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Notes for Introductory Calculus (Math 120)

Brenton LeMesurier

Section 5.4 Integration Formulas and the Net Change Theorem

References.

Now that we have seen the connection between antiderivatives and definite integrals, it is convenient to recast antiderivatives in terms of integrals, and use the notation of integrals when calculating with antiderivatives. Thus, just as \(\int_a^x f(t) \, dt\) gives one antiderivative of \(f\) (a different one for each different choice of \(a\)) we denote the general antiderivative by dropping the specific choice \(a\text{,}\) and simplifying a bit:

Definition 5.4.1.

The Indefinite Integral of \(f\) with respect to \(x\) is the most general function \(F(x)\) having \(F'(x)=f(x)\text{,}\) including an arbitrary added constant. This is denoted
\begin{equation*} \int f(x) \, dx \end{equation*}
The function inside this expression is called the integrand.
The differential “\(dx\)” is essential! For example, we can verify that
\begin{equation*} \int x \, t \, d \mathbf{x} = x^2 t/2 + C \qquad \text{while} \qquad \int x\, t\, d \mathbf{t} = x t^2/2 + C\text{.} \end{equation*}
\begin{align*} \int x^2 \, dx \amp = \frac{x^3}{3} + C\\ \int \cos x \, dx \amp = \sin x + C\\ \int \sec^2 x \, dx \amp = \tan x + C\\ \int \tan x \, dx \amp = \log|\sec x| + C\\ \int \ln x \, dx \amp = x\ln x - x + C\\ \int 2 x e^{x^2} dx \amp = e^{x^2} + C \end{align*}
How do we know that these are correct?
Differentiate the formula at right and verify that this gives the integrand, the function inside the integral expression at left.
Rather than list numerous sums and constant multiples, the list can start with the two general “combining” rules that we have for sums, differences and constant multiples of antiderivatives, rephrased as facts about integrals:
\begin{align*} \int c f(x) \, dx \amp = c \int f(x) \, dx \quad \text{for any constant c}.\\ \int f(x) \pm g(x) \, dx \amp = \int f(x) \, dx \pm \int g(x) \, dx \end{align*}
But beware: we have no rules for products or quotients or compositions of functions.

Simplify first!

As usual, it often helps to simplify the function as much as possible before looking for antiderivatives, both by using the above rules to break up sums and differences and extract constant factors and by using other algebraic rules and trigonometric facts.

Connection to Definite Integrals.

The Fundamental Theorem of Calculus gives
\begin{equation*} \int_a^b f(x) \, dx = F(b)-F(a) \end{equation*}
and we denote the difference here with the shorthand forms
\begin{equation*} \Big[ F(x) \Big]_a^b = F(x) \Big|_a^b = F(b)-F(a) \end{equation*}
so we can now use the indefinite integral notation for the antiderivative:
\begin{equation*} \int_a^b f(x) \, dx = \left[ \int f(x) \, dx \right]_a^b = \left. \int f(x) \, dx \right|_a^b \end{equation*}
(I prefer always using matching left and right brackets to avoid any possible ambiguity, but some texts use the right bracket only.)

Integrals of Derivatives and the Net Change Theorem.

The indefinite integral a function is the general antiderivative, so the indefinite integral of the derivative \(f'\) of function \(f\) is the general antiderivative of the derivative. The original function \(f\) itself is one such antiderivative, so all that remans is to add an arbitrary constant:
\begin{equation*} \int_a^b f'(x) \, dx = f(b)-f(a), = \Big[ f(x) \Big ]_a^b \end{equation*}
This says that the definite integral of the rate of change of a quantity gives the net change in the quantity.

For \(f(t)\) a Velocity, Displacement is Net Change.

For example, if function \(f\) gives position and the independent variable is time, the rate of change is velocity, \(v(t)=f'(t)\text{,}\) so the definite integral of velocity from \(a\) to \(b\) is the net change in position between times \(a\) and \(b\text{,}\) the displacement, not the total distance traveled:
The displacement between times \(a\) and \(b\) is \(\displaystyle \int_a^b v(t) \, dt = \int_a^b f'(t) \, dt = f(b)-f(a).\)
This is what we saw in Problem 2 of Section 5.1, motivating the idea of the definite integral. Geometrically, this is the difference between the area under the positive part of the graph of \(v=f'\) and the area below the negative part.

Total Distance Traveled is Total Change.

On the other hand, the total distance traveled is the “total change” of position, given by integrating the rate of change of position without regard to direction: this is speed, which is the magnitude of the velocity, \(|v(t)|\text{.}\)
The total distance traveled between times \(a\) and \(b\) is \(\displaystyle \int_a^b |v(t)| \, dt = \int_a^b |f'(t)| \, dt.\)
Geometrically, this is the total area between the graph of \(v\) and the \(t\)-axis, adding area above and below the axis. It is not given by the simple formula \(f(b)-f(a)\text{,}\) so how can it be computed?
The answer is to break the integral up into several integrals over several intervals such that on each interval, \(v=f'\) is either positive throughout or negative throughout. Then each integral is of the form either \(\int_c^d f'(t) \, dt\) or \(\int_c^d -f'(t) \, dt\text{,}\) and so each can be evaluated easily by the Net Change Theorem, as either \(f(d)-f(c)\) or \(f(c)-f(d)\text{.}\) Adding these positive pieces gives the total distance traveled.

Exercises Exercises

Study Calculus Volume 1, Section 5.4 2 ; in particular Theorem 6, Examples 23–26, 28 and 29; Checkpoints 21, 22 and 24; and Exercises 207, 209, 211 and 223.
For further practice, look at several exercises from each of the following ranges: 207-212 and 223-228.
openstax.org/books/calculus-volume-1/pages/5-4-integration-formulas-and-the-net-change-theorem
openstax.org/books/calculus-volume-1/pages/5-4-integration-formulas-and-the-net-change-theorem
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