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

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### Improper Integration

**Main formula:**

Horizontal asymptote:

Vertical asymptote:

**Hints and tips:**

It is often useful to graph the function you’re integrating to get a general idea of its shape and whether it is positive or negative.

However, you usually can’t eyeball the difference between finite and infinite area. You need to do the integral and take the limit to be sure.

Remember that infinite positive and infinite negative areas do not cancel. When we have both, we just say that the integral diverges.

Often, you can make a substitution and convert an integral into one of the form or . It is worth memorizing the chart for which values of

*p*make these integrals converge or diverge. These integrals will also be useful for examining infinite series in later sections.Watch out for “hidden ambushes” when the function is discontinuous or its denominator is zero in the middle of your interval. In these cases, split the integral up around the discontinuity and evaluate improper integrals on both sides.

If you get something that is difficult to integrate, you may need to use the Comparison Test with an easier integral (often or ). A common comparison is to use the fact that −1 ≤ sin

*x*≤ 1, and similarly for cos*x*.

### Improper Integration

_{1}

^{∞}[3/x]dx

- Replace the improper bound ∞ with k
- ∫
_{1}^{k}[3/x]dx - Integrate
- ∫
_{1}^{k}[3/x]dx = 3[ln|x|]_{1}^{k} - = 3lnk − 3ln1

_{k → ∞}3lnk = ∞

_{ − ∞}

^{0}[1/(( x + 5 )

^{2})]dx

- Replace the improperbound ∞ with k
- ∫
_{k}^{0}[1/(( x + 5 )^{2})]dx - Integrate
- ∫
_{k}^{0}[1/(( x + 5 )^{2})]dx = [ − [1/(x + 5)] ]_{k}^{0.} - = − [1/(0 + 5)] + [1/(k + 5)]

_{x → − ∞}[ [1/(k + 5)] ] − [1/5] = 0 − [1/5] = − [1/5]

_{ − ∞}

^{∞}[7/((x − 3)

^{2})]dx

- Note that [7/((x − 3)
^{2})] is an even function,and the limits are the same magnitude - ∫
_{ − ∞}^{∞}[7/((x − 3)^{2})]dx = 2∫_{0}^{∞}[7/((x − 3)^{2})]dx - Replace the improper limit ∞ with k
- 2∫
_{0}^{∞}[7/((x − 3)^{2})]dx = 2∫_{0}^{k}[7/((x − 3)^{2})]dx - Integrate
- 2∫
_{0}^{k}[7/((x − 3)^{2})]dx = 14∫_{0}^{k}[1/((x − 3)^{2})]dx - = 14[ − [1/(x − 3)] ]
_{0}^{k} - = 14[ − [1/(k − 3)] + [1/(0 − 3)] ]
- = 14[ − [1/(k − 3)] − [1/3] ]
- = 14[ − lim
_{k → ∞}[1/(k − 3)] − [1/3] ] - = 14[ 0 − [1/3] ]

_{3}

^{7}− [4/(√{x − 3} )]dx

- Replace the bound 3 with k
- ∫
_{3}^{7}− [4/(√{x − 3} )]dx = ∫_{k}^{7}− [4/(√{x − 3} )]dx - Integrate
- ∫
_{k}^{7}− [4/(√{x − 3} )]dx = − 4∫_{k}^{7}[1/(√{x − 3} )]dx - = − 4[ 2√{x − 3} ]
_{k}^{7} - = − 4[2√{7 − 3} − 2√{k − 3} ]
- = − 4[2√4 − 2√{k − 3} ]

_{ − 2}

^{k}[2/(x√{x

^{2}− 1} )]dx

- Replace the bound 1 with k
- ∫
_{ − 2}^{k}[2/(x√{x^{2}− 1} )]dx = ∫_{ − 2}^{k}[2/(x√{x^{2}− 1} )]dx - Integrate using inverse trigonmetric identity
- ∫
_{ − 2}^{k}[2/(x√{x^{2}− 1} )]dx = 2∫_{ − 2}^{k}[1/(x√{x^{2}− 1} )]dx - = 2[ sec
^{ − 1}|x| ]_{ − 2}^{k} - = 2[ sec
^{ − 1}|k| − sec^{ − 1}| − 2| ]

_{k1}sec

^{ − 1}|k| − [(π)/3] ] = 2[ 0 − [(π)/3] ] = − [(2π)/3]

