Calculus topics from before MA2051 that you'll need - integration by parts and partial fraction decomposition

Aside from remembering how to do various derivatives and integrals, in MA2051 there are some calculus topics you need to remember from earlier classes. The two that seem to give people the most grief are integration by parts and partial fraction decomposition.

Integration by parts: This is a method you need to use when you have two terms multiplied together and one is not the derivative of the other (if it were, you could use U substitution). An example is ∫xsin(x)dx.

The formula for this situation is: ∫udv = uv - ∫vdu. So you have to assign one term in the original integral to be u, and another to be dv. You'll have to then derive to find du and integrate to find v, then plug into the formula.

The piece of advice I always give is this: Pick u to be something that gets simpler when you take the derivative. For example, x becomes 1. A counterexample, sin(x) just becomes cos(x) - no easier to work with.

Let's do out an example problem - the one mentioned before, ∫xsin(x)dx. We can see, using the tip given above, that x should be chosen for u, leaving sin(x) for dv.

u = x, dv = sin(x)dx
du = 1dx, v = -cos(x)

Subbing into the formula:

∫xsin(x)dx = -xcos(x) + ∫cos(x)dx

Now there's a new integral to solve, but it's an easy one: cosine.

∫xsin(x)dx = -xcos(x) + sin(x) + C

And you're done!

Now partial fraction decomposition. You'll be doing a LOT of these when you get to Laplace transforms!

The basic idea is: sometimes you'll get a big fraction you want to integrate, such as 6/(x² + 2x - 8). How do you integrate that? You don't know how to, but what you can do is split it up into two different fractions, each of which you can integrate. And that's partial fraction decomposition.

Start by factoring the denominator: x² + 2x - 8 factors into (x+4)(x-2).

Now you set up an equality. Set the fraction equal to the sum of two fractions, each of which has just one of the terms in the denominator and a variable in the numerator. It will be these variables that you will have to find:

6/((x+4)(x-2)) = A/(x+4) + B/(x-2)

Multiply both sides by the lefthand denominator. Some cancellation will occur, and you'll have:

6 = A(x-2) + B(x+4)

Distribute:

6 = Ax - 2A + Bx + 4B

You can actually find both A and B from this equation. What, you say? Two unknowns and only one equation? It's true! Just requires a bit of ingenuity. Notice something: on the lefthand side of the equation, there is only a constant term. There are no terms containing x. Doesn't this mean that all the terms containing x on the righthand side of the equation must have to cancel out? Yes, it does! So, you can conclude:

A + B = 0, because Ax and Bx have to cancel each other out on the righthand side. If they didn't, there would be some x term on the lefthand side too!

Furthermore, you can now see that the two constant terms on the righthand side will have to add up to four, since that's what exists on the lefthand side. So:

-2A + 4B = 6

And voila, out of thin air you now have two equations for your two variables, a more familiar system of equations situation. Substitute:

A + B = 0
A = -B

-2A + 4B = 6
2B + 4B = 6
6B = 6
B = 1
A = -B
A = -1

And, going back to the original definition you wrote, you can now sub in for A and B:

6/((x+4)(x-2)) = A/(x+4) + B/(x-2)
6/((x+4)(x-2)) = -1/(x+4) + 1/(x-2)

Now instead of integrating the ugly fraction on the left side, you can integrate the two nice fractions on the right side.

∫6/((x+4)(x-2)) = ∫-1/(x+4)dx + ∫1/(x-2)dx
∫6/((x+4)(x-2)) = -ln(x+4) + ln(x-2) + C

And that's partial fraction decomposition! The rules change somewhat when the terms in the denominator are more complex than the binomials I used here, but that doesn't happen often, and when it does, you can consult the table on page 263 of the Farlow textbook for the (minor) changes that need to be made.

Happy calculating,
Rachel