View Full Version : Tricky Integral

ninguen

02-25-2012, 05:32 PM

Hi Guys,

First of all, thanks for your help. This site is awesome and I don't know what I'd do without you.

Now, on to the ugliness.

I have the following integral to solve: squiggle (x^3 - 4x^2 - 4x)^(1/2) dx

I've tried U substitution, but no dice. Too ugly.

As usual... Mathway says it can't be done. Schaum's says it can. The battle is on!

Chapter 30, This is problem 14.

I have the following integral to solve: squiggle (x^3 - 4x^2 - 4x)^(1/2) dx

Chapter 30, This is problem 14.

That does not agree with my edition.

It appears that this is not a real integral.

galactus

02-25-2012, 06:03 PM

Are you sure it is \displaystyle \int \sqrt{x^{3}-4x^{2}-4x}dx?.

If it were \displaystyle \int \sqrt{x^{3}-4x^{2}+4x}dx, then we could work with it.

But, as is, it is a nasty Elliptic Integral and beyond regular calc methods.

If it were the latter, then we could write \int \sqrt{x(x-2)^{2}}dx

ninguen

02-25-2012, 06:37 PM

That does not agree with my edition.

It appears that this is not a real integral.

Consider the region R bounded by parabola y=(8x)^(1/2) and the line x = 2.

a) find the solid generated by revolving R around the y axis.

b) find the solid generated by revolving R around the line x = 2.

Maybe I'm screwing it up. I was doing part B.

Schaum's Calculus. Chapter 30. Question 14.

ninguen

02-25-2012, 06:53 PM

Are you sure it is \displaystyle \int \sqrt{x^{3}-4x^{2}-4x}dx?.

If it were \displaystyle \int \sqrt{x^{3}-4x^{2}+4x}dx, then we could work with it.

But, as is, it is a nasty Elliptic Integral and beyond regular calc methods.

If it were the latter, then we could write \int \sqrt{x(x-2)^{2}}dx

I think you're right. \displaystyle \int \sqrt{x^{3}-4x^{2}+4x}dx

I can take it from here.

Subhotosh Khan

02-25-2012, 07:02 PM

Calculate the volume by disc method - the integration will be much simpler.

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.