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Calculus Volume 2

2.7 Integrals, Exponential Functions, and Logarithms

Calculus Volume 22.7 Integrals, Exponential Functions, and Logarithms
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 2.7.1. Write the definition of the natural logarithm as an integral.
  • 2.7.2. Recognize the derivative of the natural logarithm.
  • 2.7.3. Integrate functions involving the natural logarithmic function.
  • 2.7.4. Define the number ee through an integral.
  • 2.7.5. Recognize the derivative and integral of the exponential function.
  • 2.7.6. Prove properties of logarithms and exponential functions using integrals.
  • 2.7.7. Express general logarithmic and exponential functions in terms of natural logarithms and exponentials.

We already examined exponential functions and logarithms in earlier chapters. However, we glossed over some key details in the previous discussions. For example, we did not study how to treat exponential functions with exponents that are irrational. The definition of the number e is another area where the previous development was somewhat incomplete. We now have the tools to deal with these concepts in a more mathematically rigorous way, and we do so in this section.

For purposes of this section, assume we have not yet defined the natural logarithm, the number e, or any of the integration and differentiation formulas associated with these functions. By the end of the section, we will have studied these concepts in a mathematically rigorous way (and we will see they are consistent with the concepts we learned earlier).

We begin the section by defining the natural logarithm in terms of an integral. This definition forms the foundation for the section. From this definition, we derive differentiation formulas, define the number e,e, and expand these concepts to logarithms and exponential functions of any base.

The Natural Logarithm as an Integral

Recall the power rule for integrals:

xndx=xn+1n+1+C,n1.xndx=xn+1n+1+C,n1.

Clearly, this does not work when n=−1,n=−1, as it would force us to divide by zero. So, what do we do with 1xdx?1xdx? Recall from the Fundamental Theorem of Calculus that 1x1tdt1x1tdt is an antiderivative of 1/x.1/x. Therefore, we can make the following definition.

Definition

For x>0,x>0, define the natural logarithm function by

lnx=1x1tdt.lnx=1x1tdt.
2.24

For x>1,x>1, this is just the area under the curve y=1/ty=1/t from 11 to x.x. For x<1,x<1, we have 1x1tdt=x11tdt,1x1tdt=x11tdt, so in this case it is the negative of the area under the curve from xto1xto1 (see the following figure).

This figure has two graphs. The first is the curve y=1/t. It is decreasing and in the first quadrant. Under the curve is a shaded area. The area is bounded to the left at x=1. The area is labeled “area=lnx”. The second graph is the same curve y=1/t. It has shaded area under the curve bounded to the right by x=1. It is labeled “area=-lnx”.
Figure 2.75 (a) When x>1,x>1, the natural logarithm is the area under the curve y=1/ty=1/t from 1tox.1tox. (b) When x<1,x<1, the natural logarithm is the negative of the area under the curve from xx to 1.1.

Notice that ln1=0.ln1=0. Furthermore, the function y=1/t>0y=1/t>0 for x>0.x>0. Therefore, by the properties of integrals, it is clear that lnxlnx is increasing for x>0.x>0.

Properties of the Natural Logarithm

Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus.

Theorem 2.15

Derivative of the Natural Logarithm

For x>0,x>0, the derivative of the natural logarithm is given by

ddxlnx=1x.ddxlnx=1x.
Theorem 2.16

Corollary to the Derivative of the Natural Logarithm

The function lnxlnx is differentiable; therefore, it is continuous.

A graph of lnxlnx is shown in Figure 2.76. Notice that it is continuous throughout its domain of (0,).(0,).

This figure is a graph. It is an increasing curve labeled f(x)=lnx. The curve is increasing with the y-axis as an asymptote. The curve intersects the x-axis at x=1.
Figure 2.76 The graph of f(x)=lnxf(x)=lnx shows that it is a continuous function.

Example 2.35

Calculating Derivatives of Natural Logarithms

Calculate the following derivatives:

  1. ddxln(5x32)ddxln(5x32)
  2. ddx(ln(3x))2ddx(ln(3x))2

Solution

We need to apply the chain rule in both cases.

