Integration Problem: Starting the Integral

[ACE_99]

Homework Statement

I'm just having a bit of trouble with where to start on this integral.

$$\int[1 - (r/0.11)]^1^/^5rdr$$

The Attempt at a Solution

I've tried using integration by parts, "u" substitution and things like that but I don't seem to be getting anywhere with this. Any help would be greatly appreciated.

 

[Gib Z]

Which substitutions did you try? Its obvious the stuff under the radical is giving you trouble, so why not try substituting it away.

 

[ACE_99]

The first substitution that comes to mind is to let u = (1-(r/0.11)) so du = -1/0.11 dr. Rearranging for dr I get dr = -0.11du. If I sub this into the integral I still have an r in the equation.

From here I rearranged the expression u = (1-(r/0.11)) for r to get r = (1-u)(0.11). If I sub this into the integral I get

I = $$\int(u)^1^/^5(1-u)(0.11)(-0.11)du$$
= $$\int(u)^1^/^5(1-u)(-0.0121)du$$

It looks a bit better than before but I'm still stumped, hopefully the work I've done up to here is correct.

 

[Gib Z]

It is correct. Now just take the constant outside, and expand the brackets.

 

[ACE_99]

If I do that then the integral becomes

I = -0.0121$$\int (u^1^/^5 - u^6^/^5)$$
= -(0.0121)[(5/6)u^6/5 - (5/11)u^11/5]

So in order to get a solution I would also need to change the initial limits of integration using u = (1-(r/0.11)). If my initial limits of integration were 0 to 0.11 after subbing them into the equation for u my new limits of integration are from 1 to 0.

 

[Gib Z]

Yes that is correct. If you want to follow convention you'll have to introduce a negative factor and swap the new limits of integration to ensure the upper limit of integration is larger than the lower limit of integration.