Sign Error in Double Integral Word Problem

burt

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Aug 1, 2019
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I am working on the following problem:
1588369238279.png
I worked through the problem, but I seem to have made a sign error along the way. From what I can see, it seems that if I did make a sign error it was as I completed the integration with respect to x. But, I can't figure out what I did wrong! Any ideas?
Here is my work:

1588369507781.png
Thank you!
 
I did \(u=x^2\) and \(\frac{du}{dx}=2x\) so I ended up with \(\int \frac{e^{-u}}{2}du\)
But I do see my sign error :) - the solution to that integral is negative, not positive.
 
I did \(u=x^2\) and \(\frac{du}{dx}=2x\) so I ended up with \(\int \frac{e^{-u}}{2}du\)
But I do see my sign error :) - the solution to that integral is negative, not positive.
You need to replace everything with x's into u's. You can have nothing left over!
[math]\int e^{-x^2}dx[/math] u=x^2 and du = 2xdx or du/(2x) = dx.[/math] Then [math]\int e^{-x^2}dx[/math] becomes [math]\int e^{-u} \frac {du}{2x} =\int e^{-u} \frac {du}{2\sqrt{u}}[/math]
You lost the x in the denominator!!!!!
 
I did \(u=x^2\) and \(\frac{du}{dx}=2x\) so I ended up with \(\int \frac{e^{-u}}{2}du\)
But I do see my sign error :) - the solution to that integral is negative, not positive.
Yes, du/dx = 2x. So how does this change the integral when you make your substitution???
 
Yes, du/dx = 2x. So how does this change the integral when you make your substitution???
yes, but there is an x in the problem - so i had \(\frac{du}{2}=xdx\) and I took out both the x and the dx.
 
There was? If that is the case can you please state the correct integral, the one with the x in it.
 
Oh sorry! Now I see that x in the original integral. Sorry about all this!
 
Again I apologize for this delay. You did lose a negative sign. Can you try again and see if you get the correct results?
 
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