOREM
So, we can reformulate the FTC2 in words,
as follows.
The integral of a rate of change is
the net change:
'( ) ( ) ( )b
a
F x dx F b F a
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NET CHANGE THEOREM
This principle can be applied to all the rates
of change in the natural and social sciences
that we discussed in Section 3.7
The following are a few instances of the idea.
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If V(t) is the volume of water in a reservoir at
time t, its derivative V(t) is the rate at which
water flows into the reservoir at time t.
So,
is the change in the amount of waterin the reservoir between time t1 and time t2.
2
12 1
'( ) ( ) ( )t
tV t dt V t V t
NET CHANGE THEOREM
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If [C](t) is the concentration of the product of
a chemical reaction at time t, then the rate of
reaction is the derivative d[C]/dt.
So,
is the change in the concentration of Cfrom time t1 to time t2.
2
12 1
[ ][ ]( ) [ ]( )
t
t
d Cdt C t C t
dt
NET CHANGE THEOREM
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If the mass of a rod measured from the left
end to a point xis m(x), then the linear density
is (x) = m(x).
So,
is the mass of the segment of the rodthat lies between x= aand x= b.
( ) ( ) ( )b
ax dx m b m a
NET CHANGE THEOREM
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If the rate of growth of a population is dn/dt,
then
is the net change in population during the timeperiod from t1 to t2.
The population increases when births happenand decreases when deaths occur.
The net change takes into account both birthsand deaths.
2
12 1
( ) ( )t
t
dndt n t n t
dt
NET CHANGE THEOREM
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If C(x) is the cost of producing xunits of
a commodity, then the marginal cost is
the derivative C(x).
So,
is the increase in cost when productionis increased from x1 units to x2 units.
2
12 1
'( ) ( ) ( )x
xC x dx C x C x
NET CHANGE THEOREM
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If an object moves along a straight line
with position function s(t), then its velocity
is v(t) = s(t).
So,
is the net change of position, or displacement,of the particle during the time period from t1 to t2.
2
12 1
( ) ( ) ( )t
tv t dt s t s t
Equation 2NET CHANGE THEOREM
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In both cases, the distance is computed by
integrating |v(t)|, the speed.
Therefore,
Equation 3
2
1
| ( ) | total distance traveledt
tv t dt
NET CHANGE THEOREM
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The figure shows how both displacement and
distance traveled can be interpreted in terms
of areas under a velocity curve.
NET CHANGE THEOREM
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The acceleration of the object is
a(t) = v(t).
So,
is the change in velocity from time t1 to time t2.
2
12 1
( ) ( ) ( )t
ta t dt v t v t
NET CHANGE THEOREM
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A particle moves along a line so that its
velocity at time tis:
v(t) = t2 t 6 (in meters per second)
a) Find the displacement of the particle duringthe time period 1 t 4.
b) Find the distance traveled during this time period.
Example 6NET CHANGE THEOREM
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By Equation 2, the displacement is:
This means that the particle moved 4.5 mtoward the left.
Example 6 a
4 42
1 1
43 2
1
(4) (1) ( ) ( 6)
963 2 2
s s v t dt t t dt
t t t
NET CHANGE THEOREM
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So, from Equation 3, the distance traveled is:
4 3 4
1 1 3
3 42 2
1 3
3 43 2 3 2
1 3
( ) [ ( )] ( )
( 6) ( 6)
6 6
3 2 3 261
10.17 m6
v t dt v t dt v t dt
t t dt t t dt
t t t t t t
Example 6 bNET CHANGE THEOREM