For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Integral Test

**Main theorem and definition:**

Suppose *f (x*) is a function and you want to know if converges.

Check three conditions:

Is

*f (x*) continuous?Is

*f (x*) always positive?Is

*f (x*) decreasing? (i.e., is*f (x*) negative?)

Then if *f (x*)*dx* converges, then converges.

And if *f (x*)*dx* diverges, then diverges.

**Definition:** A __p-series__ is a series of the form . Using the Integral Test, we can show that the series converges if *p*> 1 and diverges to infinity if *p* ≤ 1.

**Hints and tips:**

It’s important to remember that the Integral Test only applies to series with

__positive__terms. However, after you learn about absolute convergence later, you may be able to use it for series with some negative terms by taking their absolute value and seeing if they are absolutely convergent.The Integral Test is one of the few two-way tests − you can use it to conclude that a series converges or that it diverges.

Don’t make the common mistake of confusing p-series with geometric series. p-series has the

*n*in the base and a constant exponent. Geometric series has a constant base and*n*in the exponent. When a geoemetric series converges, we have formulas to tell us what it converges to. When a p-series converges, we usually don’t know what it converges to.It’s worth memorizing the rule for p-series, since these are often used later in conjunction with the Comparison Test.

Remember that the Integral Test never tells you exactly what a series converges to.

### Integral Test

_{n = 1}

^{∞}[1/(n − 1)] ?

If f(n) = [1/(n − 1)], then f(x) is not continuous on the intertval [1,∞] on x = 1 Thus the Integral Test cannot be applied

_{n = 1}

^{∞}[(n

^{3})/(n + 1)] ?

If f(n) = [(n

^{3})/(n + 1)], then f(1) < f(2) and thus f(x) is increasing.

Thus the Integral Test cannot be applied.

_{}

^{}[(x

^{3})/([(x

^{4})/4] + 1)]dx

- Use substitution with u = [(x
^{4})/4] + 1 - u = [(x
^{4})/4] + 1 - du = x
^{3}dx

∫

_{}

^{}[(x

^{3})/([(x

^{4})/4] + 1)]dx = ∫

_{}

^{}[du/u] = ln|[(x

^{4})/4] + 1| + C

^{3})/([(n

^{4})/4] + 1)] converge?

- Integrate f(x) from 1 to ∞
- ∫
_{1}^{∞}[(n^{3})/([(n^{4})/4] + 1)] = ∫_{1}^{∞}[(x^{3})/([(x^{4})/4] + 1)] - Integrate the improper integral
- ∫
_{1}^{∞}[(x^{3})/([(x^{4})/4] + 1)] = lim_{k∞}∫_{1}^{k}[(x^{3})/([(x^{4})/4] + 1)] - = lim
_{k∞}[ ln|[(x^{4})/4] + 1| ]_{1}^{k} - = lim
_{k∞}[ ln|[(k^{4})/4] + 1| − ln|[1/4] + 1| ] - = ∞

_{}

^{}[(x + 61)/(e

^{x})]dx

- Setup integration by parts
- u = x + 61
- du = dx
- dv = [1/(e
^{x})] - v = − [1/(e
^{x})]

_{}

^{}[(x + 61)/(e

^{x})]dx = − [(x + 61)/(e

^{x})] + ∫

_{}

^{}[dx/(e

^{x})] = − [(x + 61)/(e

^{x})] − [1/(e

^{x})] + C = [(x + 60)/(e

^{x})] + C

^{x})] converge or diverge?

