Double integral - What are the upper and lower bounds?

gtfitzpatrick
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Homework Statement
What are the upper and lower bounds?
Relevant Equations
see picture
241600

Cant get my head around the order of the upper and lower bounds for this, Is it always the higher take away the lower?
 
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I can't get my head around the problem statement -- please type
 
gtfitzpatrick said:
View attachment 241600
Cant get my head around the order of the upper and lower bounds for this, Is it always the higher take away the lower?
The upper limit is always the larger value (i.e., more positive or less negative. For the limits on the y integration, ##-e^x < -1##, so -1 is the less negative value, and you are correct to put it at the upper limit of integration.

Also, in the future, if you post an image, please take care to post it in the correct orientation. A lot of members here don't like to have to crane their heads 90° to be able to read an image.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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