How to integrate $e^{t-u}du$

Question

I'm not quite sure how to integrate

$$\int e^{t-u}\,du$$

What particular rule do I use here?

Thanks

Answer 1

Change variable: $t-u=r$, $du=-dr$, from which you have $$ \int e^{t-u}\,du=-\int e^r\,dr=-e^r+C=-e^{t-u}+C\;. $$

Answer 2

$$ \int e^{t-u}\,du = e^t \int e^{-u}\,du = \cdots $$

Answer 3

Or you can note that, since $e^t$ is a effectively a constant, the variable of integration being $u$, we have

$$\int e^{t - u} \,du = e^t \int e^{-u}\,du = e^t (-e^{-u}) = -e^{t - u}. \tag{1}$$

The particular rules used are: i.) $\int (cf(u))\,du = c\int f(u)\,du$; and ii.) $\int e^{au}\,du = a^{-1}e^{au}$ for $a \ne 0$. There may be others.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!