If $f\left( y \right) = {e^y},$ $g\left( y \right) = y;y > 0$ and $F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $ then :

Question

If $f\left( y \right) = {e^y},$ $g\left( y \right) = y;y > 0$ and

$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $ then :

$F\left( t \right) = t{e^{ - t}}$

$F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)$

$F\left( t \right) = {e^t} - \left( {1 + t} \right)$

$F\left( t \right) = t{e^t}$.

Solution

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