- Integrate f(x) from 1 to ∞
- ∫
_{1}^{∞}[(n^{3})/([(n^{4})/4] + 1)] = ∫_{1}^{∞}[(x + 61)/(e^{x})] - Integrate the improper integral
- ∫
_{1}^{∞}[(x + 61)/(e^{x})] = lim_{k∞}∫_{1}^{k}[(x + 61)/(e^{x})] - = lim
_{k∞}[ [(x + 60)/(e^{x})] ]_{1}^{k} - = lim
_{k∞}[ [(k + 60)/(e^{x})] − [61/(e^{})] ] - = lim
_{k∞}[ [k/(e^{x})] − [1/e] ] - = 0 − [1/e]
- = − [1/e]

_{}

^{}[1/(x

^{3})]dx

∫

_{}

^{}[1/(x

^{3})]dx = − [(x

^{ − 2})/2] + C

- Determine the series formula
- 7 + [7/8] + [7/27] + [7/64] + ... = ∑[7/(n
^{3})] - Integrate f(x) from 1 to ∞
- ∫
_{1}^{∞}[7/(n^{3})] = ∫_{1}^{∞}[7/(x^{3})] - = 7∫
_{1}^{∞}[1/(x^{3})] - = 7lim
_{k∞}[ − [1/(2x^{2})] ]_{1}^{k} - = 7lim
_{k∞}[ − [1/(2k^{2})] + [1/2] ] - = [7/2]

_{n = 1}

^{∞}[1/(

^{3}√{n})] converges with P - series conditions

_{n = 1}

^{∞}[1/(

^{3}√{n})] converges with P - series conditions

- Integrate f(x) from 1 to ∞
- ∫
_{1}^{∞}[1/(^{3}√{n})] = ∫_{1}^{∞}[1/(^{3}√{x})] - = lim
_{k∞}[ [3/2]x^{[2/3]}]_{1}^{∞} - = [3/2]lim
_{k∞}[ k^{[2/3]}− 1 ] - = ∞

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Integral Test

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Important Theorem and Definition 0:05
- Three Conditions
- Converging and Diverging
- P-Series
- Lecture Example 1 2:19
- Lecture Example 2 5:08
- Lecture Example 3 6:38
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### Transcription: Integral Test

*Hi, this is educator.com.*0000

*We are trying a couple more examples of the integral test.*0002

*The first one here is the summation from n=2 to infinity of 1/n × ln(n).*0007

*You are probably used to seeing series that start at n=1.*0013

*Here I had to start it at n=2.*0018

*If you plug it n=1, to the natural log, ln(1) = 0.*0021

*If we started at n=1, we would get 1/0 and that would not make sense.*0026

*That is why I had to start at n=2.*0030

*So the integral test says you look at what you are given for a series and you convert it into a function.*0032

*f(x) = 1/x ln(x).*0037

*There are three things you have to check.*0047

*One, whether it is continuous.*0050

*This is a continuous function, because we are only looking at values of x bigger than 2.*0056

*There is no question about dividing by 0 in there.*0062

*Whether it is positive and this is positive as long as, well x is always positive.*0066

*This is positive as long as ln(x) is greater than 0, which it is,*0079

*Since we are only looking at values of x bigger than 2.*0083

*The third question is whether the function is decreasing.*0093

*To check that, we have to look a the derivative f'(x).*0098

*The derivative of that, if we think of f(x) as x × ln(x) ^{-1}.*0105

*The derivative is -x log(x) ^{-2} × the derivative of x log(x).*0112

*That is a product so we have to use the product rule.*0125

*That is x × the derivative of log(x) + log(x) times derivative of x,*0126

*The derivative of x is just 1.*0136

*This simplifies down into -1 + log(x)/x ^{2} × log(x)^{2}.*0137

*If we look at that, 1 + log(x) > 0.*0151

*x ^{2} × log(x) ^{2} > 0.*0157

*We have got numerator, denominator positive, but then we have got a big negative sign on the outside.*0162

*So, this whole thing is going to be negative.*0169

*Remember that was going to be the derivative, that means it has a negative derivative.*0173

*That means the function is decreasing.*0179

*Having checked those 3 conditions, it is ok to use the integral test.*0183

*We are going to look at the integral from 2 to infinity of 1/x log(x) dx.*0193

*That is not such a bad integral to do.*0205

*We can use u = ln(x).*0208

*Then, du = just 1/x dx.*0211

*We have got the integral from x=2 to x=infinity of 1/u du.*0217

*So, that is the integral of ln(u) evaluated from x = 2, well a value of t going to infinity.*0230

