The bilateral Laplace transform of a function
$$f\left( t \right) = \left\{ {\matrix{ {1,} & {{\rm{if}}\,a \le t \le b} \cr 0 & {{\rm{otherwise}}} \cr } } \right.$$     is

The bilateral Laplace transform of a function
$$f\left( t \right) = \left\{ {\matrix{ {1,} & {{\rm{if}}\,a \le t \le b} \cr 0 & {{\rm{otherwise}}} \cr } } \right.$$     is Correct Answer $${{{e^{ - as}} - {e^{ - bs}}} \over s}$$

Related Questions

Laplace transform of the function f(t) is given by $${\text{F}}\left( {\text{s}} \right) = {\text{L}}\left\{ {{\text{f}}\left( {\text{t}} \right)} \right\} = \int_0^\infty {{\text{f}}\left( {\text{t}} \right){{\text{e}}^{ - {\text{st}}}}{\text{dt}}{\text{.}}} $$       Laplace transform of the function shown below is given by
Transform Theory mcq question image
Consider the 5 × 5 matrix \[{\text{A}} = \left[ {\begin{array}{*{20}{c}} 1&2&3&4&5 \\ 5&1&2&3&4 \\ 4&5&1&2&3 \\ 3&4&5&1&2 \\ 2&3&4&5&1 \end{array}} \right
Given that $$F\left( s \right)$$  is the one-side Laplace transform of $$f\left( t \right),$$  the Laplace transform of $$\int_0^t {f\left( \tau \right)d\tau } $$   is