_{0}

^{∞}[(x − 4)/(x

^{2}− 3x − 4)]dx

- Factor x
^{2}− 3x − 4 to simplify - ∫
_{0}^{∞}[(x − 4)/(x^{2}− 3x − 4)]dx = ∫_{0}^{∞}[(x − 4)/((x − 4)(x + 1))]dx - = ∫
_{0}^{∞}[1/((x + 1))]dx - Replace the improper limit ∞ with k
- ∫
_{0}^{∞}[1/((x + 1))]dx = ∫_{0}^{k}[1/((x + 1))]dx - Integrate
- ∫
_{0}^{k}[1/((x + 1))]dx = [ ln|x + 1| ]_{0}^{k} - = ln|k + 1| − ln|0 + 1|
- = ln|k + 1| − ln|1|
- = ln|k + 1| − 0
- = ln|k + 1|
- = lim
_{k → ∞}ln|k + 1|

_{ − ∞}

^{1}[2/(x

^{2}+ 2x)]dx

- Replace the bound − ∞ with k
- ∫
_{ − ∞}^{1}[2/(x^{2}+ 2x)]dx = ∫_{k}^{1}[2/(x^{2}+ 2x)]dx - Integrate
- ∫
_{k}^{1}[2/(x^{2}+ 2x)]dx = ∫_{k}^{1}[1/x] − [1/(x + 2)]dx - = ∫
_{k}^{1}[1/x] − ∫_{k}^{1}[1/(x + 2)]dx - = [ ln|x| ]
_{k}^{1}− [ ln|x + 2| ]_{k}^{1} - = ln1 − lnk − ln3 + ln|k + 2|

_{k → − ∞}[ − ln3 + ln|k + 2| − lnk ] = − ln3 − ∞− ∞ = − ln3

_{0}

^{∞}[4/((x + 1)(x + 2))] dx

- Replace the bound ∞ with k
- ∫
_{0}^{∞}[4/((x + 1)(x + 2))] dx = ∫_{0}^{k}[4/((x + 1)(x + 2))] dx - Integrate
- ∫
_{0}^{k}[4/((x + 1)(x + 2))] dx = ∫_{0}^{k}[4/(x + 1)] − [4/(x + 2)] dx - = 4∫
_{0}^{k}[1/(x + 1)] − [1/(x + 2)] dx - = 4[ ln|x + 1| − ln|x + 2| ]
_{0}^{k}

_{k → ∞}ln[(k + 1)/(k + 2)] − ln[1/2] ] = 4[ 0 − ln[1/2] ] = − 4ln[1/2]

_{ − 2}

^{2}[1/(√{4 − x

^{2}} )]dx

- Note the function is symmetrical and simplify
- ∫
_{ − 2}^{2}[1/(√{4 − x^{2}} )]dx = 2∫_{0}^{2}[1/(√{4 − x^{2}} )]dx - Replace the bound 2 with k
- 2∫
_{0}^{2}[1/(√{4 − x^{2}} )]dx = 2∫_{0}^{k}[1/(√{4 − x^{2}} )]dx - Apply the inverse trignometric integration identity to integrate
- 2∫
_{0}^{k}[1/(√{4 − x^{2}} )]dx = 2[ sin^{ − 1}[x/2] ]_{0}^{k} - = 2[ sin
^{ − 1}[k/2] − sin^{ − 1}[0/2] ] - = 2[ sin
^{ − 1}[k/2] ] - = 2[ lim
_{k → 2}sin^{ − 1}[k/2] ] - = 2[ [(π)/2] ]

_{0}

^{∞}[1/(( x − 2 )

^{3})] dx

- Replace the bound ∞ with k
- ∫
_{0}^{∞}[1/(( x − 2 )^{3})] dx = ∫_{0}^{k}[1/(( x − 2 )^{3})] dx - Integrate
- ∫
_{0}^{k}[1/(( x − 2 )^{3})] dx = [ − [1/(2(x − 2)^{2})] ]_{0}^{k} - = [ − lim
_{k → ∞}[1/(2(k − 2)^{2})] + [1/(2(0 − 2)^{2})] ] - = 0 + [1/2(4)]