*That is the ln(ln(x)), because we have to change u back into the ln(x).*0244

*Again, evaluated from x=2 to x=a value that goes to infinity.*0253

*The second part of this is very easy.*0261

*- ln(ln(2)) is no problem there.*0264

*- ln(ln(2)) is just some number.*0268

*However, when we take ln(of some number going to infinity), that will go to infinity.*0272

*We actually want ln(that), but then we have ln(something going to infinity),*0277

*This whole thing diverges to infinity.*0291

*The integral diverges to infinity.*0298

*So, we can say the series 1/n log(n) diverges to infinity by the integral test.*0302

*Let us recap what happened there.*0321

*We were given the series of 1/n log(n).*0330

*We converted that into a function and checked these 3 conditions, continuous, positive, and decreasing.*0335

*The trickiest one there was decreasing.*0340

*We had to look at the derivative and work it through and check that it was negative.*0343

*Once we verified those 3 conditions, it is OK to use the integral test.*0347

*We look at the improper integral of 1/x log(x) from 2 to infinity.*0353

*That, we work out the calculus and that turns out to be ln(ln(x)).*0358

*Then when we plug in large values of x to ln(ln(x)), we get something that diverges to infinity,*0365

*So the integral test tells us that the original series diverges to infinity as well.*0374

*Let us try one more example of a problem that suggests the integral test.*0000

*Here we have the sum from n=1 to infinity of 1/n ^{2} + 1.*0005

*We would set up f(x) = 1/n ^{2} + 1.*0012

*Now, there are three conditions we need to check, continuous, positive, decreasing.*0020

*The continuous is pretty clear.*0032

*The positive is also pretty clear, no matter what x you put there that is going to be positive.*0038

*The decreasing, you should work out f'(x) and check that it is negative.*0044

*I am not going to work it out here, but it does turn out to be negative so it is ok to use the integral test.*0049

*Let us look at the integral from x=1 to x going to infinity of 1/x ^{2}+1 dx.*0062

*There are several ways to do that integral.*0074

*You can do it with a trig substitution, you would use x = tan(θ).*0077

*You could look it up in the table of integrals in the back of your calculus book.*0081

*Probably for this integral, it is one that you are going to see often enough that it is good to have the answer memorized.*0087

*The integral of 1/x ^{2}+1 = arctan(x).*0096

*That is one that if you have not memorized it yet,*0103

*You will probably see often enough as you are doing calculus, that it is worthwhile to memorize it.*0104

*Then we want to evaluate that from x=1 to a value that is going to infinity.*0109

*So, this is arctan(a number going to infinity) - arctan(1).*0121

*Now, remember what the arctangent function does.*0132

*It is the inverse of the tangent function.*0136

*There is the tangent function with asymptotes at -π/2 and π/2.*0139

*That is tan(x).*0147

*Arctan(x) says you flip that and so you get these horizontal asymptotes at -π/2 and π/2.*0149

*That is arctan(x).*0163

*When x goes to infinity, arctan(x) approaches π/2.*0168

*The limit there is π/2.*0173

*The arctan(1), well you think what angle as arctan(1) and if you remember your common values that is π/4.*0177

*So this improper integral converges to π/4.*0187

*What that tells us is that the series, 1/n ^{2}+1 converges by the integral test.*0190

*What it does not tell us is what it converges to.*0211

*Remember, the integral test does not tell you that.*0221

*I will remind you that we do not know from anything we have learned so far, what that series converges to.*0225

*Even though the integral came out to be π/4, that does not tell you anything about what the series converges to.*0235

*All we can say from the integral test is that the series converges.*0244

*To recap there, we were given a function, a series,*0249

*We convert it into a function,*0255

*We check ti see if it is continuous, positive and decreasing,*0256

*Decreasing is the challenging part, we have to check that its derivative is negative.*0259

*Once we verify those, we do the integral.*0265

*We find that the answer turns out to be a finite limit.*0269

*That does not turn out to be infinity.*0274

*That allows us to conclude that the series converges by the integral test.*0278

*But we do not know exactly what the series converges to.*0284

*This has been educator.com*0286

*Hi, this is educator.com and we are here to talk about the integral test.*0000

*The integral test is a way of determining when a series converges or diverges.*0006

