0:00

Welcome to calculus. I'm professor Ghrist.

Â We're about to begin lecture 24 on partial fractions.

Â >> The previous three lessons have solved integrals by means of running

Â differentiation rules in reverse. We're going to change our perspective now

Â and give an introduction to methods based on algebraic simplification.

Â The broadest and more generally useful such method, is called the method of

Â partial fractions. >> In this lesson we'll look at certain

Â algebraic methods for simplifying integrands.

Â Indeed certain integrals respond well to a pre-processing step.

Â This is especially true for rational functions ratios of polynomials for

Â example, if you are asked to integrate 3x squared minus 5 over x minus 2.

Â Then, naturally, you would perform the polynomial division and rewrite this as

Â 3x plus 6 with a remainder of 7 over x minus 2.

Â And while the former integral seems intimidating, the latter integral is

Â simple 3x gives 3 halves x squared. 6 integrates to 6x and 7 over x minus 2

Â yields 7 log x minus 2 and then add the constant.

Â Now, this is a trivial example but what happens when you have a higher degree

Â denominator in your rational integrand. We're going to rely on a certain result

Â from algebra, which states the following. A rational function P of x over Q of x,

Â where the degree of the denominator is greater than the degree of the numerator

Â and for which the denominator has distinct real roots, r sub i.

Â And that is, we're looking at an integrand of the form of polynomial P

Â over x minus r1 times x minus r2, times etcetera, all the way up to x minus r sub

Â n. Where Q has n distinct real roots, then

Â this can be factored or decomposed as a sum of constants A sub i over quantity x

Â minus r sub i. This fact is something that we're going

Â to take for granted, we're not going to prove it but we're going to use it

Â extensively. This integral of P over Q seems very

Â difficult in general. However, the right hand side can easily

Â integrated since all of these a sub i's are constants, then we get simply

Â logarithms. Now, what do we do to come up with this

Â decomposition? How do we determine these constants A sub

Â i. This is sometimes called the method of

Â partial fractions. Let's look at an example and simplify 3x

Â minus 1 over x squared minus 5x minus 6. We can factor the denominator as quantity

Â x plus 1 times quantity x minus 6. By our algebraic fact, we know that this

Â must be A1 over x plus 1 plus A2 over x minus 6.

Â For simplicity, let's call these constants A and B instead.

Â And now to determine A and B, we're going to multiply out the right hand side, put

Â it over a common denominator. Where we'll obtain a times quantity x

Â minus 6 plus B times quantity x plus 1 in the numerator.

Â And now we're going to expand that multiplication out and collect terms so,

Â that the first degree coefficient is A plus B and the constant coefficient is B

Â times 1 minus 6 times A. Now, these must match up with the

Â coefficient of the numerator 3x minus 1, if we have two polynomials that are the

Â same, then all of the coefficients must match up.

Â And so we see we're reduced to two equations, in two unknowns mainly A plus

Â B equals 3 and B minus 6A equals negative 1.

Â Now, we can solve such a linear system of equations for A and B.

Â Might take a bit more work than I'm willing to put on this slide.

Â So, let me show you a more direct method for computing A and B.

Â If as before, we write out 3x minus 1 equals A times x minus 6, plus B times x

Â plus 1 then, instead of expanding and collecting, we can do the following.

Â We can say, this is a true statement and it must be true for all values of x.

Â Therefore, it must be true if x equals negative 1, and happens when we

Â substitute in on the left? We get negative 4, on the right we get A

Â times a negative 7 plus B times 0. And we have eliminated that variable from

Â consideration. Solving, we obtained A is four seventh.

Â Likewise, if we substitute x equals 6 into the above equation, we get on the

Â left hand side, 17. On the right hand side, A times 0 plus B

Â times 7 we therefore, have eliminated A and we can conclude that B is seventeen

Â sevenths. And I'll let you check that these two

Â satisfy the two equations in the lower left hand corner.

Â Now, lets put this method to work. Compute the integral x squared plus 2x

Â minus 1 over 2x cubed plus 3x squared minus two x.

Â 7:24

Factoring the denominator gives us x, quantity 2x minus 1, quantity x plus 2

Â all multiplied together. Now, we hope to split these up into A

Â over x plus B over 2x minus 1 plus C over x plus 2.

Â Let's color code these and then multiply everything out and see what we get.