*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

### Improper Integration

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
- Horizontal and Vertical Asymptotes 0:04
- Example: Horizontal
- Formal Notation
- Example: Vertical
- Formal Notation
- Lecture Example 1 5:01
- Lecture Example 2 7:41
- Lecture Example 3 11:32
- Lecture Example 4 15:49
- Formulas to Remember 18:26
- Improper Integrals
- Lecture Example 5 21:34
- Lecture Example 6 (Hidden Discontinuities) 26:51
- Additional Example 7
- Additional Example 8

### 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: Improper Integration

*In this case, we are given the integral from 0 to infinity of 8 e ^{-x}.*0000

*If you graph that, e ^{-x} has the basic shape of e^{x}.*0006

*Except it is flipped around the y-axis.*0015

*The 8 multiplies it up and makes it a little bigger but it is not going to change the same basic shape.*0021

*What we are really trying to do is find that area and determine whether it is finite or infinite.*0025

*The problem actually asks a little more.*0031

*If it diverges we have to determine whether it is positive or negative infinity or neither.*0035

*If it converges, we have to find out exactly what that area is.*0039

*Let us set it up as a limit.*0044

*We take the integral from 0 to t of 8 e ^{-x}.*0047

*That is a pretty easy integral to do, if you make the substitution u = -x,*0061

*You get -8e ^{-x} from x=0 to x=t.*0065

*If you plug that in, that is -8e ^{-t} - 8e^{0}.*0073

*Sorry, subtracting a negative sign, so + 8e ^{0}.*0088

*Then we take the limit as t goes to infinity of this.*0093

*So this is minus 8e ^{-infinity} + 8e^{0} = 1.*0101

*Well e ^{-infinity}, this is 1/e^{infinity}, well actually I am not looking at the -8 part, just the e part which goes to 0.*0111

*And our answer for the whole thing there is 8.*0128

*So we say that this converges to 8.*0141

*So just to recap there, what we did was we handled the impropriety of letting x go to infinity by replacing infinity with t.*0148

*We worked through the integral and then took the limit as t goes to infinity.*0161

*We got the answer 8, and what that represents is that this area,*0166

*Even though it goes on infinitely far, the total area is only 8 square units.*0170

*So our last example is the integral from 2 to infinity of dx/x+sin(x).*0000

*Again, because we have an impropriety at infinity, we are going to integrate from 2 to t of dx/x+sin(x).*0007

*We would like just to do the integral as the next step.*0021

*The problem is that the integral of 1/x + sin(x) is something very difficult, something we do not know how to do.*0025

*How do we handle this?*0033

*The way you want to think about this is kind of ask yourself,*0035

*What does 1/x+sin(x) look like?*0037

*Now I am not asking for a detailed graph, but asking you to think about how big 1/x+sin(x) can be.*0045

*Well, x+sin(x), something I know here is that sin(x) is always between -1 and 1.*0056

*So x + sin(x) is always less than or equal to x+1.*0068

*So 1/x+sin(x), when you take 1 over something, it reverses the inequality.*0075

*That is bigger than 1/x+1.*0082

*So this integral, whatever it is, is bigger than the integral from 2 to t of dx/x+1.*0087

*That is an easy integral to do.*0096

*We can u = x+1.*0099

*And we get the ln(x+1) evaluated from x=2 to x=t.*0100

*That is the ln(t+1) - ln(2+1).*0115

*We want the limit as t goes to infinity, so this is the ln(infinity), which is infinity - ln(3),*0126

*Which is just a finite number.*0144

*So this diverges to infinity.*0148

*However, that was the integral of dx/x+1.*0152

*We are actually trying to look at the integral of dx/x+sin(x).*0158

*What we have learned though is that the integral from 2 to infinity of dx/x+sin(x),*0161

*Is bigger than or equal to the integral from 2 to infinity of dx/x+1.*0174

*That in turn is infinity, so the area under this function is infinite, this is even bigger.*0182

*So, our integral if it is even bigger than our infinite value, it diverges to infinity too.*0197

*There are a couple points I want to make about this.*0216

*One is that in order for this kind of argument to work, the inequality has to go the right way.*0219

*If this inequality had turned out to go the other way,*0225

*If this had been a greater than or equal to,*0230

*I will write this in red so that you can see this is not the actual value.*0235

*So what if that had been greater than or equal to?*0240

*That would have made this inequality less than or equal to.*0243

*Then in turn this inequality would have been less than or equal to, and this inequality would have been less than or equal to.*0247

