What's the significance of negative exponents in Laplace transformations

Integrate

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In the photo above why can't leave it as e^t(a-s)?

Why do we need to take out a negative to convert it to e^-t(s-a)
 
You can leave it as et(as)e^{t(a-s)}.

et(as)e^{t(a-s)} and et(sa)e^{-t(s-a)} are just the same thing, and will surely give you the same result.
 
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In the photo above why can't leave it as e^t(a-s)?

Why do we need to take out a negative to convert it to e^-t(s-a)
As MaxWong said, while solving the integral it does not matter which structure you use. But for convenience, it is better to make it look like e(sa)te^{-(s - a)t} for two reasons. The first reason is that the transformation answer will involve the expression (sa)(s - a) rather than (as)(a - s). The second reason is that it is some kind of a reminder to tell you that s>as > a.

Another way to see this is to look at this function, eate^{-at}. The convenient structure will be e(s+a)te^{-(s + a)t} because the answer is 1s+a\displaystyle \frac{1}{s + a} which involves the expression (s+a)(s + a).
 
As MaxWong said, while solving the integral it does not matter which structure you use. But for convenience, it is better to make it look like e(sa)te^{-(s - a)t} for two reasons. The first reason is that the transformation answer will involve the expression (sa)(s - a) rather than (as)(a - s). The second reason is that it is some kind of a reminder to tell you that s>as > a.

Another way to see this is to look at this function, eate^{-at}. The convenient structure will be e(s+a)te^{-(s + a)t} because the answer is 1s+a\displaystyle \frac{1}{s + a} which involves the expression (s+a)(s + a).
This was a very helpful answer. Thank you.
 
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