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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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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

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# Homework Help: Simple yet confusing situation. Integral.

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