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Calculus Volume 1

5.6 Integrals Involving Exponential and Logarithmic Functions

Calculus Volume 15.6 Integrals Involving Exponential and Logarithmic Functions
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 5.6.1. Integrate functions involving exponential functions.
  • 5.6.2. Integrate functions involving logarithmic functions.

Exponential and logarithmic functions are used to model population growth, cell growth, and financial growth, as well as depreciation, radioactive decay, and resource consumption, to name only a few applications. In this section, we explore integration involving exponential and logarithmic functions.

Integrals of Exponential Functions

The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, y=ex,y=ex, is its own derivative and its own integral.

Rule: Integrals of Exponential Functions

Exponential functions can be integrated using the following formulas.

exdx=ex+Caxdx=axlna+Cexdx=ex+Caxdx=axlna+C
5.21

Example 5.37

Finding an Antiderivative of an Exponential Function

Find the antiderivative of the exponential function ex.

Solution

Use substitution, setting u=x,u=x, and then du=−1dx.du=−1dx. Multiply the du equation by −1, so you now have du=dx.du=dx. Then,

exdx=eudu=eu+C=ex+C.exdx=eudu=eu+C=ex+C.
Checkpoint 5.31

Find the antiderivative of the function using substitution: x2e−2x3.x2e−2x3.

A common mistake when dealing with exponential expressions is treating the exponent on e the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent on e. This can be especially confusing when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we should always double-check to make sure we’re using the right rules for the functions we’re integrating.

Example 5.38

Square Root of an Exponential Function

Find the antiderivative of the exponential function ex1+ex.ex1+ex.

Solution

First rewrite the problem using a rational exponent:

ex1+exdx=ex(1+ex)1/2dx.ex1+exdx=ex(1+ex)1/2dx.

Using substitution, choose u=1+ex.u=1+ex.u=1+ex.u=1+ex. Then, du=exdx.du=exdx. We have (Figure 5.37)

ex(1+ex)1/2dx=u1/2du.ex(1+ex)1/2dx=u1/2du.

Then

u1/2du=u3/23/2+C=23u3/2+C=23(1+ex)3/2+C.u1/2du=u3/23/2+C=23u3/2+C=23(1+ex)3/2+C.
A graph of the function f(x) = e^x * sqrt(1 + e^x), which is an increasing concave up curve, over [-3, 1]. It begins close to the x axis in quadrant two, crosses the y axis at (0, sqrt(2)), and continues to increase rapidly.
Figure 5.37 The graph shows an exponential function times the square root of an exponential function.
Checkpoint 5.32

Find the antiderivative of ex(3ex2)2.ex(3ex2)2.

Example 5.39

Using Substitution with an Exponential Function

Use substitution to evaluate the indefinite integral 3x2e2x3dx.3x2e2x3dx.

Solution

Here we choose to let u equal the expression in the exponent on e. Let u=2x3u=2x3 and du=6x2dx..du=6x2dx.. Again, du is off by a constant multiplier; the original function contains a factor of 3x2, not 6x2. Multiply both sides of the equation by 1212 so that the integrand in u equals the integrand in x. Thus,

3x2e2x3dx=12eudu.3x2e2x3dx=12eudu.

Integrate the expression in u and then substitute the original expression in x back into the u integral:

12eudu=12eu+C=12e2x3+C.12eudu=12eu+C=12e2x3+C.
Checkpoint 5.33

Evaluate the indefinite integral 2x3ex4dx.2x3ex4dx.

As mentioned at the beginning of this section, exponential functions are used in many real-life applications. The number e is often associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. Although the derivative represents a rate of change or a growth rate, the integral represents the total change or the total growth. Let’s look at an example in which integration of an exponential function solves a common business application.

A price–demand function tells us the relationship between the quantity of a product demanded and the price of the product. In general, price decreases as quantity demanded increases. The marginal price–demand function is the derivative of the price–demand function and it tells us how fast the price changes at a given level of production. These functions are used in business to determine the price–elasticity of demand, and to help companies determine whether changing production levels would be profitable.

Example 5.40

Finding a Price–Demand Equation

Find the price–demand equation for a particular brand of toothpaste at a supermarket chain when the demand is 50 tubes per week at $2.35 per tube, given that the marginal price—demand function, p(x),p(x), for x number of tubes per week, is given as

p'(x)=−0.015e−0.01x.p'(x)=−0.015e−0.01x.

