Intuition behind Integration by parts

Polymath89

I have some problems understanding the intuition behind the integration by parts technique. I don't quite see why you solve for [itex]\int u(x)v\prime (x)[/itex], instead of one of the other parts, what makes it easier to solve for that particular term?

And in general when working with integration techniques, does an expression like du have any mathematical meaning by itself or are those substitutions just used to make integration easier?

Mark44

Integration by parts is essentially the integral counterpart of the product rule in differentiation.

If f(x) = u(x) * v(x), then f'(x) = u(x) * v'(x) + u'(x) * v(x)

If we integrate the equation above, we get
∫f'(x)dx = ∫(u(x) * v'(x) + u'(x) * v(x))dx
= ∫u(x) * v'(x) dx + ∫u'(x) * v(x)dx

Simplifying a bit, we get
f(x) = ∫u(x) * v'(x) dx + ∫u'(x) * v(x)dx

or
∫u(x) * v'(x) dx = f(x) - ∫u'(x) * v(x)dx = u(x)*v(x) - ∫u'(x) * v(x)dx

Notice that I solved for ∫u(x) * v'(x) dx. I could just as easily have solved for the other integral. So in answer to your question, it doesn't make any difference. The only criterion is that ∫u'(x) * v(x)dx should be easier to work with than the other integral.


Yes, it has meaning. When you use a substitution, say u = 3x², then du is the differential of u. In this case, du = d/dx(3x²) * dx = 6x*dx.

1MileCrash

In my experience, choose to be u the function with the simplest derivative. If you can choose a function like "2x" - make it u because when taking the integral of vdu, you want du to contribute little complexity to the integral.

As for your comment about the mathematical meaning of du, its understandable to have the feeling that it has no mathematical meaning because most people don't see the meaning until they are done learning calculus and are instead applying it.

It is a differential element of u. What does that mean? I think of it as "a piece of u so small that it can barely be said to exist at all."

Look at it this way - the derivative notation, dy/dx, is a slope. The reason why we are saying "differential element of y over a differential element of x" is because even though the rate of change of a function may not be constant, the differential elements are so small that we can take the rate of change "over the course of a differential element" as being constant because there is no room for variation. That's why for a function like x^2 it makes sense to say (dy/dx) = 6 at some point even though the rate of change of the function is continuously changing.

Polymath89

Thats was pretty helpful, thanks a lot guys.