# Homework Help: Simple yet confusing situation. Integral.

1. Jan 28, 2006

### Robokapp

Okay. basic concept. U-substitution. For an integral of an expressino that you can't "guess" or find by conventional means you use a u-substitution. To integrate correctly if you have bounds you either rewrite the finction with x after integration or you convert the bounds by plugging them back in. I'm sure you knew all this...i only try to expalin why i think about this.

Let's say we integrate y=x^2 from -1 to 1
Pretty simple...(1/3)x^3 and insert the bounds.
however...something that i encountered in a homeowrk made me wonder...what if i use a u-substitution on this fudging thing with equal yet opposite bounds?

So...i tried to integrate using the follwoing: u= x^2

so u=x^2
du/dx=2x
du=2x dx
dx= du / 2x
and since u=x^2 => x= sqrt (u)
so dx = du / 2 sqrt (u)

now my integral is int( u / 2 sqrt (u) * du)and by plugging in my bounds, it is defined from 1 to 1.
here's the problem...ignoring the whole pointless substitution and everything, my bounds are from 1 to 1!!! that gives me an answer of zero. but why? The answer is not zero. The crappy thing is perfectly continuous from 1 to 1...how could it not be so what's wrong? did i find a hole in the u-substitution?

---

if you have no idea what i just said...my question is what do you do when math causes because of your u-substitution for your boundaries to become the exact same thing? it's not homework...it's just a curiosity.

thank you for your time. i know i'm writing ponitless stuff.

~Robokapp

2. Jan 29, 2006

### Tide

You're being sloppy. Your transformation maps the negative reals onto the positive reals so you need to apply transformations to both domains by splitting the integral into two parts and paying close attention to the signs in each of them.