4.4 Indefinite Integrals and
the Net Change Theorem
Indefinite Integrals and the Net Change Theorem
In this section we introduce a notation for antiderivatives, review the formulas for antiderivatives, and use them to evaluate definite integrals.
We also reformulate FTC2 in a way that makes it easier to apply to science and engineering problems.
Indefinite Integrals
Indefinite Integrals
Both parts of the Fundamental Theorem establish
connections between antiderivatives and definite integrals.
Part 1 says that if f is continuous, then dt 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.
We need a convenient notation for antiderivatives that makes them easy to work with.
Because of the relation between antiderivatives and
integrals given by the Fundamental Theorem, the notation is traditionally used for an antiderivative of f and is called an indefinite integral.
Indefinite Integrals
Thus
For example, we can write
Indefinite Integrals
You should distinguish carefully between definite and
indefinite integrals. A definite integral is a number, whereas an indefinite integral is a function (or
family of functions).
The connection between them is given by Part 2 of the Fundamental Theorem:
if f is continuous on [a, b], then
The effectiveness of the Fundamental Theorem 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,
Indefinite Integrals
The most general antiderivative on a given interval is
obtained by adding a constant to a particular antiderivative.
Indefinite Integrals
We adopt the convention that when a formula for a
general indefinite integral is given, it is valid only on an interval.
Thus we write
with the understanding that it is valid on the interval (0, ) or on the interval ( 0).
Indefinite Integrals
This is true despite the fact that the general antiderivative of the function f(x) = 1/x2, x ≠ 0, is
Example 2
Evaluate .
Solution:
This indefinite integral isn’t immediately apparent in
Table 1, so we use trigonometric identities to rewrite the function before integrating:
Example 5
Evaluate
Solution:
First we need to write the integrand in a simpler form by carrying out the division:
Example 5 – Solution
cont’dApplications
Applications
Part 2 of the Fundamental Theorem 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
Applications
We know that 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
So we can reformulate FTC2 in words as follows.
This principle can be applied to all of the rates of change in the natural and social sciences. Here are a few instances of this idea:
Applications
So
is the change in the amount of water in the reservoir between time t1 and time t2.
• 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.
Applications
So
is the change in the concentration of C from time t1 to time t2.
• 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
Applications
• 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.)
Applications
• 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.
Applications
• 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.
This was true for the case where the object moves in the positive direction, but now we have proved that it is
always true.
Applications
• If we want to calculate the distance the object travels
during the time interval, we have to consider the intervals when v(t) ≥ 0 (the particle moves to the right) and also the intervals when v(t) ≤ 0 (the particle moves to the left).
In both cases the distance is computed by integrating | v(t)|, the speed. Therefore
total distance traveled
Applications
Figure 3 shows how both displacement and distance traveled can be interpreted in terms of areas under a velocity curve.
Figure 3
Applications
• The acceleration of the object is a(t) = v′(t), so
is the change in velocity from time t1 to time t2.
Example 6
A particle moves along a line so that its velocity at time t is v(t) = t2 – t – 6 (measured 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.
Example 6 – Solution
(a) By Equation 2, the displacement is
Example 6 – Solution
This means that the particle moved 4.5 m toward the left.
(b) Note that v(t) = t2 – t – 6 = (t – 3)(t + 2) and so v(t) ≤ 0 on the interval [1, 3] and v(t) ≥ 0 on [3, 4].
Thus, from Equation 3, the distance traveled is
cont’d
