A) \[F(t)=1-{{e}^{-t}}(1+t)\]
B) \[F(t)\,={{e}^{t}}\,-(1+t)\]
C) \[F(t)=t{{e}^{-t}}\]
D) \[F(t)=t{{e}^{-t}}\]
Correct Answer: B
Solution :
Given, \[F(t)=\int_{0}^{t}{f(t-y)\,g(y)dy}\] \[=\int_{0}^{t}{{{e}^{t-y}}.y\,dy={{e}^{t}}\int_{0}^{t}{{{e}^{-y}}ydy}}\] \[={{e}^{t}}\left[ [-y{{e}^{-y}}]_{0}^{t}-\int_{0}^{t}{1\,(-{{e}^{-y}})dy} \right]\] \[={{e}^{-t}}[(-t{{e}^{-t}}-0)-{{[{{e}^{-y}}]}_{0}}]\] \[={{e}^{t}}[(-t{{e}^{-t}}-({{e}^{-t}}-{{e}^{o}})]\] \[={{e}^{t}}[-t{{e}^{-t}}-{{e}^{-t}}+1]\] \[=[1-{{e}^{-t}}(1+t)]{{e}^{t}}={{e}^{t}}-(1+t)\]
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