If the supermarket chain sells 100 tubes per week, what price should it set?

Solution

To find the price–demand equation, integrate the marginal price–demand function. First find the antiderivative, then look at the particulars. Thus,

p(x)=−0.015e−0.01xdx=−0.015e−0.01xdx.p(x)=−0.015e−0.01xdx=−0.015e−0.01xdx.

Using substitution, let u=−0.01xu=−0.01x and du=−0.01dx.du=−0.01dx. Then, divide both sides of the du equation by −0.01. This gives

−0.015−0.01eudu=1.5eudu=1.5eu+C=1.5e−0.01x+C.−0.015−0.01eudu=1.5eudu=1.5eu+C=1.5e−0.01x+C.

The next step is to solve for C. We know that when the price is $2.35 per tube, the demand is 50 tubes per week. This means

p(50)=1.5e−0.01(50)+C=2.35.p(50)=1.5e−0.01(50)+C=2.35.

Now, just solve for C:

C=2.351.5e−0.5=2.350.91=1.44.C=2.351.5e−0.5=2.350.91=1.44.

Thus,

p(x)=1.5e−0.01x+1.44.p(x)=1.5e−0.01x+1.44.

If the supermarket sells 100 tubes of toothpaste per week, the price would be

p(100)=1.5e−0.01(100)+1.44=1.5e−1+1.441.99.p(100)=1.5e−0.01(100)+1.44=1.5e−1+1.441.99.

The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.

Example 5.41

Evaluating a Definite Integral Involving an Exponential Function

Evaluate the definite integral 12e1xdx.12e1xdx.

Solution

Again, substitution is the method to use. Let u=1x,u=1x, so du=−1dxdu=−1dx or du=dx.du=dx. Then e1xdx=eudu.e1xdx=eudu. Next, change the limits of integration. Using the equation u=1x,u=1x, we have

u=1(1)=0u=1(2)=−1.u=1(1)=0u=1(2)=−1.

The integral then becomes

12e1xdx=0−1eudu=−10eudu=eu|−10=e0(e−1)=e−1+1.12e1xdx=0−1eudu=−10eudu=eu|−10=e0(e−1)=e−1+1.

See Figure 5.38.

A graph of the function f(x) = e^(1-x) over [0, 3]. It crosses the y axis at (0, e) as a decreasing concave up curve and symptotically approaches 0 as x goes to infinity.
Figure 5.38 The indicated area can be calculated by evaluating a definite integral using substitution.
Checkpoint 5.34

Evaluate 02e2xdx.02e2xdx.

Example 5.42

Growth of Bacteria in a Culture

Suppose the rate of growth of bacteria in a Petri dish is given by q(t)=3t,q(t)=3t, where t is given in hours and q(t)q(t) is given in thousands of bacteria per hour. If a culture starts with 10,000 bacteria, find a function Q(t)Q(t) that gives the number of bacteria in the Petri dish at any time t. How many bacteria are in the dish after 2 hours?

Solution

We have

Q(t)=3tdt=3tln3+C.Q(t)=3tdt=3tln3+C.

Then, at t=0t=0 we have Q(0)=10=1ln3+C,Q(0)=10=1ln3+C, so C9.090C9.090 and we get

Q(t)=3tln3+9.090.Q(t)=3tln3+9.090.

At time t=2,t=2, we have

Q(2)=32ln3+9.090Q(2)=32ln3+9.090
=17.282.=17.282.

After 2 hours, there are 17,282 bacteria in the dish.

Checkpoint 5.35

From Example 5.42, suppose the bacteria grow at a rate of q(t)=2t.q(t)=2t. Assume the culture still starts with 10,000 bacteria. Find Q(t).Q(t). How many bacteria are in the dish after 3 hours?

Example 5.43

Fruit Fly Population Growth

Suppose a population of fruit flies increases at a rate of g(t)=2e0.02t,g(t)=2e0.02t, in flies per day. If the initial population of fruit flies is 100 flies, how many flies are in the population after 10 days?

