5.9 The Indefinite Integral; Growth and Decay Models

OBJECTIVES

When you finish this section, you should be able to:

  1. Find indefinite integrals (p. 379)
  2. Use properties of indefinite integrals (p. 380)
  3. Solve differential equations involving growth and decay (p. 382)

The Fundamental Theorem of Calculus establishes an importantrelationship between definite integrals and antiderivatives: thedefinite integral \(\int_{a}^{b}f(x)\,dx\) can be found easily if anantiderivative of \(f\) can be found. Because of this, it iscustomary to use the integral symbol \(\int \) as an instruction to find all antiderivatives of a function.

DEFINITION Indefinite Integral

The expression \(\int f(x)\,dx\), called the indefinite integral of \({\boldsymbol f}\), is defined as,\[\boxed{\bbox[5pt]{\int f(x)\, dx=F(x) +C }}\]

where \(F\) is any function for which \(\dfrac{d}{dx} F(x) =f(x)\)and \(C\) is a number, called the constant of integration.

CAUTION

In writing an indefinite integral \(\int f(x) \,dx\), remember toinclude the “\({ dx}.\)”

For example, \[\int ({x^{2}+1})\, dx=\dfrac{x^{3}}{3}+x +C \qquad {\color{#0066A7}{ \dfrac{d}{dx}\left(\dfrac{x^{3}}{3}+x+C\right) = \dfrac{3x^{2}}{3}+1+0=x^{2}+1}}\]

The process of finding either the indefinite integral \(\int f(x)dx\) or the definite integral \(\int_{a}^{b}f(x)dx\) is called integration, and in both cases the function \(f\) is called the integrand.

It is important to distinguish between the definite integral \({\int_{a}^{b}{f(x)\,dx}}\) and the indefinite integral \(\int f(x)\,dx.\) The definite integral is a number that depends on the limits of integration \(a\) and \(b.\) In contrast, the indefinite integral of \(f\) is a family of functions \(F(x) +C,\) \(C\) a constant, for which \(F^\prime (x) =f(x) .\) For example, \[\int_{0}^{2}x^{2}dx=\left[ \dfrac{x^{3}}{3}\right] _{0}^{2}=\dfrac{8}{3}\qquad \intx^{2}dx=\dfrac{x^{3}}{3}+C\]

NOTE

The definite integral \(\int_{a}^{b}f(x)\, dx\) is a number; the indefinite integral \(\int f(x)\, dx\) is a family of functions.

We summarize the antiderivatives of some importantfunctions in Table # on page 380. Each entry is a result of a differentiation formula.

NEED TO REVIEW?

Refer to Table 7 in Section 4.8, pp. xx-xx.

5.9.1 Find Indefinite Integrals

EXAMPLE 1 Finding Indefinite Integrals

Find:

  1. \(\int x^{4}\, dx\)
  2. \(\int \sqrt{x} dx\)
  3. \(\int \dfrac{\sin x}{\cos ^{2}x} dx\)

Solution (a) All the antiderivatives of \(f(x)=x^{4} \) are \(F(x) =\dfrac{x^{5}}{5}+C\), so \[\int x^{4}\, dx=\dfrac{x^{5}}{5}+C\]

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Table of Integrals
\(\int \, dx=x+C\) \(\int \sec x\tan x\,dx=\sec x+C\)
\(\int x^{a }\, dx=\dfrac{x^{a +1}}{a +1}+C;\)  \(a \neq-1\) \(\int \csc x\cot x\,dx=-\csc x+C\)
\(\int x^{-1}\, dx=\int \dfrac{1}{x}\, dx=\ln \vert x\vert +C\) \(\int \csc ^{2}x\,dx=-\cot x+C\)
\(\int e^{x}\, dx=e^{x}+C\) \(\int \dfrac{1}{\sqrt{1-x^{2}}} dx=\) \(\sin^{-1}x+C\),  \( \vert x\vert \lt1\)
\(\int a^{x}\, dx=\dfrac{a^{x}}{\ln a}+C;\)  \(a>0,\) \(a\neq 1\) \(\int \dfrac{1}{1+x^{2}}\, dx=\tan ^{-1}x+C\)
\(\int \sin x\,dx=-\cos x+C\) \(\int \dfrac{1}{x\sqrt{x^{2}-1}}\, dx=\sec ^{-1}x+C\,\),  \(\vertx\vert >1\)
\(\int \cos x\,dx=\sin x+C\) \(\int \sinh x\,dx=\cosh x+C\,\)
\(\int \sec ^{2}x\,dx=\tan x+C\) \(\int \cosh x\,dx=\sinh x+C\)

