Math Cracks – A Cool Approach to Integration by Parts


Introduction

The idea of integration by parts sounds quite scary for many Calculus students, and I think there is a good reason for that. First of all, integration by parts is a technique that involves two steps (or more) instead of one step as most students would like to. Students would want to APPLY some formula and get the answer right away, but in Calculus often times the answers come after a sequence (sometimes a long one) of steps.

First off, as a principle of the matter, one of the reasons integral Calculus is typically hard for students is the rather unfortunate notation used for integration. In fact, when calculating the indefinite integral of a function \(f\left( x \right)\), we face the following notation

\[\int{f\left( x \right)dx}\] What many students do not understand is what is really meant by the “\(dx\)” in the expression above. Clearly, there are historical reasons why the “\(dx\)” appears in the notation stated above. But, actually there is no reason to include \(dx\) or even to add \(f\left( x \right)\). When we want to compute the indefinite integral of a function \(f\), we should be able to simply write

\[\int{f}\] and that way we are stating the indefinite integral of the function \(f\).

Are these the same?

\[\int{f\left( x \right)dx}=\int{f\left( u \right)du}\]

Absolutely! That is why you see sometimes the integration variable (x or u respectively) referred as a “dummy” variable, because it does not really play any role in the integration process.

Integration By Parts as a Reverse Product Rule

After a brief introduction, now we cut to the chase. The typical integration by parts formula shown in textbooks is

\[\int{udv}=uv-\int{vdu} \,\,\,\,\,(1)\]

Then you say, “Huh? What is that?” Obviously, without giving a meaning to the above \(u\) and \(dv\), it is hard to see what is all about it. One question you can have is: Why is the integration by parts formula involving dv’s and du’s, if those don’t even play a role in the integration process, as shown in the introduction?

The answer is simple: In the context of the integration by parts formula above, \(du\) and \(dv\) are not “dummy variables”, but instead they are function. Mnemonically, the above is good to solve an integration by parts exercise, but it is no good to understand why it is actually true or why it works.

Enter the Product Rule:

The product rule says that:

\[\frac{d}{dx}\left( fg \right)=\frac{df}{dx}g+f\frac{dg}{dx} \,\,\,\,\,(2)\]

For short, I prefer to write

\[\left( fg \right)’=f’g+fg’ \,\,\,\,\,(3)\]

But WAIT! Aren’t we integrating in this article? Why do I bring up a differentiation rule?? Hum, wouldn’t it be great to have to product rule for integrals too? Wouldn’t it be great if \(\int{f’g’}=f\,g + C\)?? Unfortunately it is not, BUT there is still a product rule for integrals, only that it is slightly more complicated.

Let’s rearrange equation (3), we get:

\[fg’=\left( fg \right)’-f’g \,\,\,\,\,(4)\]

So then, if we integrate both sides of the equality above we get

\[\int{fg’}=\int{\left( \left( fg \right)’-f’g \right)}\]

which by linearity of integration leads to

\[\int{f\,g’}=f\,g-\int{f’g} \,\,\,\,\,(5)\]

And here my friends, you have your integration by part rule. Integration by parts should be seen as a cool integration tool that allows me to integrate the product of two functions. But it is a bit more restrictive, because it is the product of two functions BUT one of the functions must be a derivative of SOME function.

So, in order to fruitfully apply the integration by parts rule I need to have three things happening:

  • I’m trying to integrate the product of TWO functions.
  • One of those functions a derivative of something (so it is of the form \(g’\)).
  • I need to know how to calculate that something (I need to know who is \(g\))

If those three conditions happen, then I can use the integration by parts rule

REMEMBER: When using integration by parts you need to have the product of two functions, and one of those two functions needs to be the derivative of something that you know.

For example, let us see when you cannot apply integration by parts: Consider the following integral

\[\int{\sin \left( {{x}^{2}} \right){{e}^{{{x}^{2}}}}}\]

In this case, we are trying to integrate the product of two functions: \(\sin \left( {{x}^{2}} \right)\) and \({{e}^{{{x}^{2}}}}\), but do you know what the antiderivative of any of these two functions is? Or in other words, do you know which functions leads to any of \(\sin \left( {{x}^{2}} \right)\) or \({{e}^{{{x}^{2}}}}\) after differentiating? Well. No. Those two functions do not have elementary antiderivatives, so then integration by parts would not help in this case.

Now an example where integration by parts COULD be used:

\[\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx}\]

In this case we are trying to integrate the product of two functions: \({{x}^{2}}\) and \({{e}^{{{x}^{2}}}}\), and I know what the antiderivative of \({{x}^{2}}\) is. So I can use the rule. We have the following notation:

\[\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx}=\int{\underbrace{{{e}^{{{x}^{2}}}}}_{f\left( x \right)}\underbrace{{{x}^{2}}dx}_{g’\left( x \right)}}\]

So we have

\[\begin{aligned} & f\left( x \right)={{e}^{{{x}^{2}}}} \\ & g’\left( x \right)={{x}^{2}} \\ \end{aligned}\]

Differentiating \(f\) and integrating \(g’\) we get:

\[\begin{aligned} & f\left( x \right)=2x{{e}^{{{x}^{2}}}} \\ & g\left( x \right)=\frac{{{x}^{3}}}{3} \\ \end{aligned}\]

(Notice that the \(g\left( x \right)\) stated above is one possible antiderivative, but the rule is that I can choose ANY antiderivative, so I choose the simplest one). The integration by parts is

\[\int{f\,g’}=f\,g-\int{f’g}\]

so plugging the information we have, we get the following:

\[\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx\,}={{e}^{{{x}^{2}}}}\frac{{{x}^{3}}}{3}-\int{2x{{e}^{{{x}^{2}}}}\frac{{{x}^{3}}}{3}dx}\] \[\Rightarrow \,\,\,\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx\,}=\frac{{{e}^{{{x}^{2}}}}{{x}^{3}}}{3}-\frac{2}{3}\int{{{x}^{4}}{{e}^{{{x}^{2}}}}dx}\]

So I have used the integration by parts rule above, but actually, I landed into a harder integral to solve. This is, in order to solve \(\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx\,}\) we need to know first how to compute \(\int{{{x}^{4}}{{e}^{{{x}^{2}}}}dx}\) which is actually harder.

The moral of this story is that integration by parts is a sort of product rule for integrals, and you’re looking for a specific structure: it is the integral of the product of two functions, and one of those functions you need to know how to compute its antiderivative. If that is the case, you’re in business and you can apply the integration by parts rule.

BUT, as it could be seen in the previous example, the fact that you CAN use integration by parts DOES NOT mean that it will be useful every time.

Final Words:

How do we tie together the formula for Integration by Parts?

\[\int{f\,g’}=f\,g-\int{f’g}\]

of the “product rule for integrals” with

\[\int{udv}=uv-\int{vdu}\]

By setting

\[\begin{aligned} & u=f\left( x \right) \\ & dv=g’\left( x \right)dx \\ \end{aligned}\]

we get that \(v = g\left( x \right)\) and \(du = f’\left( x \right)dx\), which renders both equation equal.




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