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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) dy`, thenA. `F(t)=e^(t)-(1+t)`B. `F(t)=te^(t)`C. `F(t)=te^(-t)`D. `F(t)=1-e^(t)(1+t)` |
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Answer» Correct Answer - A 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)` |
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