5.5: Indefinite Integrals and the Net Change Theorem

The Net Change Theorem

The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.

Subtracting \(F(a)\) from both sides of the top equation yields the bottom one. Since they are equivalent formulas, which one we use depends on the application.

The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.

We looked at a simple example of this in The Definite Integral section. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in Figure \(\PageIndex{1}\).

Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by

\[ ∫^5_2v(t)\,dt=∫^4_240\,dt+∫^5_4−30\,dt=80−30=50 \nonumber \]

Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by

\[ ∫^5_2|v(t)|\,dt=∫^4_240\,dt+∫^5_430\,dt=80+30=110 \nonumber \]

Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.

To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.

Example \(\PageIndex{1}\): Finding Net Displacement

Given a velocity function \(v(t)=3t−5\) (in meters per second) for a particle in motion from time \(t=0\) to time \(t=3,\) find the net displacement of the particle.

Solution

Applying the net change theorem, we have

\[ ∫^3_0(3t−5)\,dt=\left(\frac{3t^2}{2}−5t\right)\bigg|^3_0=\left[\frac{3(3)^2}{2}−5(3)\right]−0=\frac{27}{2}−15=\frac{27}{2}−\frac{30}{2}=−\frac{3}{2} \nonumber \]

The net displacement is \( \boxed{−\frac{3}{2} \text{ meters}}\)

Applying the Net Change Theorem

The net change theorem can be applied to the flow and consumption of fluids, as shown below.

Example \(\PageIndex{4}\): Chapter Opener: Iceboats

As we saw at the beginning of the chapter, top iceboat racers can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed.

Suppose Andrew takes his iceboat out one morning when a light \(5\)-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function \(v(t)=20t+5.\) For the second half hour of Andrew’s outing, the wind remains steady at \(15\) mph. In other words, the wind speed is given by

\[ v(t)=\begin{cases}20t+5 & \text{for } 0≤t≤\dfrac{1}{2}\\15 & \text{for } \dfrac{1}{2}≤t≤1\end{cases} \nonumber \]

Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after \(1\) hour?

Solution

To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then

\[ ∫^1_02v(t)\,dt \nonumber \]

Substituting the expressions we were given for \(v(t)\), we get

\[ \begin{align*} \text{Distance} &=∫^1_02v(t)\,dt \\[4pt] &=∫^{\frac{1}{2}}_02v(t)\,dt+∫^1_{\frac{1}{2}}2v(t)\,dt \\[4pt]
&=∫^{\frac{1}{2}}_02(20t+5)\,dt+∫^1_{\frac{1}{2}}2(15)\,dt \\[4pt]
&=∫^{\frac{1}{2}}_0(40t+10)\,dt+∫^1_{\frac{1}{2}}30\,dt \\[4pt]
&=\big(20t^2+10t\big)\bigg|^{\frac{1}{2}}_0+\big(30t\big)\bigg|^1_{\frac{1}{2}} \\[4pt]
&=\left(\frac{20}{4}+5\right)−0+(30−15) \\[4pt]
&=25 \end{align*}\]

Andrew is \(\boxed{25 \text{ miles}}\) from his starting point after 1 hour.

Integrating Even and Odd Functions

We saw in Functions and Graphs that an even function is a function in which \(f(−x)=f(x)\) for all \(x\) in the domain—that is, the graph of the curve is unchanged when \(x\) is replaced with \(−x\). The graphs of even functions are symmetric about the \(y\)-axis. An odd function is one in which \(f(−x)=−f(x)\) for all \(x\) in the domain, and the graph of the function is symmetric about the origin.

Integrals of even functions, when the limits of integration are from \(−a\) to \(a\), involve two equal areas, because they are symmetric about the \(y\)-axis. Integrals of odd functions, when the limits of integration are similarly \([−a,a],\) evaluate to zero because the areas above and below the \(x\)-axis are equal.

Example \(\PageIndex{5}\): Integrating an Even Function

Integrate the even function \(\displaystyle ∫^2_{−2}\left(3x^8−2\right)\,dx\) and verify that the integration formula for even functions holds.

