$${{a - b} over s}$$
$${{{e^z}left( {a - b} ight)} over s}$$
$${{{e^{ - as}} - {e^{ - bs}}} over s}$$
$${{{e^{ - left( {a - b} ight)}}} over s}$$
Answer is Right!
Answer is Wrong!
The correct answer is $\boxed{{e^{ – \left( {a – b} \right)}}} \over s}$.
The bilateral Laplace transform of a function $f(t)$ is defined as
$$Lf(t): s = \int_0^\infty f(t) e^{-st} dt$$
In this case, $f(t)$ is a step function that is $1$ for $a \le t \le b$ and $0$ otherwise. The Laplace transform of a step function is
$$L\delta(t-a): s = {1 \over s} e^{-as}$$
where $\delta(t)$ is the Dirac delta function. The Laplace transform of a product of two functions is the convolution of their Laplace transforms. In this case, the Laplace transform of $f(t)$ is
$$Lf(t): s = \int_a^b {1 \over s} e^{-st} dt = {e^{-as} – e^{-bs} \over s}$$
Therefore, the bilateral Laplace transform of $f(t)$ is
$$Lf(t): s = {e^{-\left( {a – b} \right)}} \over s}$$