This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 118, LECTURE 2: Integration by Parts 1 Integration By Parts Consider trying to take the integral of integraldisplay xe x dx. We could try to find a substitution but would quickly grow frustrated — there is no substitution we can make which would simplify the integral. Instead we can rely on information we know for the product rule for differ entiation. If we let u and v be functions of x , we know from the product rule for differentiation that d dx [ uv ] = du dx v + u dv dx . We want to use this to tell us something about integration, not differentia tion, but what can we do? Well, let’s just integrate over x and see what we get. Since integration undoes differentiation, we obtain uv = integraldisplay du dx v dx + integraldisplay u dv dx dx. We can now simplify the differentials ( dx ) and rearrange to get integraldisplay u dv = uv integraldisplay v du. (1) This is the Integration by Parts formula and it is one of the primary tools for integration. It can also be stated for definite integrals as integraldisplay b a u dv = [ uv ] b a integraldisplay b a v du. (2) It is probably not immediately obvious how this helps us, since we still have to evaluate the integral on the righthand side. The trick is that, if we make appropriate choices for the parts of the original integral — u and dv — it may turn out upon differentiating u and integrating dv to get du and 1 v , respectively, that the integral on the righthand side is easier to evaluate than the one the left. So, ideally, we replace the original integral with a simpler one! Let’s see if the above integral can be solved in this matter. Our first task is to select u and dv from the expression xe x dx . One of these quantities will be differentiated, and one integrated, in order to form our new quantity to be integrated. It turns out the appropriate choice for this problem is u = x dv = e x dx du = dx v = e x . Now using formula (1) we have integraldisplay xe x dx = xe x integraldisplay e x dx....
View
Full
Document
This note was uploaded on 04/14/2009 for the course ENGINEERIN MATH 118 taught by Professor Soares during the Spring '09 term at Waterloo.
 Spring '09
 Soares

Click to edit the document details