(b) All the antiderivatives of \(f(x) =\sqrt{x}=x^{1/2} \) are \(F(x) =\dfrac{x^{3/2}}{\dfrac{3}{2}}+C=\dfrac{2x^{3/2}}{3}+C\), so \[\int \sqrt{x} dx=\dfrac{2x^{3/2}}{3}+C\]

(c) No antiderivative in Table # corresponds to \(f(x)=\dfrac{\sin x}{\cos ^{2}x}\), so we begin by using trigonometric identities to rewrite \(\dfrac{\sin x}{\cos ^{2}x}\) in a form whose antiderivative is recognizable. \[\dfrac{\sin x}{\cos ^{2}x}=\dfrac{\sin x}{\cos x\cdot \cos x}=\dfrac{1}{\cosx}\cdot \dfrac{\sin x}{\cos x}=\sec x\tan x\]

NEED TO REVIEW?

Trigonometric identities are discussed in Appendix A.4, pp. A-32 to A-35.

Then\[\int \dfrac{\sin x}{\cos ^{2}x}\, dx=\int \sec x\tan x dx=\sec x+C \]

NOW WORK

Problems 5 and 7.

5.9.2 Use Properties of Indefinite Integrals

Since the definite integral and the indefinite integral are closely related,properties of indefinite integrals are very similar to those of definiteintegrals:

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Property (1) is a consequence of the definition of \(\int f(x)\,dx.\) For example,\[\dfrac{d}{dx} \int \sqrt{x^{2}+1}\,dx=\sqrt{x^{2}+1} \qquad \dfrac{d}{dt}\int e^{t}\cos t\,dt=e^{t}\cos t\]

The proof of property (2) follows directly from properties ofderivatives, and is left as an exercise. See Problem 68.

IN WORDS

The indefinite integral of a sum of two functions equals the sumof the indefinite integrals.

To prove property (3), differentiate the right side of the (3). \[\begin{eqnarray*}{\dfrac{{d}}{{dx}}}{\left[ {k\int {f(x)\,dx}}\right] } \underset{\underset{{\color{#0066A7}{\hbox{Constant Multiple Rule}}}}{\color{#0066A7}{\uparrow}}} {=} k{\left[ {{\dfrac{{d}}{{dx}}}\int {f(x)\,dx}}\right] } \underset{\underset{{\color{#0066A7}{\hbox{Property (1)}}}}{\color{#0066A7}{\uparrow}}} {=}k\,f(x)\end{eqnarray*}\]

IN WORDS

To find the indefinite integral of a constant \(k\) times afunction \(f\), find the indefinite integral of \(f\) and thenmultiply by \(k\).

EXAMPLE 2 Using Properties of the Indefinite Integral

\[\begin{eqnarray*}\int (2x^{1/3}+5x^{-1})\,dx &=& \int 2x^{1/3}\,dx+\int \dfrac{5}{x}\,dx=2\intx^{1/3}\,dx+5\int \dfrac{1}{x}\,dx \\&=& 2\cdot {\dfrac{{x^{4/3}}}{{{\dfrac{{4}}{{3}}}}}}+{5\ln \,}\vert x\vert +C={\dfrac{{3x^{4/3}}}{{2}}}+5\ln\,\vert x\vert +C \end{eqnarray*}\]

NOW WORK

Problem 13.

Sometimes an appropriate algebraic manipulation is required before integrating.