Solution

The symmetry appears in the graphs in Figure \(\PageIndex{4}\). Graph (a) shows the region below the curve and above the \(x\)-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above the curve and below the \(x\)-axis. The signed area of this region is negative. Both views illustrate the symmetry about the \(y\)-axis of an even function. We have

\[ \begin{align*} ∫^2_{−2}\left(3x^8−2\right)\,dx &=\left(\frac{x^9}{3}−2x\right)\bigg|^2_{−2} \\[4pt]
&=\left[\frac{2^9}{3}−2(2)\right]−\left[\frac{(−2)^9}{3}−2(−2)\right] \\[4pt]
&= \left(\frac{512}{3}−4\right)−\left(−\frac{512}{3}+4\right) \\[4pt]
&=\boxed{\frac{1000}{3}} \end{align*}\]

To verify the integration formula for even functions, we can calculate the integral from \(0\) to \(2\) and double it, then check to make sure we get the same answer.

\[ ∫^2_0\left(3x^8−2\right)\,dx=\left(\frac{x^9}{3}−2x\right)\bigg|^2_{0}=\frac{512}{3}−4=\frac{500}{3} \nonumber \]

Since \( 2⋅\dfrac{500}{3}=\dfrac{1000}{3},\) we have verified the formula for even functions in this particular example.

Example \(\PageIndex{6}\): Integrating an Odd Function

Evaluate the definite integral of the odd function \(−5 \sin (x)\) over the interval \([−π,π]\)

Solution

The graph is shown in Figure \(\PageIndex{5}\). We can see the symmetry about the origin by the positive area above the \(x\)-axis over \([−π,0]\), and the negative area below the \(x\)-axis over \([0,π].\) we have

\[ ∫^π_{−π}−5\sin (x) \,dx =−5(−\cos(x))\bigg|^π_{−π} =5\cos (x)\,\bigg|^π_{−π} =5\cos (π)−5\cos(−π) =−5−(−5)=\boxed{0} \nonumber \]

Indefinite Integrals

We now look at the formal notation used to represent anti-derivatives and examine some of their properties. These properties allow us to find anti-derivatives of more complicated functions. Given a function \(f\), we use the notation \(f′(x)\) or \(\dfrac{df}{dx}\) to denote the derivative of \(f\). Here we introduce notation for anti-derivatives. If \(F\) is an anti-derivative of \(f\), we say that \(F(x)+C\) is the most general anti-derivative of \(f\) and write

\[\int f(x)\,dx=F(x)+C \nonumber \]

The symbol \(\int \) is called an integral sign, and \(\int f(x)\,dx\) is called the indefinite integral of \(f\).

Given the terminology introduced in this definition, the act of finding the anti-derivatives of a function \(f\) is usually referred to as integrating \(f\).

For a function \(f\) and an anti-derivative \(F\), the functions \(F(x)+C\), where \(C\) is any real number, is often referred to as the family of anti-derivatives of \(f\). For example, since \(x^2\) is an anti-derivative of \(2x\) and any anti-derivative of \(2x\) is of the form \(x^2+C,\) we write

\[\int 2x\,dx=x^2+C \nonumber \]

The collection of all functions of the form \(x^2+C,\) where \(C\) is any real number, is known as the family of anti-derivatives of \(2x\). The figure below shows a graph of this family of anti-derivatives.

Evaluating indefinite integrals for any function is the same as finding this family of anti-derivatives. This notation is used since we calculate anti-derivatives while solving definite integrals, the only difference is that we do not get numbers to plug in once we are done.

When claiming that

\[\int f(x)\, dx=F(x)+C \nonumber\]

it is important to check whether this statement is correct by verifying that \(F′(x)=f(x).\)

We have already discussed the properties of the anti-derivative that are inherited from differentiation rules. These properties are summarized below:

From this theorem, we can evaluate any integral involving a sum, difference, or constant multiple of functions with anti-derivatives that are known. Evaluating integrals involving products, quotients, or compositions is more complicated (as we saw in Example \(\PageIndex{8b}\) above). We look at and address integrals involving these more complicated functions in Introduction to Integration. In the next example, we examine how to use this theorem to calculate the indefinite integrals of several functions.

Key Concepts

• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positive \(x\)-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.
• The indefinite integral represents the family of anti-derivatives of a function

Key Equations

• Net Change Theorem \[F(b)=F(a)+∫^b_aF'(x)\,dx \quad \text{ or equivalently, }\quad ∫^b_aF'(x)\,dx=F(b)−F(a) \nonumber \]

Glossary

indefinite integral

the most general anti-derivative of \(f(x)\) is the indefinite integral of \(f\); we use the notation \(\int f(x)dx\) to denote the indefinite integral of \(f\)

net change theorem

if we know the rate of change of a quantity, the net change theorem says the future quantity is equal to the initial quantity plus the integral of the rate of change of the quantity