Skip to Content
OpenStax Logo
Calculus Volume 1

5.5 Substitution

Calculus Volume 15.5 Substitution
Buy book
  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

  • 5.5.1. Use substitution to evaluate indefinite integrals.
  • 5.5.2. Use substitution to evaluate definite integrals.

The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.

At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to see? We are looking for an integrand of the form f[g(x)]g(x)dx.f[g(x)]g(x)dx. For example, in the integral (x23)32xdx,(x23)32xdx, we have f(x)=x3,g(x)=x23,f(x)=x3,g(x)=x23, and g'(x)=2x.g'(x)=2x. Then,

f[g(x)]g(x)=(x23)3(2x),f[g(x)]g(x)=(x23)3(2x),

and we see that our integrand is in the correct form.

The method is called substitution because we substitute part of the integrand with the variable u and part of the integrand with du. It is also referred to as change of variables because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.

Theorem 5.7

Substitution with Indefinite Integrals

Let u=g(x),,u=g(x),, where g(x)g(x) is continuous over an interval, let f(x)f(x) be continuous over the corresponding range of g, and let F(x)F(x) be an antiderivative of f(x).f(x). Then,

f[g(x)]g(x)dx=f(u)du=F(u)+C=F(g(x))+C.f[g(x)]g(x)dx=f(u)du=F(u)+C=F(g(x))+C.
5.19

Proof

Let f, g, u, and F be as specified in the theorem. Then

ddxF(g(x))=F(g(x))g(x)=f[g(x)]g(x).ddxF(g(x))=F(g(x))g(x)=f[g(x)]g(x).

Integrating both sides with respect to x, we see that

f[g(x)]g(x)dx=F(g(x))+C.f[g(x)]g(x)dx=F(g(x))+C.

If we now substitute u=g(x),u=g(x), and du=g'(x)dx,du=g'(x)dx, we get

f[g(x)]g(x)dx=f(u)du=F(u)+C=F(g(x))+C.f[g(x)]g(x)dx=f(u)du=F(u)+C=F(g(x))+C.

Returning to the problem we looked at originally, we let u=x23u=x23 and then du=2xdx.du=2xdx. Rewrite the integral in terms of u:

(x23)u3(2xdx)du=u3du.(x23)u3(2xdx)du=u3du.

Using the power rule for integrals, we have

u3du=u44+C.u3du=u44+C.

Substitute the original expression for x back into the solution:

u44+C=(x23)44+C.u44+C=(x23)44+C.

We can generalize the procedure in the following Problem-Solving Strategy.

Problem-Solving Strategy: Integration by Substitution
  1. Look carefully at the integrand and select an expression g(x)g(x) within the integrand to set equal to u. Let’s select g(x).g(x). such that g(x)g(x) is also part of the integrand.
  2. Substitute u=g(x)u=g(x) and du=g(x)dx.du=g(x)dx. into the integral.
  3. We should now be able to evaluate the integral with respect to u. If the integral can’t be evaluated we need to go back and select a different expression to use as u.
  4. Evaluate the integral in terms of u.
  5. Write the result in terms of x and the expression g(x).g(x).

Example 5.30

Using Substitution to Find an Antiderivative

Use substitution to find the antiderivative of 6x(3x2+4)4dx.6x(3x2+4)4dx.

Solution

The first step is to choose an expression for u. We choose u=3x2+4.u=3x2+4. because then du=6xdx.,du=6xdx., and we already have du in the integrand. Write the integral in terms of u:

6x(3x2+4)4dx=u4du.6x(3x2+4)4dx=u4du.

Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Now we can evaluate the integral with respect to u:

u4du=u55+C=(3x2+4)55+C.u4du=u55+C=(3x2+4)55+C.

Analysis

We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for C of 1, we let y=15(3x2+4)5+1.y=15(3x2+4)5+1. We have

y=15(3x2+4)5+1,y=15(3x2+4)5+1,

so

y=(15)5(3x2+4)46x=6x(3x2+4)4.y=(15)5(3x2+4)46x=6x(3x2+4)4.

This is exactly the expression we started with inside the integrand.