EXAMPLE 3 Using Properties of the Indefinite Integral

  1. \[\begin{eqnarray*}\int \left( \dfrac{12}{x^{5}}+\dfrac{1}{\sqrt{x}}\right)dx&=&12\int \dfrac{1}{x^{5}}\,dx+\int \dfrac{1}{\sqrt{x}}\,dx=12\intx^{-5}\,dx+\int x^{-1/2}\,dx \\&=& 12\left( \dfrac{x^{-4}}{{-}4}\right) +\dfrac{x^{1/2}}{\dfrac{1}{2}}+C = -\dfrac{3}{x^{4}}+2\sqrt{x}+C\end{eqnarray*}\]
  2. \[\begin{eqnarray*}\int \dfrac{x^{2}+6}{x^{2}+1} dx & \underset{\underset{{\color{#0066A7}{\hbox{Algebra}}}}{\color{#0066A7}{\uparrow}}}{=}& \int \dfrac{( x^{2}+1) +5}{x^{2}+1} dx =\int \left[ \dfrac{x^{2}+1}{x^{2}+1}+\dfrac{5}{x^{2}+1}\right] dx \\&=&\int \left[ 1+\dfrac{5}{x^{2}+1}\right] dx \underset{\underset{{\color{#0066A7}{\hbox{Sum Property}}}}{\color{#0066A7}{\uparrow }}} {=} \int dx+\int \dfrac{5}{x^{2}+1} dx \\&=&\int dx+5\int \dfrac{1}{x^{2}+1} dx=x+5 \tan ^{-1}x+C \end{eqnarray*}\]

NOW WORK

Problem 33.

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5.9.3 Solve Differential Equations Involving Growth and Decay

There are situations in science and nature, such as radioactive decay, population growth, and interest paid on an investment, in which a quantity \(A \) varies with time \(t\) in such a way that the rate of change of \(A\) with respect to \(t\) is proportional to \(A\) itself. These situations can be modeled by the differential equation \[\boxed{\bbox[#FAF8ED,5pt]{\dfrac{{\it dA}}{dt}=kA }}\tag{4}\]

where \(k\neq 0\) is a real number:

Suppose that the initial amount \(A_{0}\) of the substance is known, giving us the boundary condition, or initial condition, \(A=A(0) =A_{0}\) when \(t=0.\)

We solve differential equations of the form \(\dfrac{\itdA}{dt}={\it kA}\) by writing \(\dfrac{{\it dA}}{dt}={\it kA}\) as\(\dfrac{{\it dA}}{A}=k\,dt.\)Footnote # Then we integrate both sides of the equation, on the left with respect to \(A\) and on the right with respect to \(t.\)\[\begin{eqnarray*}\int \dfrac{1}{A} {\it dA} &=&\int k\,dt \\\ln \left\vert A\right\vert &=&kt+C \\\ln A &=&k\,t+C\qquad A \gt 0\end{eqnarray*}\]

*This technique for solving a differential equation, called separating the variables, is discussed in moredetail in Chapter 16.

The initial condition requires that \(A=A_{0}\) when \(t=0\). Then \(\ln A_{0}=C\), so\[\begin{eqnarray*}\ln A &=&kt+\ln A_{0} \\\ln A-\ln A_{0} &=&kt \\\ln \dfrac{A}{A_{0}} &=&kt \\\dfrac{A}{A_{0}} &=&e^{kt} \\A &=&A_{0}e^{kt}\\\end{eqnarray*}\]

The solution to the differential equation \(\dfrac{{\it dA}}{dt}=kA\) is \[\boxed{\bbox[#FAF8ED,5pt]{A=A_{0}e^{kt}}}\tag{5}\]

where \(A_{0}\) is the initial amount.

Functions \(A=A(t) \) whose rates of change are \(\dfrac{{\it dA}}{dt}=kA\) are said to follow the exponential law, or the law of uninhibited growth or decay—or in a business context, the law of continuously compounded interest. Figure #, on page 383, shows the graphs of the function \(A(t) =A_{0}e^{kt}\) for both \(k>0\) and \(k\lt0\).

383

NOTE

Example 4 is a model of uninhibited growth; it accuratelyreflects growth in early stages. After a time, growth no longer continues ata rate proportional to the number present. Factors, such as disease, lack ofspace, and dwindling food supply, begin to affect the rate of growth.

EXAMPLE 4 Solving a Differential Equation for Growth

Assume that a colony of bacteria grows at a rate proportional to the number of bacteria present. If the number of bacteria doubles in \(5\) hours (h), how long will it take for the number of bacteria to triple?

