INDEFINITE INTEGRALS
Both parts of the FTC establish
connections between antiderivatives
and definite integrals.
Part 1 says that if, f is continuous, then is an antiderivative of f.
Part 2 says that can be found by evaluating F(b) – F(a), where F is an antiderivative of f.
( )x
af t dt
( )b
af x dx
INDEFINITE INTEGRALS
We need a convenient notation for
antiderivatives that makes them easy
to work with.
Due to the relation given by the FTC between
antiderivatives and integrals, the notation
∫ f(x) dx is traditionally used for an
antiderivative of f and is called an indefinite
integral.
Thus, ∫ f(x) dx = F(x) means F’(x) = f(x)
INDEFINITE INTEGRAL
INDEFINITE INTEGRALS
For example, we can write
Thus, we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C).
3 32 2because
3 3
x d xx dx C C x
dx
INDEFINITE VS. DEFINITE INTEGRALS
You should distinguish carefully between
definite and indefinite integrals.
A definite integral is a number.
An indefinite integral ∫ f(x) dx is a function (or family of functions).
( )b
af x dx
The connection between them is given
by the FTC2.
If f is continuous on [a, b], then
( ) ( )bb
a af x dx f x dx
INDEFINITE VS. DEFINITE INTEGRALS
INDEFINITE INTEGRALS
The effectiveness of the FTC depends
on having a supply of antiderivatives
of functions.
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
dx C x
dx
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
xk dx kx C x dx C n
n
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
INDEFINITE INTEGRALS
We write
with the understanding that it is valid on
the interval (0, ∞) or on the interval (-∞, 0).
2
1 1dx C
xx
INDEFINITE INTEGRALS
This is true despite the fact that the general
antiderivative of the function f(x) = 1/x2,
x ≠ 0, is:
1
2
1if 0
( )1
if 0
C xxF x
C xx
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.
INDEFINITE INTEGRALS
Evaluate
This indefinite integral isn’t immediately apparent in Table 1.
So, we use trigonometric identities to rewrite the function before integrating:
2
cos 1 cos
sin sinsin
csc cot csc
d d
d C
2
cos
sind
INDEFINITE INTEGRALS
Evaluate
Using FTC2 and Table 1, we have:
3 3
0( 6 )x x dx
34 23 3
00
4 2 4 21 14 4
814
( 6 ) 64 2
3 3 3 0 3 0
27 0 0 6.75
x xx x dx
INDEFINITE INTEGRALS
Find
12
0( 12sin ) x x dx
The FTC gives:
This is the exact value of the integral.
INDEFINITE INTEGRALS
212 12
00
212
12sin 12( cos )2
(12) 12(cos12 cos0)
72 12cos12 12
60 12cos12
xx x dx x
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
The figure shows the graph of the integrand
in the example.
We know that the value of the integral can be interpreted as the sum of the areas labeled with a plus sign minus the area labeled with a minus sign.
INDEFINITE INTEGRALS
INDEFINITE INTEGRALS
Evaluate
First, we need to write the integrand in a simpler form by carrying out the division:
2 29
21
2 1t t tdt
t
2 29 9 1 2 221 1
2 1(2 )
t t tdt t t dt
t
Then,9 1 2 2
1
93 2 1
32 1
93 2
1
3 2 3 22 1 2 13 9 3 1
1 2 49 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
APPLICATIONS
The FTC2 says that, if f is continuous on
[a, b], then
where F is 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
We know F ’(x) represents the rate of change
of y = F(x) with respect to x and F(b) – F(a) is
the change in y when x changes from a to b.
Note that y could, for instance, increase, then decrease, then increase again.
Although y might change in both directions, F(b) – F(a) represents the net change in y.
APPLICATIONS
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
aF x dx F b F a
NET CHANGE THEOREM
This principle can be applied to all the rates
of change in the natural and social sciences.
The following are a few instances of the idea.
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 water in the reservoir between time t1 and time t2.