Checkpoint 5.25

Use substitution to find the antiderivative of 3x2(x33)2dx.3x2(x33)2dx.

Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.

Example 5.31

Using Substitution with Alteration

Use substitution to find the antiderivative of zz25dz.zz25dz.

Solution

Rewrite the integral as z(z25)1/2dz.z(z25)1/2dz. Let u=z25u=z25 and du=2zdz.du=2zdz. Now we have a problem because du=2zdzdu=2zdz and the original expression has only zdz.zdz. We have to alter our expression for du or the integral in u will be twice as large as it should be. If we multiply both sides of the du equation by 12.12. we can solve this problem. Thus,

u=z25du=2zdz12du=12(2z)dz=zdz.u=z25du=2zdz12du=12(2z)dz=zdz.

Write the integral in terms of u, but pull the 1212 outside the integration symbol:

z(z25)1/2dz=12u1/2du.z(z25)1/2dz=12u1/2du.

Integrate the expression in u:

12u1/2du=(12)u3/232+C=(12)(23)u3/2+C=13u3/2+C=13(z25)3/2+C.12u1/2du=(12)u3/232+C=(12)(23)u3/2+C=13u3/2+C=13(z25)3/2+C.
Checkpoint 5.26

Use substitution to find the antiderivative of x2(x3+5)9dx.x2(x3+5)9dx.

Example 5.32

Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral sintcos3tdt.sintcos3tdt.

Solution

We know the derivative of costcost is sint,sint, so we set u=cost.u=cost. Then du=sintdt.du=sintdt. Substituting into the integral, we have

sintcos3tdt=duu3.sintcos3tdt=duu3.

Evaluating the integral, we get

duu3=u−3du=(12)u−2+C.duu3=u−3du=(12)u−2+C.

Putting the answer back in terms of t, we get

sintcos3tdt=12u2+C=12cos2t+C.sintcos3tdt=12u2+C=12cos2t+C.
Checkpoint 5.27

Use substitution to evaluate the integral costsin2tdt.costsin2tdt.

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, u should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of u. This technique should become clear in the next example.

Example 5.33

Finding an Antiderivative Using u-Substitution

Use substitution to find the antiderivative of xx1dx.xx1dx.

Solution

If we let u=x1,u=x1, then du=dx.du=dx. But this does not account for the x in the numerator of the integrand. We need to express x in terms of u. If u=x1,u=x1, then x=u+1.x=u+1. Now we can rewrite the integral in terms of u:

xx1dx=u+1udu=u+1udu=(u1/2+u−1/2)du.xx1dx=u+1udu=u+1udu=(u1/2+u−1/2)du.

Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result. Thus,

(u1/2+u−1/2)du=23u3/2+2u1/2+C=23(x1)3/2+2(x1)1/2+C=(x1)1/2[23(x1)+2]+C=(x1)1/2(23x23+63)=(x1)1/2(23x+43)=23(x1)1/2(x+2)+C.(u1/2+u−1/2)du=23u3/2+2u1/2+C=23(x1)3/2+2(x1)1/2+C=(x1)1/2[23(x1)+2]+C=(x1)1/2(23x23+63)=(x1)1/2(23x+43)=23(x1)1/2(x+2)+C.
Checkpoint 5.28

Use substitution to evaluate the indefinite integral cos3tsintdt.cos3tsintdt.

Substitution for Definite Integrals

Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.

Theorem 5.8

Substitution with Definite Integrals

Let u=g(x)u=g(x) and let gg be continuous over an interval [a,b],[a,b], and let f be continuous over the range of u=g(x).u=g(x). Then,

abf(g(x))g(x)dx=g(a)g(b)f(u)du.abf(g(x))g(x)dx=g(a)g(b)f(u)du.

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if F(x)F(x) is an antiderivative of f(x),f(x), we have

f(g(x))g(x)dx=F(g(x))+C.f(g(x))g(x)dx=F(g(x))+C.

Then

abf[g(x)]g(x)dx=F(g(x))|x=ax=b=F(g(b))F(g(a))=F(u)|u=g(a)u=g(b)=g(a)g(b)f(u)du,abf[g(x)]g(x)dx=F(g(x))|x=ax=b=F(g(b))F(g(a))=F(u)|u=g(a)u=g(b)=g(a)g(b)f(u)du,
5.20

and we have the desired result.