*Then we would have something less than or equal to an infinite area.*0253

*That is not something that we can say is finite or infinite.*0258

*If something is less than infinity, well it might be finite, or it could still be infinite.*0264

*So what is very important for this kind of argument to work, is that the inequality must go the correct way.*0267

*The second point I wanted to make about this is that when you use this kind of argument,*0288

*You are not figuring out the exact value of the integral.*0301

*This kind of argument will never tell you the exact value of the integral,*0305

*Even when you get an integral that converges.*0307

*This never tells you the exact value of the integral.*0311

*Now, in this problem, we got kind of lucky because we were not asked what the exact value of the integral was.*0329

*We were only asked to determine whether the integral converges or diverges.*0336

*In that sense, that is sometimes a tip-off that you want to use this comparison idea.*0342

*If the problem does not ask you what the exact value is,*0347

*If it just asks you whether it converges or diverges,*0353

*That is a little hint that you might want to use this idea of comparison.*0359

*Today we are going to talk about improper integration.*0000

*The idea of improper integration is the function that you are trying to integrate has either a horizontal or a vertical asymptote.*0005

*For example, if you have a horizontal asymptote, that means something like this,*0016

*Where the curve goes on forever and gets closer and closer to a horizontal line.*0020

*What you might be trying to do is find the area under the curve but this is an area that goes on forever.*0027

*The formal notation for this is to take the integral from x=a and we say the integral from a to infinity of f(x) dX.*0038

*The way you really do that since we do not really know what it means to plug in infinity for a function is we cut off this integral at a certain point which we call T.*0048

*Then we just find the integral from x=a to x=t.*0065

*So, we are finding the area under the curve from x=a to x=t, and then we let t get bigger and bigger to go off to infinity.*0070

*We take the limit as t goes to infinity.*0081

*What can happen is it can turn out that that can approach a finite limit or it can diverge to infinity.*0085

*The same way that the limits that you learned about in Calculus 1 can approach a finite limit or they can diverge to infinity, or negative infinity.*0094

*When we do some examples here, we will be asking whether these integrals converge to a finite limit or if so what the limit is,*0104

*Or whether they diverge to either negative infinity, or positive infinity, or neither.*0112

*The other kind of integral we will be looking at is a vertical asymptote.*0117

*That means that the function you are trying to integrate maybe goes up to infinity, in other words it approaches a vertical line.*0129

*If it has a vertical asymptote at b, we might want to try to find the total area under that curve between a and b.*0134

*We try to find that area but the reason the function probably goes up to infinity at b,*0148

*Is because we are dividing by 0 at b and it is not legitimate to just plug in a value at b for the function.*0155

*We cut off the function a little bit short of the vertical asymptote, so if there is b, and there is a,*0168

*What we do is we cut off the function a little bit short.*0179

*We call that value t.*0183

*We cut it somewhere short of the vertical asymptote where the function is still defined and it does not blow up to infinity and we find the area between a and t.*0185

*Then what we do is we gradually move t closer and closer to b,*0196

*That this area approaches the area under the entire curve that we are looking at.*0200

*Again, this area can turn out to be finite or infinite,*0213

*So, we will talk about the integral converging to a finite number or diverging to infinity or negative infinity, or neither.*0217

*Again, we handle it with limits, we take the limit as t goes to b.*0226

*Since t is coming towards b from the left hand side,*0230

*That is why we put a little negative sign down there to show that t is coming to b from the negative side.*0234

*We can also talk about functions that approach vertical asymptotes from the right hand side.*0242

*We can ask ourselves what is the area under that function between a and b where a is the vertical asymptote.*0254

*Then what we would say is that the integral from a to b of f(x) dX,*0261

*We would look at the integral and now we would have t be just a little bit to the right of a,*0270

*And t would approach a from the right, and we would integrate from t to b of f(x) dX.*0277

*Then we would take the limit as T gets closer and closer to a, but this time t is on the right-hand side of a.*0285

*I am going to put a little plus sign there to show that t is approaching a from the positive side.*0294

*Let us try some examples and see how it works out.*0299

*Our first example is to look at the integral from 0 to infinity of 1/(x+1).*0304

