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  4. If f(y)=ey , g(y) = y; y>0 and F(t) = ∫ limitst0 f(t -y)g(y)dy, then

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

Continuity and Differentiability

Solution:

$F\left(t\right) = \int\limits^{t}_{0} f\left(t-y\right)g\left(y\right)dy $
$ = \int\limits^{t}_{0}e^{t-y} ydy = e^{t} \int\limits^{t}_{0}e^{-y} ydy$
$ = e^{t} \left[-ye^{-y} - e^{-y}\right]^{t}_{0} = -e^{t} \left[ye^{-y} +e^{-y}\right]^{t}_{0} $
$=-e^{t} \left[te^{-t} + e^{-t} -0-1\right]=-e^{t} \left[\frac{t+1-e^{t}}{e^{t}}\right] $
$= e^{t} -\left(1+t\right) $