  1. ddxln(5x32)=15x25x32ddxln(5x32)=15x25x32
  2. ddx(ln(3x))2=2(ln(3x))·33x=2(ln(3x))xddx(ln(3x))2=2(ln(3x))·33x=2(ln(3x))x
Checkpoint 2.35

Calculate the following derivatives:

  1. ddxln(2x2+x)ddxln(2x2+x)
  2. ddx(ln(x3))2ddx(ln(x3))2

Note that if we use the absolute value function and create a new function ln|x|,ln|x|, we can extend the domain of the natural logarithm to include x<0.x<0. Then (d/(dx))ln|x|=1/x.(d/(dx))ln|x|=1/x. This gives rise to the familiar integration formula.

Theorem 2.17

Integral of (1/u) du

The natural logarithm is the antiderivative of the function f(u)=1/u:f(u)=1/u:

1udu=ln|u|+C.1udu=ln|u|+C.

Example 2.36

Calculating Integrals Involving Natural Logarithms

Calculate the integral xx2+4dx.xx2+4dx.

Solution

Using uu-substitution, let u=x2+4.u=x2+4. Then du=2xdxdu=2xdx and we have

xx2+4dx=121udu12ln|u|+C=12ln|x2+4|+C=12ln(x2+4)+C.xx2+4dx=121udu12ln|u|+C=12ln|x2+4|+C=12ln(x2+4)+C.
Checkpoint 2.36

Calculate the integral x2x3+6dx.x2x3+6dx.

Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.

Theorem 2.18

Properties of the Natural Logarithm

If a,b>0a,b>0 and rr is a rational number, then

  1. ln1=0ln1=0
  2. ln(ab)=lna+lnbln(ab)=lna+lnb
  3. ln(ab)=lnalnbln(ab)=lnalnb
  4. ln(ar)=rlnaln(ar)=rlna

Proof

  1. By definition, ln1=111tdt=0.ln1=111tdt=0.
  2. We have
    ln(ab)=1ab1tdt=1a1tdt+aab1tdt.ln(ab)=1ab1tdt=1a1tdt+aab1tdt.

    Use u-substitutionu-substitution on the last integral in this expression. Let u=t/a.u=t/a. Then du=(1/a)dt.du=(1/a)dt. Furthermore, when t=a,u=1,t=a,u=1, and when t=ab,u=b.t=ab,u=b. So we get
    ln(ab)=1a1tdt+aab1tdt=1a1tdt+1abat·1adt=1a1tdt+1b1udu=lna+lnb.ln(ab)=1a1tdt+aab1tdt=1a1tdt+1abat·1adt=1a1tdt+1b1udu=lna+lnb.
  3. Note that
    ddxln(xr)=rxr1xr=rx.ddxln(xr)=rxr1xr=rx.

    Furthermore,
    ddx(rlnx)=rx.ddx(rlnx)=rx.

    Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have
    ln(xr)=rlnx+Cln(xr)=rlnx+C

    for some constant C.C. Taking x=1,x=1, we get
    ln(1r)=rln(1)+C0=r(0)+CC=0.ln(1r)=rln(1)+C0=r(0)+CC=0.

    Thus ln(xr)=rlnxln(xr)=rlnx and the proof is complete. Note that we can extend this property to irrational values of rr later in this section.
    Part iii. follows from parts ii. and iv. and the proof is left to you.

Example 2.37

Using Properties of Logarithms

Use properties of logarithms to simplify the following expression into a single logarithm:

ln92ln3+ln(13).ln92ln3+ln(13).

Solution

We have

ln92ln3+ln(13)=ln(32)2ln3+ln(3−1)=2ln32ln3ln3=ln3.ln92ln3+ln(13)=ln(32)2ln3+ln(3−1)=2ln32ln3ln3=ln3.

Checkpoint 2.37

Use properties of logarithms to simplify the following expression into a single logarithm:

ln8ln2ln(14).ln8ln2ln(14).

Defining the Number e

Now that we have the natural logarithm defined, we can use that function to define the number e.e.

Definition

The number ee is defined to be the real number such that

lne=1.lne=1.