Example 5.34

Using Substitution to Evaluate a Definite Integral

Use substitution to evaluate 01x2(1+2x3)5dx.01x2(1+2x3)5dx.

Solution

Let u=1+2x3,u=1+2x3, so du=6x2dx.du=6x2dx. Since the original function includes one factor of x2 and du=6x2dx,du=6x2dx, multiply both sides of the du equation by 1/6.1/6. Then,

du=6x2dx16du=x2dx.du=6x2dx16du=x2dx.

To adjust the limits of integration, note that when x=0,u=1+2(0)=1,x=0,u=1+2(0)=1, and when x=1,u=1+2(1)=3.x=1,u=1+2(1)=3. Then

01x2(1+2x3)5dx=1613u5du.01x2(1+2x3)5dx=1613u5du.

Evaluating this expression, we get

1613u5du=(16)(u66)|13=136[(3)6(1)6]=1829.1613u5du=(16)(u66)|13=136[(3)6(1)6]=1829.
Checkpoint 5.29

Use substitution to evaluate the definite integral −10y(2y23)5dy.−10y(2y23)5dy.

Example 5.35

Using Substitution with an Exponential Function

Use substitution to evaluate 01xe4x2+3dx.01xe4x2+3dx.

Solution

Let u=4x3+3.u=4x3+3. Then, du=8xdx.du=8xdx. To adjust the limits of integration, we note that when x=0,u=3,x=0,u=3, and when x=1,u=7.x=1,u=7. So our substitution gives

01xe4x2+3dx=1837eudu=18eu|37=e7e38134.568.01xe4x2+3dx=1837eudu=18eu|37=e7e38134.568.
Checkpoint 5.30

Use substitution to evaluate 01x2cos(π2x3)dx.01x2cos(π2x3)dx.

Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in Example 5.36.

Example 5.36

Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate 0π/2cos2θdθ.0π/2cos2θdθ.

Solution

Let us first use a trigonometric identity to rewrite the integral. The trig identity cos2θ=1+cos2θ2cos2θ=1+cos2θ2 allows us to rewrite the integral as

0π/2cos2θdθ=0π/21+cos2θ2dθ.0π/2cos2θdθ=0π/21+cos2θ2dθ.

Then,

0π/2(1+cos2θ2)dθ=0π/2(12+12cos2θ)dθ=120π/2dθ+120π/2cos2θdθ.0π/2(1+cos2θ2)dθ=0π/2(12+12cos2θ)dθ=120π/2dθ+120π/2cos2θdθ.

We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let u=2θ.u=2θ. Then, du=2dθ,du=2dθ, or 12du=dθ.12du=dθ. Also, when θ=0,u=0,θ=0,u=0, and when θ=π/2,u=π.θ=π/2,u=π. Expressing the second integral in terms of u, we have

120π/2dθ+120π/2cos2θdθ=120π/2dθ+12(12)0πcosudu=θ2|θ=0θ=π/2+14sinu|u=0u=θ=(π40)+(00)=π4.120π/2dθ+120π/2cos2θdθ=120π/2dθ+12(12)0πcosudu=θ2|θ=0θ=π/2+14sinu|u=0u=θ=(π40)+(00)=π4.

Section 5.5 Exercises

254.

Why is u-substitution referred to as change of variable?

255.

2. If f=gh,f=gh, when reversing the chain rule, ddx(gh)(x)=g(h(x))h(x),ddx(gh)(x)=g(h(x))h(x), should you take u=g(x)u=g(x) or u=h(x)?u=h(x)?

In the following exercises, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form f(u)du.f(u)du.

256.

xx+1dx=215(x+1)3/2(3x2)+C;u=x+1xx+1dx=215(x+1)3/2(3x2)+C;u=x+1

257.

