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Professor Murray

Professor Murray

Improper Integration

Slide Duration:

Table of Contents

I. Advanced Integration Techniques
Integration by Parts

24m 52s

Important Equation
Where It Comes From (Product Rule)
Why Use It?
Lecture Example 1
Lecture Example 2
Shortcut: Tabular Integration
Lecture Example 3
Mnemonic: LIATE
Ln, Inverse, Algebra, Trigonometry, e
Additional Example 4
Additional Example 5
Integration of Trigonometric Functions

25m 30s

Important Equation
Powers (Odd and Even)
What To Do
Lecture Example 1
Lecture Example 2
Half-Angle Formulas
Both Powers Even
Lecture Example 3
Lecture Example 4
Additional Example 5
Additional Example 6
Trigonometric Substitutions

30m 9s

Important Equations
How They Work
Remember: du and dx
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Partial Fractions

41m 22s

Why Use It?
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Integration Tables


Using Tables
Match Exactly
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule

22m 36s

Trapezoidal Rule
Graphical Representation
How They Work
Why a Trapezoid?
Lecture Example 1
Midpoint Rule
Why Midpoints?
Lecture Example 2
Left/Right Endpoint Rule
Left Endpoint
Right Endpoint
Lecture Example 3
Additional Example 4
Additional Example 5
Simpson's Rule

21m 8s

Important Equation
Estimating Area
Difference from Previous Methods
General Principle
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Improper Integration

44m 18s

Horizontal and Vertical Asymptotes
Example: Horizontal
Formal Notation
Example: Vertical
Formal Notation
Lecture Example 1
Lecture Example 2
Lecture Example 3
Lecture Example 4
Formulas to Remember
Improper Integrals
Lecture Example 5
Lecture Example 6 (Hidden Discontinuities)
Additional Example 7
Additional Example 8
II. Applications of Integrals, part 2

23m 20s

Important Equation
Why It Works
Common Mistake
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Surface Area of Revolution

28m 53s

Important Equation
Surface Area
Relation to Arclength
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Hydrostatic Pressure

24m 37s

Important Equation
Main Idea
Different Forces
Weight Density Constant
Variables (Depth and Width)
Lecture Example 1
Additional Example 2
Additional Example 3
Center of Mass

25m 39s

Important Equation
Main Idea
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
III. Parametric Functions
Parametric Curves

22m 26s

Important Equations
Slope of Tangent Line
Arc length
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Polar Coordinates

30m 59s

Important Equations
Polar Coordinates in Calculus
Arc length
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
IV. Sequences and Series

31m 13s

Definition and Theorem
Monotonically Increasing
Monotonically Decreasing
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5

31m 46s

Important Definitions
Sigma Notation
Sequence of Partial Sums
Converging to a Limit
Diverging to Infinite
Geometric Series
Common Ratio
Sum of a Geometric Series
Test for Divergence
Not for Convergence
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Integral Test

23m 26s

Important Theorem and Definition
Three Conditions
Converging and Diverging
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Comparison Test

22m 44s

Important Tests
Comparison Test
Limit Comparison Test
Lecture Example 1
Lecture Example 2
Lecture Example 3
Lecture Example 4
Additional Example 5
Additional Example 6
Alternating Series

25m 26s

Main Theorems
Alternation Series Test (Leibniz)
How It Works
Two Conditions
Never Use for Divergence
Estimates of Sums
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Ratio Test and Root Test

33m 27s

Theorems and Definitions
Two Common Questions
Absolutely Convergent
Conditionally Convergent
Missing Case
Ratio Test
Root Test
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
Power Series

38m 36s

Main Definitions and Pattern
What Is The Point
Radius of Convergence Pattern
Interval of Convergence
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
V. Taylor and Maclaurin Series
Taylor Series and Maclaurin Series

30m 18s

Taylor and Maclaurin Series
Taylor Series
Maclaurin Series
Taylor Polynomial
Lecture Example 1
Lecture Example 2
Lecture Example 3
Lecture Example 4
Additional Example 5
Additional Example 6
Taylor Polynomial Applications

50m 50s

Main Formulas
Alternating Series Error Bound
Taylor's Remainder Theorem
Lecture Example 1
Lecture Example 2
Lecture Example 3
Additional Example 4
Additional Example 5
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For more information, please see full course syllabus of College Calculus: Level II
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Lecture Comments (8)

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??