To put it another way, the area under the curve y=1/ty=1/t between t=1t=1 and t=et=e is 11 (Figure 2.77). The proof that such a number exists and is unique is left to you. (Hint: Use the Intermediate Value Theorem to prove existence and the fact that lnxlnx is increasing to prove uniqueness.)

This figure is a graph. It is the curve y=1/t. It is decreasing and in the first quadrant. Under the curve is a shaded area. The area is bounded to the left at x=1 and to the right at x=e. The area is labeled “area=1”.
Figure 2.77 The area under the curve from 11 to ee is equal to one.

The number ee can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and Maclaurin Series). Its approximate value is given by

e2.71828182846.e2.71828182846.

The Exponential Function

We now turn our attention to the function ex.ex. Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by expx.expx. Then,

exp(lnx)=xforx>0andln(expx)=xfor allx.exp(lnx)=xforx>0andln(expx)=xfor allx.

The following figure shows the graphs of expxexpx and lnx.lnx.

This figure is a graph. It has three curves. The first curve is labeled exp x. It is an increasing curve with the x-axis as a horizontal asymptote. It intersects the y-axis at y=1. The second curve is a diagonal line through the origin. The third curve is labeled lnx. It is an increasing curve with the y-axis as an vertical axis. It intersects the x-axis at x=1.
Figure 2.78 The graphs of lnxlnx and expx.expx.

We hypothesize that expx=ex.expx=ex. For rational values of x,x, this is easy to show. If xx is rational, then we have ln(ex)=xlne=x.ln(ex)=xlne=x. Thus, when xx is rational, ex=expx.ex=expx. For irrational values of x,x, we simply define exex as the inverse function of lnx.lnx.

Definition

For any real number x,x, define y=exy=ex to be the number for which

lny=ln(ex)=x.lny=ln(ex)=x.
2.25

Then we have ex=exp(x)ex=exp(x) for all x,x, and thus

elnx=xforx>0andln(ex)=xelnx=xforx>0andln(ex)=x
2.26

for all x.x.

Properties of the Exponential Function

Since the exponential function was defined in terms of an inverse function, and not in terms of a power of e,e, we must verify that the usual laws of exponents hold for the function ex.ex.

Theorem 2.19

Properties of the Exponential Function

If pp and qq are any real numbers and rr is a rational number, then

  1. epeq=ep+qepeq=ep+q
  2. epeq=epqepeq=epq
  3. (ep)r=epr(ep)r=epr

Proof

Note that if pp and qq are rational, the properties hold. However, if pp or qq are irrational, we must apply the inverse function definition of exex and verify the properties. Only the first property is verified here; the other two are left to you. We have

ln(epeq)=ln(ep)+ln(eq)=p+q=ln(ep+q).ln(epeq)=ln(ep)+ln(eq)=p+q=ln(ep+q).

Since lnxlnx is one-to-one, then

epeq=ep+q.epeq=ep+q.

As with part iv. of the logarithm properties, we can extend property iii. to irrational values of r,r, and we do so by the end of the section.

We also want to verify the differentiation formula for the function y=ex.y=ex. To do this, we need to use implicit differentiation. Let y=ex.y=ex. Then

lny=xddxlny=ddxx1ydydx=1dydx=y.lny=xddxlny=ddxx1ydydx=1dydx=y.

Thus, we see

ddxex=exddxex=ex

as desired, which leads immediately to the integration formula

exdx=ex+C.exdx=ex+C.

We apply these formulas in the following examples.

Example 2.38

Using Properties of Exponential Functions

Evaluate the following derivatives:

  1. ddte3tet2ddte3tet2
  2. ddxe3x2ddxe3x2

Solution

We apply the chain rule as necessary.

  1. ddte3tet2=ddte3t+t2=e3t+t2(3+2t)ddte3tet2=ddte3t+t2=e3t+t2(3+2t)
  2. ddxe3x2=e3x26xddxe3x2=e3x26x

Checkpoint 2.38

Evaluate the following derivatives:

  1. ddx(ex2e5x)ddx(ex2e5x)
  2. ddt(e2t)3ddt(e2t)3

Example 2.39

Using Properties of Exponential Functions

Evaluate the following integral: 2xex2dx.2xex2dx.