For x>1:x2x1dx=215x1(3x2+4x+8)+C;u=x1x>1:x2x1dx=215x1(3x2+4x+8)+C;u=x1

258.

x4x2+9dx=112(4x2+9)3/2+C;u=4x2+9x4x2+9dx=112(4x2+9)3/2+C;u=4x2+9

259.

x4x2+9dx=144x2+9+C;u=4x2+9x4x2+9dx=144x2+9+C;u=4x2+9

260.

x(4x2+9)2dx=18(4x2+9);u=4x2+9x(4x2+9)2dx=18(4x2+9);u=4x2+9

In the following exercises, find the antiderivative using the indicated substitution.

261.

(x+1)4dx;u=x+1(x+1)4dx;u=x+1

262.

(x1)5dx;u=x1(x1)5dx;u=x1

263.

(2x3)−7dx;u=2x3(2x3)−7dx;u=2x3

264.

(3x2)−11dx;u=3x2(3x2)−11dx;u=3x2

265.

xx2+1dx;u=x2+1xx2+1dx;u=x2+1

266.

x1x2dx;u=1x2x1x2dx;u=1x2

267.

(x1)(x22x)3dx;u=x22x(x1)(x22x)3dx;u=x22x

268.

(x22x)(x33x2)2dx;u=x3=3x2(x22x)(x33x2)2dx;u=x3=3x2

269.

cos3θdθ;u=sinθcos3θdθ;u=sinθ (Hint:cos2θ=1sin2θ)(Hint:cos2θ=1sin2θ)

270.

sin3θdθ;u=cosθsin3θdθ;u=cosθ (Hint:sin2θ=1cos2θ)(Hint:sin2θ=1cos2θ)

In the following exercises, use a suitable change of variables to determine the indefinite integral.

271.

x(1x)99dxx(1x)99dx

272.

t(1t2)10dtt(1t2)10dt

273.

(11x7)−3dx(11x7)−3dx

274.

(7x11)4dx(7x11)4dx

275.

cos3θsinθdθcos3θsinθdθ

276.

sin7θcosθdθsin7θcosθdθ

277.

cos2(πt)sin(πt)dtcos2(πt)sin(πt)dt

278.

sin2xcos3xdxsin2xcos3xdx (Hint:sin2x+cos2x=1)(Hint:sin2x+cos2x=1)

279.

tsin(t2)cos(t2)dttsin(t2)cos(t2)dt

280.

t2cos2(t3)sin(t3)dtt2cos2(t3)sin(t3)dt

281.

x2(x33)2dxx2(x33)2dx

282.

x31x2dxx31x2dx

283.

y5(1y3)3/2dyy5(1y3)3/2dy

284.

cosθ(1cosθ)99sinθdθcosθ(1cosθ)99sinθdθ

285.

(1cos3θ)10cos2θsinθdθ(1cos3θ)10cos2θsinθdθ

286.

(cosθ1)(cos2θ2cosθ)3sinθdθ(cosθ1)(cos2θ2cosθ)3sinθdθ

287.

(sin2θ2sinθ)(sin3θ3sin2θ)3cosθdθ(sin2θ2sinθ)(sin3θ3sin2θ)3cosθdθ

In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.

288.

[T] y=3(1x)2y=3(1x)2 over [0,2][0,2]

289.

[T] y=x(1x2)3y=x(1x2)3 over [−1,2][−1,2]

290.

[T] y=sinx(1cosx)2y=sinx(1cosx)2 over [0,π][0,π]

291.

[T] y=x(x2+1)2y=x(x2+1)2 over [−1,1][−1,1]

In the following exercises, use a change of variables to evaluate the definite integral.

292.

01x1x2dx01x1x2dx

293.

01x1+x2dx01x1+x2dx

294.

02t25+t2dt02t25+t2dt

295.

01t21+t3dt01t21+t3dt

296.

0π/4sec2θtanθdθ0π/4sec2θtanθdθ

297.

0π/4sinθcos4θdθ0π/4sinθcos4θdθ

In the following exercises, evaluate the indefinite integral f(x)dxf(x)dx with constant C=0C=0 using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it equal to the definite integral F(x)=axf(t)dt,F(x)=axf(t)dt, with a the left endpoint of the given interval.

298.

