Thread: Definite Integration - Double integration by parts

I need to find the integral from 0 to 1 of the function below. I have tried using integration by parts and then another integration by parts with a substitution, but I am unable to come up with the correct answer.

x[(1-x)^(1/7)] dx

Sorry I cannot make it easier to read.

Originally Posted by cheme

I need to find the integral from 0 to 1 of the function below. I have tried using integration by parts and then another integration by parts with a substitution, but I am unable to come up with the correct answer.

x[(1-x)^(1/7)] dx

You could use integration by parts, but it's probably easier just to make the substitution u = 1 – x.

How would substitution work? When I find du from u I will get -dx. If I don't get something like xdx, how would that take care of the x in front of the (1-x)^(1/7)

If u= 1- x, then x+ u= 1 and so x= 1- u, of course.

Also, the limits of integration get changed, going from 1 to 0 instead of 0 to 1. When you flip them back again, that introduces a minus sign, which cancels with the one in –dx.

Just to make I get this correct...

Substitution would leave me with: -(1-u)[1-(1-u)]^(1/7) du
Simplification and switching the limits of integration will give: (1-u)(u)^(1/7) du

Originally Posted by cheme

Just to make I get this correct...

Substitution would leave me with: -(1-u)[1-(1-u)]^(1/7) du
Simplification and switching the limits of integration will give: (1-u)(u)^(1/7) du

Yes, and (1- u)u^{1/7}= u^{1/7}- u^{8/7}.

Originally Posted by HallsofIvy

Yes, and (1- u)u^{1/7}= u^{1/7}- u^{8/7}.

Thank you. I think I will try substituting whenever I can from now on rather than jumping into integration by parts.

Originally Posted by cheme

I think I will try substituting whenever I can from now on rather than jumping into integration by parts.

In an integration by parts, you have a product of two functions, one of which is going to get integrated and the other one differentiated (in the hope that this will simplify the original integral). In general, a function becomes simpler when differentiated, but more complicated when integrated. So in an integration by parts, it's convenient if the function that gets differentiated is one that becomes very much simpler when differentiated. Typical such functions are and . If you see one of these in an integral, then it's worth seeing whether integration by parts will pay off. Otherwise, substitution is likely to be a more useful technique.

Substitution is often unpopular with students, because it's not an automatic process: you have to think about what substitution to make, and that's a skill that only comes with practice. But it's worth persevering with it, because it is definitely the most effective technique for the majority of integrals.