*The way it often works is you will be given a series in this form.*0013

*f(n) is just some expression in terms of n.*0017

*What you do is convert that into a function of x.*0020

*Then there are 3 things you have to check before you can apply the integral test.*0025

*You have to check whether it is a continuous function,*0029

*You have to check whether it is always positive.*0033

*The integral test only works for series with positive terms.*0036

*You have to check whether it is decreasing.*0040

*In other words, is the derivative negative?*0043

*The derivative has to be negative to pass the integral test.*0046

*If you checked all three of those conditions, the integral test tells you that you can look at the integral of f(x) from 1 to infinity,*0050

*And if it converges, then your series converges.*0060

*If it diverges, then your series diverges.*0064

*We will try some examples of that in a moment.*0069

*In the meantime, let me give you 1 important definition that comes out of the integral test.*0073

*We are going to talk about p series.*0077

*That is a series of the form 1/n ^{p} where p is a constant value.*0080

*P series lend themselves very nicely to integral tests.*0088

*Because, we figured out back in the integration section on improper integrals.*0096

*We figured out when the integral of dx/x ^{p} converges.*0106

*What we figured out was that the integral from 1 to infinity of dx/x ^{p} converges exactly when p > 1 and diverges to infinity if p < or = 1.*0109

*We will use that result to look at some series today.*0127

*You will use it also in your calculus homework.*0131

*Let us try out some examples of the integral test.*0135

*First one is to determine whether the series 1/sqrt(n) converges or diverges.*0140

*That one we can write it as 1/n ^{1/2}.*0148

*n = one to infinity.*0157

*That is now a p series.*0162

*With p = 1/2, 1/2 < or = to 1, so the series diverges to positive infinity.*0169

*Let me mention something about p series, because we will be seeing several examples of them.*0198

*We have also seen another kind of series called a geometric series.*0204

*These are frequently mixed up by students.*0210

*Let me mention what the difference is.*0211

*This n ^{1/2} is a p series because the variable n is in the base, and the constant 1/2 is in the exponent, so that is a p series.*0213

*A geometric series, the general form of a p series is 1/n ^{p}.*0231

*A geometric series, if you had something like 1/2 ^{n}, where the constant is the base and the variable is in the exponent,*0236

*That is a geometric series.*0250

*We saw some examples of those earlier.*0253

*The general form of a geometric series would be the sum of r ^{n},*0258

*Where r is the constant.*0262

*Remember in the p series, the exponent is the constant, and in the geometric series, the base is the constant.*0269

*Those two are very frequently mixed up by students.*0277

*They are different kinds of series and the rules for determining when they converge or diverge look similar on the surface.*0280

*You really have to remember separate rules for geometric rules and p series.*0288

*You have to be careful not to label a geometric series as a p series, or a p series as a geometric series.*0297

*This one is the series 1/n ^{3/2}.*0309

*Again, this is a p series, because the constant is in the exponent.*0315

*Here p is 3/2, which is bigger than 1.*0324

*So, the series converges.*0332

*Let me mention another feature of the integral test in a p series.*0345

*Which is that the integral test and the rules for when a p series converge,*0351

*None of those tell you what the series converges to.*0357

*We do not know from anything we have learned so far what the series converges to.*0367

*We do not know the limit of the series.*0378

*In that sense, the integral test and the p series rules are limited.*0384

*They will tell you whether the series converges or diverges, but if a series does converge, it will not tell you what it converges to.*0389

*Let us try another example here.*0398

*The sum of n ^{2}/e^{n}.*0402

*This one is a little more complicated and is not just a simple p series.*0405

*Let us look at f(x) = x ^{2}/x^{n}.*0409

*We want to use the integral test to tell whether or not the series converges or diverges.*0420

*It will be easier to integrate this thing if we look at it as x ^{2}e^{-x}.*0424