*The key thing here is that if you graph 1/(x+1), well, at 0 it is 1, and as x approaches infinity it goes down to 0.*0310

*We are trying to find that area there and see whether that area is finite or infinite.*0325

*The way we set this up is we take the integral from 0 to t, of 1/(x+1) dX.*0331

*We will go ahead and do that integral and then we will take the limit as t goes to infinity.*0341

*We are cutting this integral off at some value of t that will then let the t get bigger and bigger.*0348

*The integral of 1/(x+1) is a pretty easy integral if you let u = x+1 and put the substitution through there, then it turns into ln(x+1).*0356

*Evaluated from x=0 to x=t, and so we get ln(t+1) - ln(0+1), so ln(1) but that goes away to 0.*0367

*We get ln(t+1) and we want to take the limit of that as t goes to infinity.*0390

*If you remember, ln(x), y = ln(x).*0401

*As you plug in bigger and bigger numbers, ln(x) goes to infinity.*0411

*If we plug in bigger and bigger values for t, this limit goes to infinity.*0417

*What we say about this integral is that it diverges to positive infinity.*0424

*This represents the fact that there is actually infinite area under that curve.*0435

*Again, the way you handle these improper integrals is you convert the infinity into a t and then you do the integration*0444

*Then you take the limit as t goes to infinity and you see whether it turns out to be a finite or an infinite area.*0453

*Let us try another example.*0463

*Again, the same function actually, 1/(x+1).*0464

*But, this time we are taking the integral from -1 to 0 so the point here is that if you plug in x=-1 to this function,*0471

*It blows up because you get 0 in the denominator and because this function has a vertical asymptote at x=-1.*0488

*We are trying to figure out whether the area under that function as it approaches that vertical asymptote is finite or infinite.*0500

*Again, we handle this with limits.*0509

*We will take a value t that is just to the right of -1,*0512

*And we will integrate from t to 0, instead of from -1 to 0 because you cannot plug in -1 to that function.*0515

*We will look at dX/(x+1) and then we will take the limit as t gets closer and closer to -1.*0524

*I will put a little positive sign to indicate that t is approaching -1 from the positive side.*0535

*Again, we will take the integral and we get ln(abs(x+1)) evaluated from x=t up to x=0.*0543

*That is ln, if you plug in x=0, you get ln(1) - ln(t+1), so ln(1) goes to 0.*0555

*This leaves us with -ln(t+1).*0573

*Now, let us see what happens when we take values of t that are closer and closer to -1.*0581

*Well, if t goes to -1, that means that t+1 is going to 0, so what we are trying to do is take the ln, I should keep my absolute values here*0589

*Take the ln of something that is going to 0 from the positive side.*0610

*If you remember what the graph of ln(x) looks like, when you go to 0 from the positive side, ln(x) goes down to -infinity.*0615

*This is minus -infinity so we say the whole integral diverges to infinity.*0631

*Which represents the fact that the area under that curve is infinite.*0649

*Here we had a vertical asymptote and again the way to handle the vertical asymptote was to replace the problem value of x.*0659

*The problem value of x was -1, with t.*0668

*Go ahead and work through the integral and then take the limit as t approaches -1,*0672

*And I should have put a little plus in here to show that it is approaching -1 from the positive side.*0679

*We take that limit and we see whether it comes out to be a finite or infinite area.*0685

*Our next example is a little different.*0694

*We are going to look at cos(x).*0697

*We are going to look at the integral from 0 to infinity of cos(x).*0699

*For this one, just like the others, it is very helpful to draw a graph, so I will draw a graph of cos(x).*0703

*It starts at 1 and then it goes down to 0, -1, up to 1, down to -1, and so on.*0714

*So that is y=cos(x).*0722

*Now what we are really doing here is we are finding the area under the curve of cos(x),*0728

*Where we count area above the x axis as positive and below the x axis as negative.*0735

*We are really trying to add up all of those areas where this counts positive and this counts negative, positive, negative, and so on.*0743

*You can see easily just by looking at the graph here that there is going to be infinite positive area and also infinite negative area.*0758

*What we are really trying to do with this integral is add up infinitely many positive things and infinitely many negative things.*0784

*So, a positive infinite and a negative infinity.*0792

*A common question at this point is to ask whether the positive infinity and negative infinities cancel each other.*0797

*The answer is that there is no mathematically meaningful way to give rules for positive and negative infinities to cancel each other.*0807

