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If f(y)=e^y,g(y)=y,y>0, and F(t)=int0^t...

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

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)`

Text Solution

Verified by Experts

The correct Answer is:
B
`F(t)=int_(0)^(t)f(t-y)g(y)dy`
`int_(0)^(t)e^(t-y)*ydy=e^(t)int_(0)^(t)e^(-y)ydy`
`=e^(t)[(-ye^(-y))_(0)^(t)-(e^(-y))_(0)^(t)]`
`=e^(t)[-te^(-t)-e^(-t)+1]`
`=e^(t)-(1+t)`
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