If `f(y)=e^y,g(y)=y,y>0, and F(t)=int_0^t f(t-y)g(y) dy`, then

Options

A. `F(t)=e^(t)-(1+t)`
B. `F(t)=te^(t)`
C. `F(t)=te^(-t)`
D. `F(t)=1-e^(t)(1+t)`

Answer

A

Explanation

We have `f(y)=e^(y),g(y)=y,ygt0`
`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) y dy`
`=e^(t) ([-ye^(-y)]_(0)^(t)+int_(0)^(t) e^(-y) dy)`
`=e^(t) (-te^(-t) -[e^(-y)]_(0)^(t))`
`=e^(t) (-te^(-t)-e^(-1) +1)`
`=e^(t)-(1+t)`