Solution Let \(N(t) \) be the number of bacteriapresent at time \(t\). Then the assumption that this colony of bacteria growsat a rate proportional to the number present can be modeled by \[\dfrac{dN}{dt}=kN\]

where \(k\) is a positive constant of proportionality. To find \(k,\) we writethe differential equation as \(\dfrac{dN}{N}=kdt\) and integrate both sides. This differential equation is of form (4), and its solution is given by (5).So, we have\[N(t) =N_{0}e^{kt}\]

where \(N_{0}\) is the initial number of bacteria in the colony. Since thenumber of bacteria doubles to \(2N_{0}\) in 5h, \[\begin{eqnarray*}N(5) =N_{0}\,e^{5k} &=& 2N_{0} \\e^{5k} &=&2 \\k &=&\dfrac{1}{5}\ln 2\end{eqnarray*}\]

The time \(t\) required for this colony to triple obeys the equation \[\begin{eqnarray*}N(t) &=&3N_{0} \\N_{0}e^{kt} &=&3N_{0} \\e^{kt} &=&3 \\t &=&\dfrac{1}{k}\ln 3 \underset{\underset{{\color{#0066A7}{\hbox{\( k=\tfrac{1}{5}\ln 2\)}}}}{\color{#0066A7}{\uparrow }}}{=}5\dfrac{\ln 3}{\ln 2}\approx 8 \end{eqnarray*}\]

The number of bacteria will triple in about 8h.

NOW WORK

Problem 51.

For a radioactive substance, the rate of decay is proportional to the amount of substance present at a given time \(t.\) That is, if \(A=A(t) \) represents the amount of a radioactive substance at time \(t\), then\[\boxed{\bbox[#FAF8ED,5pt]{\dfrac{{\it dA}}{dt}=kA }}\]

384

where \(k\lt0\) and depends on the radioactive substance. The half-lifeof a radioactive substance is the time required for half of the substance todecay.

ORIGINS

Willard F. Libby (1908-1980) grew up in California and went tocollege and graduate school at UC Berkeley. Libby was a physical chemist whotaught at the University of Chicago and later at UCLA. While at Chicago, hedeveloped the methods for using natural carbon-14 to date archaeologicalartifacts. Libby won the Nobel Prize in Chemistry in 1960 for thiswork.

Carbon dating, a method for determining the age of an artifact, uses thefact that all living organisms contain two kinds of carbon: carbon-12 (astable carbon) and a small proportion of carbon-14 (a radioactive isotope).When an organism dies, the amount of carbon-12 present remains unchanged,while the amount of carbon-14 begins to decrease. This change in the amountof carbon-14 present relative to the amount of carbon-12 present makes itpossible to calculate how long ago the organism died.

EXAMPLE 5 Solving a Differential Equation for Decay

The skull of an animal found in an archaeological dig contains about 20% of the original amount of carbon-14. If the half-life of carbon-14 is 5730 years, how long ago did the animal die?

Solution Let \(A=A(t) \) be the amount of carbon-14 present in the skull at time \(t\). Then \(A\) satisfies the differential equation \(\dfrac{{\it dA}}{dt}=kA\), whose solution is\[A=A_{0}e^{kt}\]

where \(A_{0}\) is the amount of carbon-14 present at time \(t=0\). To determinethe constant \(k\), we use the fact that when \(t=5730,\) half of the originalamount \(A_{0}\) remains. \[\begin{eqnarray*}\dfrac{1}{2}A_{0} &=&A_{0}e^{5730k} \\\dfrac{1}{2} &=&e^{5730k} \\5730k &=&\ln \dfrac{1}{2} = -\ln 2 \\ k &=&-\dfrac{\ln 2}{5730} \end{eqnarray*}\]

The relationship between the amount \(A\) of carbon-14 present andthe time \(t\) is \[A(t) =A_{0}e^{\left( -\ln 2/5730\right) t}\]

In this skull, 20% of the original amount of carbon-14 remains, so\(A(t) =0.20A_{0}.\)\[\begin{eqnarray*}0.20A_{0} &=&A_{0}e^{( -\ln 2/5730) t} \\0.20 &=&e^{( -\ln 2/5730) t}\end{eqnarray*}\]

Now, we take the natural logarithm of both sides. \[\begin{eqnarray*}\ln 0.20 &=&-\dfrac{\ln 2}{5730}\cdot t \\t &=&-5730\cdot \dfrac{\ln 0.20}{\ln 2}\approx 13{,}300\end{eqnarray*}\]

The animal died approximately 13,300 years ago.

NOW WORK

Problem 59.