Why Integrate from 0 to I for Inductor Energy Calculation

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why they integrate in from 0 till I in order to find the energy on the inductor

http://i27.tinypic.com/11rfbyf.jpg
 
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I'd think because as you turn on the circuit, the current through the conductor rises from 0 until it reaches the set current I, so the limits are from 0 to I.
 
look in this example
http://i28.tinypic.com/2h3ykxx.jpg
here they pick intervals as time (not current like in the first example)
why??
 
nabliat said:
look in this example
http://i28.tinypic.com/2h3ykxx.jpg
here they pick intervals as time (not current like in the first example)
why??

because they are integrating with respect to time, so the limits must be associated with time. If they were integrating with respect to I, then they'd need limits associated with the current.
 
why in the first case they integrate with respect to current
and in the second with respect to time

those cases should be calculated in the same way
??
 
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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