Solution

Let G(t)G(t) represent the number of flies in the population at time t. Applying the net change theorem, we have

G(10)=G(0)+0102e0.02tdt=100+[20.02e0.02t]|010=100+[100e0.02t]|010=100+100e0.2100122.G(10)=G(0)+0102e0.02tdt=100+[20.02e0.02t]|010=100+[100e0.02t]|010=100+100e0.2100122.

There are 122 flies in the population after 10 days.

Checkpoint 5.36

Suppose the rate of growth of the fly population is given by g(t)=e0.01t,g(t)=e0.01t, and the initial fly population is 100 flies. How many flies are in the population after 15 days?

Example 5.44

Evaluating a Definite Integral Using Substitution

Evaluate the definite integral using substitution: 12e1/xx2dx.12e1/xx2dx.

Solution

This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as a power of x, then bring the x2 in the denominator up to the numerator using a negative exponent. We have

12e1/xx2dx=12ex−1x−2dx.12e1/xx2dx=12ex−1x−2dx.

Let u=x−1,u=x−1, the exponent on e. Then

du=x−2dxdu=x−2dx.du=x−2dxdu=x−2dx.

Bringing the negative sign outside the integral sign, the problem now reads

eudu.eudu.

Next, change the limits of integration:

u=(1)−1=1u=(2)−1=12.u=(1)−1=1u=(2)−1=12.

Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange the limits. Thus,

11/2eudu=1/21eudu=eu|1/21=ee1/2=ee.11/2eudu=1/21eudu=eu|1/21=ee1/2=ee.
Checkpoint 5.37

Evaluate the definite integral using substitution: 121x3e4x−2dx.121x3e4x−2dx.

Integrals Involving Logarithmic Functions

Integrating functions of the form f(x)=x−1f(x)=x−1 result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as f(x)=lnxf(x)=lnx and f(x)=logax,f(x)=logax, are also included in the rule.

Rule: Integration Formulas Involving Logarithmic Functions

The following formulas can be used to evaluate integrals involving logarithmic functions.

x−1dx=ln|x|+Clnxdx=xlnxx+C=x(lnx1)+Clogaxdx=xlna(lnx1)+Cx−1dx=ln|x|+Clnxdx=xlnxx+C=x(lnx1)+Clogaxdx=xlna(lnx1)+C
5.22

Example 5.45

Finding an Antiderivative Involving lnxlnx

Find the antiderivative of the function 3x10.3x10.

Solution

First factor the 3 outside the integral symbol. Then use the u−1 rule. Thus,

3x10dx=31x10dx=3duu=3ln|u|+C=3ln|x10|+C,x10.3x10dx=31x10dx=3duu=3ln|u|+C=3ln|x10|+C,x10.

See Figure 5.39.

A graph of the function f(x) = 3 / (x – 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity.
Figure 5.39 The domain of this function is x10.x10.
Checkpoint 5.38

Find the antiderivative of 1x+2.1x+2.

Example 5.46

Finding an Antiderivative of a Rational Function

Find the antiderivative of 2x3+3xx4+3x2.2x3+3xx4+3x2.

Solution

This can be rewritten as (2x3+3x)(x4+3x2)−1dx.(2x3+3x)(x4+3x2)−1dx. Use substitution. Let u=x4+3x2,u=x4+3x2, then du=4x3+6x.du=4x3+6x. Alter du by factoring out the 2. Thus,

du=(4x3+6x)dx=2(2x3+3x)dx12du=(2x3+3x)dx.du=(4x3+6x)dx=2(2x3+3x)dx12du=(2x3+3x)dx.

Rewrite the integrand in u:

(2x3+3x)(x4+3x2)−1dx=12u−1du.(2x3+3x)(x4+3x2)−1dx=12u−1du.

Then we have

12u−1du=12ln|u|+C=12ln|x4+3x2|+C.12u−1du=12ln|u|+C=12ln|x4+3x2|+C.

Example 5.47

Finding an Antiderivative of a Logarithmic Function

Find the antiderivative of the log function log2x.log2x.

Solution

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

log2xdx=xln2(lnx1)+C.log2xdx=xln2(lnx1)+C.
Checkpoint 5.39

Find the antiderivative of log3x.log3x.

Example 5.48 is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.

Example 5.48

Evaluating a Definite Integral

Find the definite integral of 0π/2sinx1+cosxdx.0π/2sinx1+cosxdx.