*I want to look at the improper integral from 1 to infinity of x ^{2}e^{-x} dx.*0431

*That is a classic integral to solve by integration by parts.*0444

*I am going to use the trick of tabular integration that we learned in the integration by parts lecture to integrate this.*0447

*If you do not remember the tabular integration trick, you might want to go back and look at th integration by parts lecture.*0460

*I will also go through it slowly here.*0468

*If we take the derivatives of x ^{2}, we get 2x, 2 and then 0.*0472

*If we take the derivatives of e ^{-x}, the first integral is -e^{-x}, the integral of that is just e^{-x} again.*0479

*The integral of that is -e ^{-x}.*0488

*Then we write these diagonal lines and the signs +, -, +.*0493

*Then we multiply along the diagonal lines to get the answer x ^{2}e^{-x} - 2xe^{-x} -2e^{-x}.*0500

*We want to evaluate that from x=1 to x=some value t which will go to infinity.*0516

*This tabular integration is a shorthand way of doing integration by parts.*0527

*You can also do integration by parts by using u and dv.*0535

*You should get the same answer here.*0542

*If you plug that in, e ^{-x}, this is the same as saying x^{2}/e^{-x}.*0543

*As we plug in infinite values there, larger and larger values of x,*0551

*The e ^{-x} dominates the x^{2}.*0560

*When we take the limit as t goes to -infinity,*0568

*Since the e ^{-x} dominates the x^{2}, we actually get 0 on each of these terms.*0569

*0-0-0, and now if we plug in x=1.*0575

*We get - a -, so that is a + 1/e + 2/e +2/e.*0581

*If you add those up to get 5/e, which is a finite number.*0604

*The conclusion we can draw from that is the series that we were given n ^{2}/e^{n} converges by the integral test.*0613

*That is our answer.*0644

*If this integral had diverged to infinity, then we would have said the series diverges to infinity.*0645

*The integral test is nice in that it can give you either answer.*0654

*Unlike some of the other tests.*0657

*If this integral had diverged to infinity, we would have said the series diverged to infinity.*0660

*Where the integral test falls, and does not give us an answer, is it does not tell us exactly what the series converges to.*0667

*Even this 5/e is attempting to say, well that is the value of the interval, that must be the value of the series.*0674

*That is not safe.*0682

*This 5/e, we do not actually get that much from the numerical value 5/e.*0684

*The important thing is just that it is finite, and so it tells use the series converges.*0689

*We do not know using what we have learned so far what it converges to.*0698

*So, that is how the integral test works.*0713

*You take the series that you are given, convert it into a function, work out the integral from 1 to infinity of that function.*0717

*Then, if you get a finite answer, you can say the series converges.*0727

*If you get an infinite answer, you can say the series diverges.*0733

*We will try a couple more examples of that later on.*0736

1 answer

Last reply by: Dr. William Murray

Mon Oct 21, 2013 10:31 PM

Post by Wael Saleh on October 20, 2013

Great job, As you know there are many types of tests, Therefore, I have problem !! Are there any certain signals in question describe which kind of tests should I use it to solve problem?

Best regards,

1 answer

Last reply by: Dr. William Murray

Sun Apr 28, 2013 11:12 AM

Post by Megan Kell on April 10, 2012

I'm so confused... I understand the p-series method and the integral method, but what do they have to do to each other? and at the beginning of this lecture, he talks about the three points you must identify before doing the integral test, but yet he only covers that in additional example IV, although he should have demonstrated this in Example 3 as well..? The thought process in this lecture is not very clearly executed.

1 answer

Last reply by: Dr. William Murray

Sun Apr 28, 2013 11:04 AM

Post by Jaspreet Singh on March 15, 2010

There's a typo in the notes, for number 3 it should be "Is f'(x) negative"

1 answer

Last reply by: Dr. William Murray

Sun Apr 28, 2013 10:44 AM

Post by John Haddad on February 10, 2010

Tabular integration is a great technique for Ex 3, unfortunately my university disallows it's use on exams; but it'll come in handy for multiple choice problems.