Absolute value in exponential, signal energy?

Question

How can this give this result? Isn't the absolute of $(e^(-2*t))$ always 1?

Answer 1

No. The absolute value of $e$ to a purely imaginary power has absolute value one; i.e.

$$ |e^{iy}| = 1 $$

for any real $y$, but for a general complex number $z = x + iy$,

$$ |e^z| = |e^{x + iy}| = |e^xe^{iy}| = |e^x||e^{iy}| = |e^x| $$

In particular, $e^x$ for real $x$ takes values in $(0,\infty)$, so $|e^x| = e^x$, but it certainly isn't $1$ (unless $x = 0$). Other values: $e^{-1} \simeq 0.368, e^1 \simeq 2.72, e^2 \simeq 7.39$, etc.

The antiderivative of $e^{cx}$ is $\dfrac{1}{c}e^{cx}$, this is how you would go about evaluating the given integral.