Solution

Using uu-substitution, let u=x2.u=x2. Then du=−2xdx,du=−2xdx, and we have

2xex2dx=eudu=eu+C=ex2+C.2xex2dx=eudu=eu+C=ex2+C.

Checkpoint 2.39

Evaluate the following integral: 4e3xdx.4e3xdx.

General Logarithmic and Exponential Functions

We close this section by looking at exponential functions and logarithms with bases other than e.e. Exponential functions are functions of the form f(x)=ax.f(x)=ax. Note that unless a=e,a=e, we still do not have a mathematically rigorous definition of these functions for irrational exponents. Let’s rectify that here by defining the function f(x)=axf(x)=ax in terms of the exponential function ex.ex. We then examine logarithms with bases other than ee as inverse functions of exponential functions.

Definition

For any a>0,a>0, and for any real number x,x, define y=axy=ax as follows:

y=ax=exlna.y=ax=exlna.

Now axax is defined rigorously for all values of x. This definition also allows us to generalize property iv. of logarithms and property iii. of exponential functions to apply to both rational and irrational values of r.r. It is straightforward to show that properties of exponents hold for general exponential functions defined in this way.

Let’s now apply this definition to calculate a differentiation formula for ax.ax. We have

ddxax=ddxexlna=exlnalna=axlna.ddxax=ddxexlna=exlnalna=axlna.

The corresponding integration formula follows immediately.

Theorem 2.20

Derivatives and Integrals Involving General Exponential Functions

Let a>0.a>0. Then,

ddxax=axlnaddxax=axlna

and

axdx=1lnaax+C.axdx=1lnaax+C.

If a1,a1, then the function axax is one-to-one and has a well-defined inverse. Its inverse is denoted by logax.logax. Then,

y=logaxif and only ifx=ay.y=logaxif and only ifx=ay.

Note that general logarithm functions can be written in terms of the natural logarithm. Let y=logax.y=logax. Then, x=ay.x=ay. Taking the natural logarithm of both sides of this second equation, we get

lnx=ln(ay)lnx=ylnay=lnxlnalogax=lnxlna.lnx=ln(ay)lnx=ylnay=lnxlnalogax=lnxlna.

Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base a.a. Again, let y=logax.y=logax. Then,

dydx=ddx(logax)=ddx(lnxlna)=(1lna)ddx(lnx)=1lna·1x=1xlna.dydx=ddx(logax)=ddx(lnxlna)=(1lna)ddx(lnx)=1lna·1x=1xlna.
Theorem 2.21

Derivatives of General Logarithm Functions

Let a>0.a>0. Then,

ddxlogax=1xlna.ddxlogax=1xlna.

Example 2.40

Calculating Derivatives of General Exponential and Logarithm Functions

Evaluate the following derivatives:

  1. ddt(4t·2t2)ddt(4t·2t2)
  2. ddxlog8(7x2+4)ddxlog8(7x2+4)

Solution

We need to apply the chain rule as necessary.

  1. ddt(4t·2t2)=ddt(22t·2t2)=ddt(22t+t2)=22t+t2ln(2)(2+2t)ddt(4t·2t2)=ddt(22t·2t2)=ddt(22t+t2)=22t+t2ln(2)(2+2t)
  2. ddxlog8(7x2+4)=1(7x2+4)(ln8)(14x)ddxlog8(7x2+4)=1(7x2+4)(ln8)(14x)
Checkpoint 2.40

Evaluate the following derivatives:

  1. ddt4t4ddt4t4
  2. ddxlog3(x2+1)ddxlog3(x2+1)

Example 2.41

Integrating General Exponential Functions

Evaluate the following integral: 323xdx.323xdx.

Solution

Use u-substitutionu-substitution and let u=−3x.u=−3x. Then du=−3dxdu=−3dx and we have

323xdx=3·2−3xdx=2udu=1ln22u+C=1ln22−3x+C.323xdx=3·2−3xdx=2udu=1ln22u+C=1ln22−3x+C.
Checkpoint 2.41

Evaluate the following integral: x22x3dx.x22x3dx.

Section 2.7 Exercises

For the following exercises, find the derivative dydx.dydx.