2
12 1'( ) ( ) ( )
t
tV t dt V t V t
NET CHANGE THEOREM
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 C from time t1 to time t2.
2
12 1
[ ][ ]( ) [ ]( )
t
t
d Cdt C t C t
dt
NET CHANGE THEOREM
If the mass of a rod measured from the left
end to a point x is m(x), then the linear density
is ρ(x) = m’(x).
So,
is the mass of the segment of the rod that lies between x = a and x = b.
( ) ( ) ( )b
ax dx m b m a
NET CHANGE THEOREM
If the rate of growth of a population is dn/dt,
then
is the net change in population during the time
period from t1 to t2.
The population increases when births happen and decreases when deaths occur.
The net change takes into account both births and deaths.
2
12 1( ) ( )
t
t
dndt n t n t
dt
NET CHANGE THEOREM
If C(x) is the cost of producing x units of
a commodity, then the marginal cost is
the derivative C’(x).
So,
is the increase in cost when production is increased from x1 units to x2 units.
2
12 1'( ) ( ) ( )
x
xC x dx C x C x
NET CHANGE THEOREM
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
NET CHANGE THEOREM
If we want to calculate the distance the object
travels during that time interval, we have to
consider the intervals when:
v(t) ≥ 0 (the particle moves to the right)
v(t) ≤ 0 (the particle moves to the left)
NET CHANGE THEOREM
In both cases, the distance is computed by
integrating |v(t)|, the speed.
Therefore,
2
1
| ( ) | total distance traveledt
tv t dt
NET CHANGE THEOREM
The figure shows how both displacement and
distance traveled can be interpreted in terms
of areas under a velocity curve.
NET CHANGE THEOREM
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
A particle moves along a line so that its
velocity at time t is:
v(t) = t2 – t – 6 (in meters per second)
a) Find the displacement of the particle during the time period 1 ≤ t ≤ 4.
b) Find the distance traveled during this time period.
NET CHANGE THEOREM
By Equation 2, the displacement is:
This means that the particle moved 4.5 m toward the left.
Example 6 a
4 4 2
1 1
43 2
1
(4) (1) ( ) ( 6)
96
3 2 2
s s v t dt t t dt
t tt
NET CHANGE THEOREM
Note that
v(t) = t2 – t – 6 = (t – 3)(t + 2)
Thus, v(t) ≤ 0 on the interval [1, 3] and v(t) ≥ 0 on [3, 4]
NET CHANGE THEOREM
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 63 2 3 2
6110.17 m
6
v t dt v t dt v t dt
t t dt t t dt
t t t tt t
NET CHANGE THEOREM
We can use a Ti-calculator to find this definite integral.Use math 9 on Ti-83 & 84.
The figure shows the power consumption in
San Francisco for a day in September. P is measured in megawatts. t is measured in hours starting at midnight.
Estimate the
energy used
on that day.
Power is the rate of change of energy : P(t) = E’(t)
Power is the rate of change of energy:
P(t) = E’(t)
So, by the Net Change Theorem,
is the total amount of energy used that day.
24 24
0 0( ) '( ) (24) (0)P t dt E t dt E E
NET CHANGE THEOREM
We approximate the value of
the integral using the Midpoint Rule
with 12 subintervals and ∆t = 2,
as follows.
NET CHANGE THEOREM
The energy used was approximately 15,840 megawatt-hours.
24
0( )
[ (1) (3) (5) ... (21) (23)]
(440 400 420 620 790 840 850
840 810 690 670 550)(2)
15,840
P t dt
P P P P P t
NET CHANGE THEOREM
How did we know what
units to use for energy in
the example?
NET CHANGE THEOREM
The integral is defined as the limit
of sums of terms of the form P(ti*) ∆t.
Now, P(ti*) is measured in megawatts and
∆t is measured in hours. So, their product is measured in megawatt-hours. The same is true of the limit.
24
0( )P t dt
NET CHANGE THEOREM
In general, the unit of measurement for
is the product of the unit for f(x) and
the unit for x.
( )b
af x dx
NET CHANGE THEOREM