Solution

We need substitution to evaluate this problem. Let u=1+cosx,,u=1+cosx,, so du=sinxdx.du=sinxdx. Rewrite the integral in terms of u, changing the limits of integration as well. Thus,

u=1+cos(0)=2u=1+cos(π2)=1.u=1+cos(0)=2u=1+cos(π2)=1.

Then

0π/2sinx1+cosx=21u−1du=12u−1du=ln|u||12=[ln2ln1]=ln2.0π/2sinx1+cosx=21u−1du=12u−1du=ln|u||12=[ln2ln1]=ln2.

Section 5.6 Exercises

In the following exercises, compute each indefinite integral.

320.

e2xdxe2xdx

321.

e−3xdxe−3xdx

322.

2xdx2xdx

323.

3xdx3xdx

324.

12xdx12xdx

325.

2xdx2xdx

326.

1x2dx1x2dx

327.

1xdx1xdx

In the following exercises, find each indefinite integral by using appropriate substitutions.

328.

lnxxdxlnxxdx

329.

dxx(lnx)2dxx(lnx)2

330.

dxxlnx(x>1)dxxlnx(x>1)

331.

dxxlnxln(lnx)dxxlnxln(lnx)

332.

tanθdθtanθdθ

333.

cosxxsinxxcosxdxcosxxsinxxcosxdx

334.

ln(sinx)tanxdxln(sinx)tanxdx

335.

ln(cosx)tanxdxln(cosx)tanxdx

336.

xex2dxxex2dx

337.

x2ex3dxx2ex3dx

338.

esinxcosxdxesinxcosxdx

339.

etanxsec2xdxetanxsec2xdx

340.

elnxdxxelnxdxx

341.

eln(1t)1tdteln(1t)1tdt

In the following exercises, verify by differentiation that lnxdx=x(lnx1)+C,lnxdx=x(lnx1)+C, then use appropriate changes of variables to compute the integral.

342.

lnxdxlnxdx (Hint:lnxdx=12xln(x2)dx)(Hint:lnxdx=12xln(x2)dx)

343.

x2ln2xdxx2ln2xdx

344.

lnxx2dxlnxx2dx (Hint:Setu=1x.)(Hint:Setu=1x.)

345.

lnxxdxlnxxdx (Hint:Setu=x.)(Hint:Setu=x.)

346.

Write an integral to express the area under the graph of y=1ty=1t from t=1t=1 to ex and evaluate the integral.

347.

Write an integral to express the area under the graph of y=ety=et between t=0t=0 and t=lnx,t=lnx, and evaluate the integral.

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

348.

tan(2x)dxtan(2x)dx

349.

sin(3x)cos(3x)sin(3x)+cos(3x)dxsin(3x)cos(3x)sin(3x)+cos(3x)dx

350.

xsin(x2)cos(x2)dxxsin(x2)cos(x2)dx

351.

xcsc(x2)dxxcsc(x2)dx

352.

ln(cosx)tanxdxln(cosx)tanxdx

353.

ln(cscx)cotxdxln(cscx)cotxdx

354.

exexex+exdxexexex+exdx

In the following exercises, evaluate the definite integral.

355.

121+2x+x23x+3x2+x3dx121+2x+x23x+3x2+x3dx

356.

0π/4tanxdx0π/4tanxdx

357.

0π/3sinxcosxsinx+cosxdx0π/3sinxcosxsinx+cosxdx

358.

π/6π/2cscxdxπ/6π/2cscxdx

359.

π/4π/3cotxdxπ/4π/3cotxdx

In the following exercises, integrate using the indicated substitution.

360.

xx100dx;u=x100xx100dx;u=x100

361.

y1y+1dy;u=y+1y1y+1dy;u=y+1

362.

1x23xx3dx;u=3xx31x23xx3dx;u=3xx3

363.

sinx+cosxsinxcosxdx;u=sinxcosxsinx+cosxsinxcosxdx;u=sinxcosx

364.

e2x1e2xdx;u=e2xe2x1e2xdx;u=e2x

365.

ln(x)1(lnx)2xdx;u=lnxln(x)1(lnx)2xdx;u=lnx

In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R50 and solve for the exact area.

366.

[T] y=exy=ex over [0,1][0,1]

367.

[T] y=exy=ex over [0,1][0,1]

368.

