JEE MAIN - Mathematics (2003 - No. 51)
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 :
$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $$ then :
$$F\left( t \right) = t{e^{ - t}}$$
$$F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)$$
$$F\left( t \right) = {e^t} - \left( {1 + t} \right)$$
$$F\left( t \right) = t{e^t}$$.
Explanation
$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy$$
$$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$$
$$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$$
$$ = - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t$$
$$ = - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]$$
$$ = - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]$$
$$ = {e^t} - \left( {1 + t} \right)$$
$$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$$
$$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$$
$$ = - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t$$
$$ = - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]$$
$$ = - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]$$
$$ = {e^t} - \left( {1 + t} \right)$$
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