Home : Resources : Integral Evaluation
Helps to know: Notes About Math

Ok, now its time to show that integral who's boss.  This section is a continuation of a discussion on solving integrals intuitively.  We want to evaluate the integral

Integral{ dL . x / (L^2 + x^2)^3/2 }

We know that the antiderivative should resemble

L / (x * (L^2 + x^2)^1/2 )

Why?  Well, we know that when we take the derivative of something to the -(1/2) power, we get something to the -(3/2) power.  The expression inside our integral is to the -(3/2) power, so its antiderivative should be to the -(1/2) power.   However, the antiderivative won't be this exactly but something similar.

Now, lets try some extreme conditions; this lets us see how to modify our integral.  Suppose we let x get really large (in the original function).  It will overpower L^2 and make it effectively zero.  As x gets infinitely large, L^2 doesn't matter at all.  Using these steps our integral becomes

Integral{dL . x / (x^2)^3/2  } =  (x/x^3) * Integral(dL) = L / x^2

(Remember that x is a just a constant [we are integrating L]
so we can put x outside the integral)

Thus, we want an x^2 on the bottom of the real antiderivative.  Looking at our test antiderivative, we see that we get only an x on the denominator when x overpowers L.  That means we need to add an x on the bottom of our test antiderivative.  We also need an L on the top, because it isn't in our test antiderivative.  Another reason is that our units would be unbalanced; we want units of (1/meter), just like the original.  We can't add an x on top because it would just divide out.  The only other thing wtih units of length that we could add is L.  Putting in these components, our test antiderivative becomes

L / (x * (L^2 + x^2)^1/2 )

Now for the checks.  When x gets huge, it reduces to   L/(x^2)  as we predicted.  What about units?  We have meters on top.  On the bottom we have (meters * meters)   [see the notes page for a more complete analysis of units].  This is (meters/meters^2) or (1/meter).  These are the same units as the function in the original integral so we are good to go.  If you take the derivative of this function you'll see that it works out to our original function, as we expect. 

This method of units is a easy way to see if you are on the right track, and can even help you solve the integral (esepcially in insane cases like these.  This antiderivative wasn't even in the back on my textbook).  Congratulations for making it this far, I hope you have learned something for your trouble :).

Send questions, comments, corrections, and suggestions to kazad@princeton.edu.
Last modified: 8/7/01