[T] (2x+1)ex2+x6dx(2x+1)ex2+x6dx over [−3,2][−3,2]

299.

[T] cos(ln(2x))xdxcos(ln(2x))xdx on [0,2][0,2]

300.

[T] 3x2+2x+1x3+x2+x+4dx3x2+2x+1x3+x2+x+4dx over [−1,2][−1,2]

301.

[T] sinxcos3xdxsinxcos3xdx over [π3,π3][π3,π3]

302.

[T] (x+2)ex24x+3dx(x+2)ex24x+3dx over [−5,1][−5,1]

303.

[T] 3x22x3+1dx3x22x3+1dx over [0,1][0,1]

304.

If h(a)=h(b)h(a)=h(b) in abg'(h(x))h(x)dx,abg'(h(x))h(x)dx, what can you say about the value of the integral?

305.

Is the substitution u=1x2u=1x2 in the definite integral 02x1x2dx02x1x2dx okay? If not, why not?

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

306.

0πcos2(2θ)sin(2θ)dθ0πcos2(2θ)sin(2θ)dθ

307.

0πtcos(t2)sin(t2)dt0πtcos(t2)sin(t2)dt

308.

01(12t)dt01(12t)dt

309.

0112t(1+(t12)2)dt0112t(1+(t12)2)dt

310.

0πsin((tπ2)3)cos(tπ2)dt0πsin((tπ2)3)cos(tπ2)dt

311.

02(1t)cos(πt)dt02(1t)cos(πt)dt

312.

π/43π/4sin2tcostdtπ/43π/4sin2tcostdt

313.

Show that the average value of f(x)f(x) over an interval [a,b][a,b] is the same as the average value of f(cx)f(cx) over the interval [ac,bc][ac,bc] for c>0.c>0.

314.

Find the area under the graph of f(t)=t(1+t2)af(t)=t(1+t2)a between t=0t=0 and t=xt=x where a>0a>0 and a1a1 is fixed, and evaluate the limit as x.x.

315.

Find the area under the graph of g(t)=t(1t2)ag(t)=t(1t2)a between t=0t=0 and t=x,t=x, where 0<x<10<x<1 and a>0a>0 is fixed. Evaluate the limit as x1.x1.

316.

The area of a semicircle of radius 1 can be expressed as −111x2dx.−111x2dx. Use the substitution x=costx=cost to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.

317.

The area of the top half of an ellipse with a major axis that is the x-axis from x=−1x=−1 to a and with a minor axis that is the y-axis from y=by=b to b can be written as aab1x2a2dx.aab1x2a2dx. Use the substitution x=acostx=acost to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

318.

[T] The following graph is of a function of the form f(t)=asin(nt)+bsin(mt).f(t)=asin(nt)+bsin(mt). Estimate the coefficients a and b, and the frequency parameters n and m. Use these estimates to approximate 0πf(t)dt.0πf(t)dt.

A graph of a function of the given form over [0, 2pi], which has six turning points. They are located at just before pi/4, just after pi/2, between 3pi/4 and pi, between pi and 5pi/4, just before 3pi/2, and just after 7pi/4 at about 3, -2, 1, -1, 2, and -3. It begins at the origin and ends at (2pi, 0). It crosses the x axis between pi/4 and pi/2, just before 3pi/4, pi, just after 5pi/4, and between 3pi/2 and 4pi/4.
319.

[T] The following graph is of a function of the form f(x)=acos(nt)+bcos(mt).f(x)=acos(nt)+bcos(mt). Estimate the coefficients a and b and the frequency parameters n and m. Use these estimates to approximate 0πf(t)dt.0πf(t)dt.

The graph of a function of the given form over [0, 2pi]. It begins at (0,1) and ends at (2pi, 1). It has five turning points, located just after pi/4, between pi/2 and 3pi/4, pi, between 5pi/4 and 3pi/2, and just before 7pi/4 at about -1.5, 2.5, -3, 2.5, and -1. It crosses the x axis between 0 and pi/4, just before pi/2, just after 3pi/4, just before 5pi/4, just after 3pi/2, and between 7pi/4 and 2pi.
Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction
Citation information

© Mar 30, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.