*What you do in a situation like this is you say that the integral diverges but not to infinity or negative infinity.*0815

*You might be tempted to say that all of the positives and the negative infinities cancel each other and the whole integral is equal to 0.*0844

*Or you might be tempted to say that, well, these two areas cancel each other and this area cancels with the next one, so all of these areas cancel*0852

*Then you are just left with this area and you can find the value of that area.*0862

*There are all sorts of ways that you might think you could resolve this.*0868

*Because those all contradict each other, we do not sort of put the stamp of legitimacy on any of those.*0873

*We just say the whole thing diverges but not to positive infinity or negative infinity.*0881

*We just say it diverges.*0886

*You could also do this one by actually calculating the integral from 0 to 1 of cos(x) dx.*0888

*Which would give you sin(x) evaluated from 0 to 1.*0896

*That in turn gives you sin(t) - sin(0), which gives you sin(t) and then you are finding your limit as t goes to infinity of sin(t).*0905

*Again, we know that since sin oscillates forever, this does not exist.*0923

*It is not positive infinity, it is not negative infinity, it is not 0, we just say the whole thing does not exist.*0927

*It diverges.*0943

*OK, let us do another example.*0950

*Again this is one that if we look at the graph it is something we can resolve quickly.*0953

*We are looking at the integral from negative infinity to infinity of x ^{2} dx.*0958

*In this case we actually have improprieties, places where the intervals are improper at both ends.*0963

*If we graph this one, y = x ^{2}, it looks like that.*0971

*What we are trying to do is find the area under this curve as we go to infinity in both the positive and negative directions.*0979

*Now, clearly if you look at this area, the area is infinite.*0991

*Both on the positive and the negative sides.*0998

*Here is another common mistake that Calculus 2 students make*1000

*They will look at this area over here and say wait, is that not negative area?*1004

*That is a little mistake because remember, negative area is when the area is below the x axis.*1012

*This area is not below the x axis, it is to the left of the y axis, but it is still above the x axis.*1020

*That is not negative area.*1029

*It is still positive area because it is above the x axis.*1035

*If you simply look at the graph here, we are looking at positive infinite area + another batch of positive infinite area.*1041

*If you put those together we have two positive infinities.*1064

*Remember we learned in the last example that you cannot cancel a positive and a negative infinity, but what about 2 positive infinities?*1066

*You can add those up and say that the answer will be a positive infinity.*1074

*You can say that it diverges to positive infinity.*1081

*You do not just say this one diverges, you say that it diverges to positive infinity.*1088

*Because, if you look at these two areas we definitely have positive infinite area, positive infinite area, we put them together and we have got positive infinite area.*1095

*There are a few integrals that come up very, very often when you are looking at improper integrals.*1109

*The most important ones are the integral from 0 to 1 of dX/(x ^{p}).*1115

*The reason that is improper is because when x=0, you have 0 in the denominator here so that blows up that x=0, so x=0 is improper there.*1124

*The other one that comes up very often is the integral from 1 to infinity of dx/(x ^{p}), again that is improper because of the infinity there.*1139

*These integrals come up so often that it is worth working them out once and then probably memorizing the answers.*1153

*Especially this one, the integral of 1 to infinity of dx/x ^{p} is worth remembering.*1162

*It is worth remembering because it comes up later on in Calculus 2.*1177

*When you start looking at infinite series.*1183

*When you start looking at infinite series we are going to be using something called the integral test to determine whether infinite series converge or diverge.*1188

*The integral test says instead of looking at a series, you look at an integral.*1202

*The series that we are going to start with is 1/n ^{p}.*1210

*We will convert that into the integral of 1/x ^{p} dx.*1215

*Then we will remember what this integral was in order to determine whether the series is convergent or divergent.*1222

*It is worth remembering these.*1230

*Each of these integrals, you can evaluate yourself.*1233

*I am not going to work through the details there.*1236

*But you can evaluate yourself and it turns out that it depends on the values of p.*1239

*What values of p determine whether these integrals converge or diverge.*1245

*You want to remember that the first integral, if p < 1, it converges.*1250

*If p = 1, it diverges.*1258

*If p > 1, it also diverges.*1261

*The second integral, if p < 1 it diverges.*1265

*Equal to 1, it diverges, and p > 1 it converges.*1269

*That seems like a lot to remember.*1272

*You kind of remember, if p = 1, they both diverge, but if p < 1 or p > 1, they kind of flip flop back and forth there.*1276