[T] y=ln(x)y=ln(x) over [1,2][1,2]

369.

[T] y=x+1x2+2x+6y=x+1x2+2x+6 over [0,1][0,1]

370.

[T] y=2xy=2x over [−1,0][−1,0]

371.

[T] y=2xy=2x over [0,1][0,1]

In the following exercises, f(x)0f(x)0 for axb.axb. Find the area under the graph of f(x)f(x) between the given values a and b by integrating.

372.

f(x)=log10(x)x;a=10,b=100f(x)=log10(x)x;a=10,b=100

373.

f(x)=log2(x)x;a=32,b=64f(x)=log2(x)x;a=32,b=64

374.

f(x)=2x;a=1,b=2f(x)=2x;a=1,b=2

375.

f(x)=2x;a=3,b=4f(x)=2x;a=3,b=4

376.

Find the area under the graph of the function f(x)=xex2f(x)=xex2 between x=0x=0 and x=5.x=5.

377.

Compute the integral of f(x)=xex2f(x)=xex2 and find the smallest value of N such that the area under the graph f(x)=xex2f(x)=xex2 between x=Nx=N and x=N+10x=N+10 is, at most, 0.01.

378.

Find the limit, as N tends to infinity, of the area under the graph of f(x)=xex2f(x)=xex2 between x=0x=0 and x=5.x=5.

379.

Show that abdtt=1/b1/adttabdtt=1/b1/adtt when 0<ab.0<ab.

380.

Suppose that f(x)>0f(x)>0 for all x and that f and g are differentiable. Use the identity fg=eglnffg=eglnf and the chain rule to find the derivative of fg.fg.

381.

Use the previous exercise to find the antiderivative of h(x)=xx(1+lnx)h(x)=xx(1+lnx) and evaluate 23xx(1+lnx)dx.23xx(1+lnx)dx.

382.

Show that if c>0,c>0, then the integral of 1/x1/x from ac to bc (0<a<b)(0<a<b) is the same as the integral of 1/x1/x from a to b.

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition ln(x)=1xdtt,ln(x)=1xdtt, using properties of the definite integral and making no further assumptions.

383.

Use the identity ln(x)=1xdttln(x)=1xdtt to derive the identity ln(1x)=lnx.ln(1x)=lnx.

384.

Use a change of variable in the integral 1xy1tdt1xy1tdt to show that lnxy=lnx+lnyforx,y>0.lnxy=lnx+lnyforx,y>0.

385.

Use the identity lnx=1xdtxlnx=1xdtx to show that ln(x)ln(x) is an increasing function of x on [0,),[0,), and use the previous exercises to show that the range of ln(x)ln(x) is (,).(,). Without any further assumptions, conclude that ln(x)ln(x) has an inverse function defined on (,).(,).

386.

Pretend, for the moment, that we do not know that exex is the inverse function of ln(x),ln(x), but keep in mind that ln(x)ln(x) has an inverse function defined on (,).(,). Call it E. Use the identity lnxy=lnx+lnylnxy=lnx+lny to deduce that E(a+b)=E(a)E(b)E(a+b)=E(a)E(b) for any real numbers a, b.

387.

Pretend, for the moment, that we do not know that exex is the inverse function of lnx,lnx, but keep in mind that lnxlnx has an inverse function defined on (,).(,). Call it E. Show that E'(t)=E(t).E'(t)=E(t).

388.

The sine integral, defined as S(x)=0xsinttdtS(x)=0xsinttdt is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large x. Show that for k1,|S(2πk)S(2π(k+1))|1k(2k+1)π.k1,|S(2πk)S(2π(k+1))|1k(2k+1)π. (Hint:sin(t+π)=sint)(Hint:sin(t+π)=sint)

389.

[T] The normal distribution in probability is given by p(x)=1σ2πe(xμ)2/2σ2,p(x)=1σ2πe(xμ)2/2σ2, where σ is the standard deviation and μ is the average. The standard normal distribution in probability, ps,ps, corresponds to μ=0andσ=1.μ=0andσ=1. Compute the left endpoint estimates R10andR100R10andR100 of −1112πex2/2dx.−1112πex2/2dx.

390.

[T] Compute the right endpoint estimates R50andR100R50andR100 of −35122πe(x1)2/8.−35122πe(x1)2/8.

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