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
  • 1k [3/x]dx
  • Integrate
  • 1k [3/x]dx = 3[ln|x|]1k
  • = 3lnk − 3ln1
= 3lnk = limk → ∞ 3lnk = ∞
− ∞0 [1/(( x + 5 )2)]dx
  • Replace the improperbound ∞ with k
  • k0 [1/(( x + 5 )2)]dx
  • Integrate
  • k0 [1/(( x + 5 )2)]dx = [ − [1/(x + 5)] ]k0.
  • = − [1/(0 + 5)] + [1/(k + 5)]
= limx → − ∞ [ [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∫0k [7/((x − 3)2)]dx
  • Integrate
  • 2∫0k [7/((x − 3)2)]dx = 14∫0k [1/((x − 3)2)]dx
  • = 14[ − [1/(x − 3)] ]0k
  • = 14[ − [1/(k − 3)] + [1/(0 − 3)] ]
  • = 14[ − [1/(k − 3)] − [1/3] ]
  • = 14[ − limk → ∞ [1/(k − 3)] − [1/3] ]
  • = 14[ 0 − [1/3] ]
= [( − 14)/3]
37− [4/(√{x − 3} )]dx
  • Replace the bound 3 with k
  • 37− [4/(√{x − 3} )]dx = ∫k7− [4/(√{x − 3} )]dx
  • Integrate
  • k7− [4/(√{x − 3} )]dx = − 4∫k7 [1/(√{x − 3} )]dx
  • = − 4[ 2√{x − 3} ]k7
  • = − 4[2√{7 − 3} − 2√{k − 3} ]
  • = − 4[2√4 − 2√{k − 3} ]
= − 4[4 − 0] = − 4[4] = − 16
− 2k [2/(x√{x2 − 1} )]dx
  • Replace the bound 1 with k
  • − 2k [2/(x√{x2 − 1} )]dx = ∫ − 2k [2/(x√{x2 − 1} )]dx
  • Integrate using inverse trigonmetric identity
  • − 2k [2/(x√{x2 − 1} )]dx = 2∫ − 2k [1/(x√{x2 − 1} )]dx
  • = 2[ sec − 1|x| ] − 2k
  • = 2[ sec − 1|k| − sec − 1| − 2| ]
= 2[ limk1 sec − 1|k| − [(π)/3] ] = 2[ 0 − [(π)/3] ] = − [(2π)/3]
0 [(x − 4)/(x2 − 3x − 4)]dx
  • Factor x2 − 3x − 4 to simplify
  • 0 [(x − 4)/(x2 − 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 = ∫0k [1/((x + 1))]dx
  • Integrate
  • 0k [1/((x + 1))]dx = [ ln|x + 1| ]0k
  • = ln|k + 1| − ln|0 + 1|
  • = ln|k + 1| − ln|1|
  • = ln|k + 1| − 0
  • = ln|k + 1|
  • = limk → ∞ ln|k + 1|
= ∞
− ∞1 [2/(x2 + 2x)]dx
  • Replace the bound − ∞ with k
  • − ∞1 [2/(x2 + 2x)]dx = ∫k1 [2/(x2 + 2x)]dx
  • Integrate
  • k1 [2/(x2 + 2x)]dx = ∫k1 [1/x] − [1/(x + 2)]dx
  • = ∫k1 [1/x] − ∫k1 [1/(x + 2)]dx
  • = [ ln|x| ]k1 − [ ln|x + 2| ]k1
  • = ln1 − lnk − ln3 + ln|k + 2|
= limk → − ∞ [ − 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 = ∫0k [4/((x + 1)(x + 2))] dx
  • Integrate
  • 0k [4/((x + 1)(x + 2))] dx = ∫0k [4/(x + 1)] − [4/(x + 2)] dx
  • = 4∫0k [1/(x + 1)] − [1/(x + 2)] dx
  • = 4[ ln|x + 1| − ln|x + 2| ]0k
= 4[ limk → ∞ ln[(k + 1)/(k + 2)] − ln[1/2] ] = 4[ 0 − ln[1/2] ] = − 4ln[1/2]
− 22 [1/(√{4 − x2} )]dx
  • Note the function is symmetrical and simplify
  • − 22 [1/(√{4 − x2} )]dx = 2∫02 [1/(√{4 − x2} )]dx
  • Replace the bound 2 with k
  • 2∫02 [1/(√{4 − x2} )]dx = 2∫0k [1/(√{4 − x2} )]dx
  • Apply the inverse trignometric integration identity to integrate
  • 2∫0k [1/(√{4 − x2} )]dx = 2[ sin − 1[x/2] ]0k
  • = 2[ sin − 1[k/2] − sin − 1[0/2] ]
  • = 2[ sin − 1[k/2] ]
  • = 2[ limk → 2 sin − 1[k/2] ]
  • = 2[ [(π)/2] ]
= π
10.∫0 [1/(( x − 2 )3)] dx
  • Replace the bound ∞ with k
  • 0 [1/(( x − 2 )3)] dx = ∫0k [1/(( x − 2 )3)] dx
  • Integrate
  • 0k [1/(( x − 2 )3)] dx = [ − [1/(2(x − 2)2)] ]0k
  • = [ − limk → ∞ [1/(2(k − 2)2)] + [1/(2(0 − 2)2)] ]
  • = 0 + [1/2(4)]
= [1/8]

*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.


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

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 ex.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 - 8e0.0073

Sorry, subtracting a negative sign, so + 8e0.0088

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

So this is minus 8e-infinity + 8e0 = 1.0101

Well e-infinity, this is 1/einfinity, 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 integration0444

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 here0589

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 cancel0852

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 x2 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 = x2, 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 make1000

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/(xp).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/(xp), 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/xp 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/np.1210

We will convert that into the integral of 1/xp 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)x1/3.1297

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/u1/3 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 = u1/3 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/u1/3 du.1428

Plus the integral from 1 to 5 of 1/u1/3 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/u1/3.1476

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

I will just write it as 01 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 xp,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 × 53/2.1584

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/x2 - 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/x2 - 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/x2 - 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 denominator2090

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


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