Integration by Parts

At some point or another during your Calculus course you might have stumbled upon some very strange integrals with mixed terms of the form \int \! e^x ln x \, \mathrm{d}x or maybe even \int \! x^3 sin x \,\mathrm{d}x . Giving these a look, you can quickly conclude that ordinary elementary integration will not do since there is more than one function, and integration by sight is clearly out of the question since one term isn’t the derivative of the other. The next step to help us evaluate such integrals is by the incredibly useful, integration by parts.

This is presented in the following format for indefinite integrals:

\int \! u \, \mathrm{d}v = uv - \int \! v \,\mathrm{d}u

But for most, this is as far as it gets when it comes to the origin of this formula. There is no meaning or explanation to why or how it comes to be.

Let’s try and figure this out together step by step…

The first step in tackling such a problem is to first understand what you’re trying to find. In this case, we are interested in asking the question “What is the integral of a product?”. Although we would like to eventually come to answer this question by some means, you can recall that this sounds very similar to an equivalent question done for derivatives, i.e. What is the derivative of a product?”. But we already know what the answer to that question is! For derivatives, we can evaluate products using the product rule.

The product rule says that if we have 2 differentiable functions f(x) and g(x) over some domain, then

(f(x) g(x))'=f'(x)g(x)+f(x)g'(x) 

Given that integrals and derivatives are complements of one another, we can determine that this is a good place to start tackling the problem.
Using the product rule, we may try to convert it to integral form by taking the integrals on both sides to obtain:

\int \! (f(x) g(x))' \, \mathrm{d}x=\int \! f'(x)g(x) \, \mathrm{d}x+\int \! f(x)g'(x) \, \mathrm{d}x

Keeping in mind that the integral of a derivative of a function is the function itself i.e. \int \! f'(x)  \mathrm{d}x = f(x)  and that \mathrm{d}f = f'(x)  \mathrm{d}x , then we may simplify the above expression to be:

f(x) g(x)=\int \! g(x) \, \mathrm{d}f+\int \! f(x) \, \mathrm{d}g

Substituting f=u  and g=v ,

u v=\int \! v \, \mathrm{d}u+\int \! u \, \mathrm{d}v

and with a bit of playing round we obtain:

\int \! u \, \mathrm{d}v = uv - \int \! v \,\mathrm{d}u

TA-DA! Almost like magic we are able to handily find an equivalent version of the product rule used from derivatives, for integrals! Surprisingly, this is very useful to break down those very strange mixed integrals of the sort we had in the beginning.

If you are more interested in applying such a result with definite integrals, then you need not worry! The result can be achieved in exactly the same method, just slightly adjusted to give:

\int_a^b \! u \, \mathrm{d}v = uv\big|_a^b - \int_a^b \! v \,\mathrm{d}u

Next time we will be discussing some handy tips on how to better solve such problems analytically and even look at cases which don’t behave well when integrated by parts!

Thanks for reading!

-Sinthorel

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