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.
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