If f(y)=ey,g(y)=y,y>0 and F(t)=∫0t f(t−y)g(y)dy, then F(t)=

F(t)=∫0t et−y⋅ydy by def. of f and g=∫0t et−(t−y)(t−y)dy Prop. IV =∫0t ey⋅(t−y)dy. Integrate by parts =ey(t−y)0t−∫ey(−1)dy=(0−t)+ey0t=−t+et−1=et⋅−(1+t)⇒(2)

If f(y)=ey,g(y)=y,y>0 and F(t)=∫0t f(t−y)g(y)dy, then F(t)=

F(t)=∫0t et−y⋅ydy by def. of f and g=∫0t et−(t−y)(t−y)dy Prop. IV =∫0t ey⋅(t−y)dy. Integrate by parts =ey(t−y)0t−∫ey(−1)dy=(0−t)+ey0t=−t+et−1=et⋅−(1+t)⇒(2)

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If $f (y) = e^{y}, g (y) = y, y > 0$ and $F (t) = \int_{0}^{t} f (t - y) g (y) d y$ , then $F (t) =$

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$t e^{- t}$

$e^{t} [- (1 + t)]$

$t e^{t}$

$1 - e^{- t} (1 + t)$

answer is B.

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Detailed Solution

$\begin{array}{l} F (t) = \int_{0}^{t} e^{t - y} \cdot y d y by def. of f and g \\ = \int_{0}^{t} e^{t - (t - y)} (t - y) d y Prop. IV \\ = \int_{0}^{t} e^{y} \cdot (t - y) d y . Integrate by parts \\ = {[e^{y} (t - y)]}_{0}^{t} - \int e^{y} (- 1) d y \\ = (0 - t) + {[e^{y}]}_{0}^{t} = - t + (e^{t} - 1) \\ = e^{t} \cdot - (1 + t) \Rightarrow (2) \end{array}$