Â A times 2x minus 1 times x plus 2, plus B times x times x plus 2, plus C times x

Â times 2x minus 1 must equal the numerator of the integrand x squared plus 2x minus

Â 1. Let's use our direct method of

Â substitution. First we'll substitute in x equals 0.

Â When we do that, we eliminate all of the terms except the one involving A.

Â We obtain negative 2A equals minus 1. Solving, we get a is a half.

Â Likewise, if we substitute in x equals one half, we eliminate all the terms

Â except for the B, which we solve to get B as one fifth.

Â What's our last root? If we substitute in x equals negative 2,

Â then we obtain a value for C of negative one tenth.

Â 9:03

Substituting these coefficients back in to the integrand gives us for our

Â integral, the integral of one half over x plus, one fifth over 2x minus 1, minus

Â one tenth over x plus 2. These are all going to be in logarithms.

Â We're going to get one half log of x, plus one tenth log of 2x minus 1 being

Â careful with that one half coefficient, minus one tenth log of x plus 2.

Â Let's put this to use in a differential equation.

Â Consider the deflection x, as a function of time t associated to a thin beam that

Â we've applied force to. Question is how does this deflection

Â evolve? Well, if the load is proportional to a

Â positive constant lambda squared, then one simple model for this beam would be

Â dx dt equal lambda x squared minus x cubed.

Â We can factor that as x times lambda minus x times lambda plus x.

Â And now applying the separation method, we get on the left dx over x times lambda

Â minus x times lambda plus x. On the right, dt.

Â Integrating both sides prompts the use of a partial fractions method.

Â I'll leave it to you, to show that the coefficients in front of the terms are

Â respectively, 1 over lambda squared, 1 over 2 lambda squared and negative 1 over

Â 2 lambda squared. So, that when we perform the integral, we

Â get a collection of natural logs with coefficients in front namely, log of x

Â minus one half log of lambda minus x, minus one half log of lambda plus x

Â equals lambda squared t plus a constant. Where I've moved the lambda squared's to

Â the right hand side for simplicity. Now, what we would do next is try to

Â solve for x as a function of t. This looks like it's going to involve a

Â lot of algebra. And I'm not really in the mood so, let's

Â use linearization and solve for the equilibrium to see if we can understand

Â what's happening. If we plot x dot versus x, we see that

Â there are three roots. One root at zero, one root at negative

Â lambda and one root at positive lambda. By looking at the sign of x dot or by

Â linearizing the right hand side of the equation, we can see that zero is going

Â to be an unstable equilibrium. That makes physical sense, if you're

Â compressing the beam if it were perfectly symmetric, then it would not deflect.

Â However, that situation is inherently unstable and if you start off with just a

Â little bit of deflection in either direction, you will quickly move to one

Â of the stable equilibria at lambda or negative lambda.

Â Depending on which side you buckle. Now, there are some complications

Â involved with applying the method of partial fractions.

Â We've assumed existence of real distinct roots.

Â That is, a bit of a luxury, in general you don't have that.

Â Sometimes you have repeated roots, something of the form of polynomial over

Â x minus root r to the nth power. This decomposes but it decomposes into

Â something a bit more complicated. There are n terms, each is a constant

Â over x minus r to some power ranging from one up to N.

Â Now all of these individual terms are Integrable, but some might take a little

Â more work than others. What is potentially worse is complex

Â roots, where you don't necessarily have real roots.

Â Let's say, we look at a polynomial over a quadratic, ax squared plus bx plus c.

Â Then if the discriminant b squared minus four ac is negative, we have complex

Â roots. This does break up, but it breaks up into

Â a linear polynomial in the numerator and then the quadratic in the denominator.

Â And if you have repeated complex routes, well then it gets even more difficult

Â still. All of these examples are doable, but

Â very cumbersome. And I'm not going to test you on these in

Â this course, in the end it's really just algebra.

Â 14:43

And in the end, it's really just roots, the roots control everything whether

Â they're real, whether they're complex, whether they're distinct or repeated.

Â As we saw in the example of the buckling beam, the roots have physical

Â interpretations in differential equation models.

Â Always pay attention to your roots, not just in calculus when trying to solve an

Â integral, but in differential equations or even later still when you take linear

Â algebra, you will see that the root influences everything.

Â >> We've seen in this lesson how algebraic methods can be incredibly

Â helpful. And incredibly complicated.

Â We're not going to descend any further into that underworld of integration

Â techniques. Rather, we're going to turn our gaze to a

Â new character in our story. That, of the definite integral.

Â