*We will use these on the next example to help us determine whether an interval converges or diverges.*1286

*Here we have been given the integral of from 6 to 11 of 11/(x-6)x ^{1/3}.*1297

*We have been asked to determine whether it converges or diverges.*1305

*The first thing to look at with this is to look at it and see if we can make a substitution to make it simpler.*1309

*We can.*1316

*Let us do u = x - 6.*1317

*Then, whenever you make a substitution, you also have to change the differential.*1320

*So, du, if u is just x - 6, that is very easy.*1329

*Let us also change the limits.*1331

*If x = 1, then u will be equal to 5 because that is 11 - 6.*1334

*If x = 6, then u = 0.*1340

*This integral converts into the integral from 0 = u to u = 5 of 11.*1347

*I am just going to write the 11 on the outside.*1356

*1/u ^{1/3} du.*1360

*Now it is a little more obvious why this is an improper integral, because u = 0, you will get 0 in the denominator.*1363

*That is obviously an impropriety there.*1374

*Actually you might have been able to notice that from the original integral there.*1378

*When x = 6, you might plug in x = 6 to the denominator you get 0 in the denominator and that is clearly improper.*1380

*I am not going to worry about the 11 for the time being because it is not going to affect whether the thing converges or diverges.*1391

*What we are going to do is instead of 0, I am going to change the 0 to t.*1400

*We are going to integrate from t to 5, of 1 over, now u to the cube root = u ^{1/3} dU.*1407

*Then we take the limit of that as t goes to 0 from the positive side.*1420

*Now I can split this integral up from t to 1 of 1/u ^{1/3} du.*1428

*Plus the integral from 1 to 5 of 1/u ^{1/3} du.*1445

*Now the point about that is that this is not an improper integral.*1455

*This completely proper and it will not determine whether the whole thing converges or diverges.*1460

*This will give us a finite number.*1469

*This is the improper part right here.*1473

*What we have here is the integral of du/u ^{1/3}.*1476

*That is exactly the integral that we were looking at on the previous page.*1482

*I will just write it as 0 ^{1} for now, with a p value of 1/3.*1492

*On the previous page, we learned that the integral from 0 to 1 of dx over x ^{p},*1498

*If p < 1, converges.*1505

*That tells us that this whole integral converges, or at least this part of it converges.*1514

*Then, the second part with some finite number,*1523

*I have not really incorporated the 11 but we can have an 11 times the whole thing.*1531

*An 11 multiplied times a whole number is not going to make it more likely to go to infinity or less likely.*1536

*So, the whole thing goes to it converges to a finite number.*1545

*In this case, the problem only asked us whether the integral converges or diverges.*1558

*At this point, we are finished.*1562

*In fact, this integral is completely feasible.*1566

*You could if you wanted, go back and evaluate the integral.*1572

*What you should get is 3/2 × 5 ^{3/2}.*1584

*That is what you get if you plug in the whole integral and then you multiply the whole thing by 11.*1593

*That is what you get if you evaluate the whole integral.*1600

*The important thing here is that is a finite integral, so we would say the integral converges.*1603

*There is a very common sort of ambush in Calculus 2, where they will give you a sort of innocent-seeming problem.*1613

*It turns out what the problem has is hidden discontinuities.*1622

*Let me give you an example of that.*1628

*The integral from 1 to 4 of dx/x ^{2} - 5x + 6.*1630

*This is an example that is not that hard to integrate using the techniques we learned in earlier lectures.*1636

*In particular, if you use partial fractions.*1645

*If you use partial fractions on this integral, you get 1/x ^{2} - 5x + 6.*1656

*Well, remember what we learned in partial fractions was to factor the denominator.*1671

*So, 1/(x-2) × (x-3).*1675

*If you do the partial fractions work, I will not spell out the details now, but we learned how to do that in a previous lecture.*1683

*You get 1/(x - 3) - 1/(x - 2).*1688

*So it is very tempting to do that partial fractions work and then integrate it.*1697

