Why the at term on the exponential turned positive

Firepanda
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Firstly, I don't get why the at term on the exponential turned positive (red arrow).. can someone explain that please?

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And how do I start on this? How do I split it up such that I can do it for t>0 and t=<0?

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Do I just integrate e^2t between -inf and 0 and integrate e^-t between 0 and inf?


Thanks!
 
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What is |t| when t is negative?
 
dx said:
What is |t| when t is negative?

positive, ah I see now, thanks

still stuck on 2nd though

edit: actually why would that change the sign of the a?
 
Just split the integral into two parts, one on (-∞,0) and the other on (0,∞).
 
edit: actually why would that change the sign of the a?

You should really draw f(t)=e^{|t|}. And then give a function that represents f(t) in the first quadrant and f(t) in the second quadrant.
 
|t| = -t by definition when t is negative.
 
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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