295.

y=ln(2x)y=ln(2x)

296.

y=ln(2x+1)y=ln(2x+1)

297.

y=1lnxy=1lnx

For the following exercises, find the indefinite integral.

298.

dt3tdt3t

299.

dx1+xdx1+x

For the following exercises, find the derivative dy/dx.dy/dx. (You can use a calculator to plot the function and the derivative to confirm that it is correct.)

300.

[T] y=ln(x)xy=ln(x)x

301.

[T] y=xln(x)y=xln(x)

302.

[T] y=log10xy=log10x

303.

[T] y=ln(sinx)y=ln(sinx)

304.

[T] y=ln(lnx)y=ln(lnx)

305.

[T] y=7ln(4x)y=7ln(4x)

306.

[T] y=ln((4x)7)y=ln((4x)7)

307.

[T] y=ln(tanx)y=ln(tanx)

308.

[T] y=ln(tan(3x))y=ln(tan(3x))

309.

[T] y=ln(cos2x)y=ln(cos2x)

For the following exercises, find the definite or indefinite integral.

310.

01dx3+x01dx3+x

311.

01dt3+2t01dt3+2t

312.

02xdxx2+102xdxx2+1

313.

02x3dxx2+102x3dxx2+1

314.

2edxxlnx2edxxlnx

315.

2edxx(lnx)22edxx(lnx)2

316.

cosxdxsinxcosxdxsinx

317.

0π/4tanxdx0π/4tanxdx

318.

cot(3x)dxcot(3x)dx

319.

(lnx)2dxx(lnx)2dxx

For the following exercises, compute dy/dxdy/dx by differentiating lny.lny.

320.

y=x2+1y=x2+1

321.

y=x2+1x21y=x2+1x21

322.

y=esinxy=esinx

323.

y=x−1/xy=x−1/x

324.

y=e(ex)y=e(ex)

325.

y=xey=xe

326.

y=x(ex)y=x(ex)

327.

y=xx3x6y=xx3x6

328.

y=x−1/lnxy=x−1/lnx

329.

y=elnxy=elnx

For the following exercises, evaluate by any method.

330.

510dtt5x10xdtt510dtt5x10xdtt

331.

1eπdxx+−2−1dxx1eπdxx+−2−1dxx

332.

ddxx1dttddxx1dtt

333.

ddxxx2dttddxxx2dtt

334.

ddxln(secx+tanx)ddxln(secx+tanx)

For the following exercises, use the function lnx.lnx. If you are unable to find intersection points analytically, use a calculator.

335.

Find the area of the region enclosed by x=1x=1 and y=5y=5 above y=lnx.y=lnx.

336.

[T] Find the arc length of lnxlnx from x=1x=1 to x=2.x=2.

337.

Find the area between lnxlnx and the x-axis from x=1tox=2.x=1tox=2.

338.

Find the volume of the shape created when rotating this curve from x=1tox=2x=1tox=2 around the x-axis, as pictured here.

This figure is a surface. It has been generated by revolving the curve ln x about the x-axis. The surface is inside of a cube showing it is 3-dimensinal.
339.

[T] Find the surface area of the shape created when rotating the curve in the previous exercise from x=1x=1 to x=2x=2 around the x-axis.

If you are unable to find intersection points analytically in the following exercises, use a calculator.

340.

Find the area of the hyperbolic quarter-circle enclosed by x=2andy=2x=2andy=2 above y=1/x.y=1/x.

341.

[T] Find the arc length of y=1/xy=1/x from x=1tox=4.x=1tox=4.

342.

Find the area under y=1/xy=1/x and above the x-axis from x=1tox=4.x=1tox=4.

For the following exercises, verify the derivatives and antiderivatives.

343.

ddxln(x+x2+1)=11+x2ddxln(x+x2+1)=11+x2

344.

ddxln(xax+a)=2a(x2a2)ddxln(xax+a)=2a(x2a2)

345.

ddxln(1+1x2x)=1x1x2ddxln(1+1x2x)=1x1x2

346.

ddxln(x+x2a2)=1x2a2ddxln(x+x2a2)=1x2a2

347.

dxxln(x)ln(lnx)=ln(ln(lnx))+Cdxxln(x)ln(lnx)=ln(ln(lnx))+C

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