*You integrate it and you get ln(x-3) - ln(x-2).*1703

*Then you think, OK, I can plug in my bounds, x=1 and x=4.*1713

*You can plug in those numbers just fine.*1717

*And, you get a nice numerical answer.*1721

*That is the temptation in this kind of problem.*1724

*That is exactly the trap that your Calculus 2 teacher is trying to make you fall into.*1734

*You work all the way through this problem and you get a nice numerical answer and you present it as your answer.*1741

*The fact is that that is flat wrong.*1747

*What is wrong with that?*1752

*Well, what is wrong with that is that if you look at the function we had to integrate there, 1/(x-2) × (x-3).*1754

*That blows up at x=2 and x=3.*1765

*Because if you plug in x=2 or x=3, you get 0 in the denominator and the thing explodes.*1779

*So, this integral, which looked very tame and safe and looked like a fairly easy Calculus 2 integral actually is an improper integral.*1786

*Since we are talking about it in the section in improper integrals.*1795

*Maybe it is obvious to look for that, but if this comes in a swarm of other integrals where you are using all kinds of other techniques,*1799

*It is very easy to overlook things like this.*1809

*Instead of just solving it, using the generic partial fractions idea,*1812

*What you have to do is you have to split this up into pieces at the discontinuities.*1819

*So, you split it up at the integral from 1 to 2 of dx over the denominator + the integral from 2 to 3 of dx over the denominator + the integral from 3 to 4 of dx over the denominator.*1828

*You split it up at these two places, at 2 and 3, because those are the two discontinuities.*1851

*In fact, this middle integral, is now discontinuous at both ends because 1 end is 2 and 1 end is 3.*1860

*This middle integral is discontinuous in two different place.*1868

*We are going to split that up from 2 to 2.5 of dx over the denominator + the integral from 2.5 to 3 of dx over the denominator.*1872

*So, then we look at these 4 different improper integrals.*1886

*Four improper integrals.*1899

*You solve all four of them.*1907

*If any one of them diverges.*1909

*Or if more than one diverges, then we say the whole original integral from 1 to 4 diverges.*1920

*Now, each one of these 4 integrals you could work on using partial fractions.*1939

*So, using that technique that I outlined up here,*1947

*Solve each one using partial fractions.*1953

*It turns out that all 4 of those integrals diverge.*1972

*So, we say the original integral from 1 to 4 diverges, as our answer.*1987

*This is really quite dangerous because this depends on your noticing that there are these hidden discontinuities between the bounds of integration.*2008

*In order not to fall into this trap, what you have to do is look at the thing being integrated and ask yourself when does that blow up.*2023

*In this case that blows up at x=2 and x=3.*2031

*2 and 3 would be in the bounds of integration.*2035

*They are between 1 and 4 and so that is why we have this problem.*2038

*If we were asked to integrate the same function from 4 to 6 of dx/x ^{2} - 5x + 6,*2044

*We would have no problems here and would not have to split it up.*2059

*It would not be improper.*2064

*We could go back and use this regular technique of partial fractions and just plug in the bounds and we would not have to worry about limits at all.*2067

*The reason there is that the place where it blows up, x=2 and x=3, is not in between 4 and 6.*2077

*There would be no discontinuities in our region of interactions.*2084

*That is a very dangerous kind of integral.*2088

*You kind of have to watch out for when you see a denominator*2090

*Ask yourself, where is the denominator 0, where does that make you function blow up.*2095

*If that is inside your region of integration, then you have to split up the integral and work on each part.*2100

*If any one of them diverges.*2107

*Then you say the whole integral diverges.*2109

*We will do some more examples later on.*2114

*This has been Will Murray for educator.com.*2116

1 answer

Last reply by: Dr. William Murray

Thu Nov 20, 2014 3:26 PM

Post by David Llewellyn on November 19, 2014

In lecture example 6, wouldn't you get a hint that there was something wrong when, when doing the initial partial fractions, you tried to evaluate ln(x-2) at x=1 or ln(-1)?

1 answer

Last reply by: Dr. William Murray

Sun Feb 23, 2014 8:17 AM

Post by Wick Cunningham on February 18, 2014

Hello. I am having trouble integrating from 0 to infinity of e^x/(1+e^x)

3 answers

Last reply by: Dr. William Murray

Mon Nov 26, 2012 6:06 PM

Post by yoon nam on October 9, 2012

for some reason the video keeps on stopping after the formal notation after horizontal example.... please fix it??