If f(y) = e^y,g(y) = y for y 0,and F(t) = i | vedclass

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(B) Given $F(t) = \int_{0}^{t} f(t - y) g(y) dy$. Substituting the given functions $f(y) = e^y$ and $g(y) = y$: $F(t) = \int_{0}^{t} e^{t - y} y dy =

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$F(t) = e^t - (1 + t)$

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(B) Given $F(t) = \int_{0}^{t} f(t - y) g(y) dy$. Substituting the given functions $f(y) = e^y$ and $g(y) = y$: $F(t) = \int_{0}^{t} e^{t - y} y dy = e^t \int_{0}^{t} y e^{-y} dy$. Using integration by parts for $\int y e^{-y} dy$: Let $u = y$ and $dv = e^{-y} dy$. Then $du = dy$ and $v = -e^{-y}$. $\int y e^{-y} dy = -y e^{-y} - \int (-e^{-y}) dy = -y e^{-y} - e^{-y}$. Evaluating the definite integral from $0$ to $t$: $F(t) = e^t [ -y e^{-y} - e^{-y} ]_{0}^{t} = e^t [ (-t e^{-t} - e^{-t}) - (0 - 1) ]$. $F(t) = e^t [ -t e^{-t} - e^{-t} + 1 ] = -t - 1 + e^t$. Thus,$F(t) = e^t - (1 + t)$.

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

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