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

  • Option 1)

    F(t)=e^{t}-(1+t)\;

  • Option 2)

    \; \; F(t)=t\; e\, ^{t}\;

  • Option 3)

    \; \; \; F(t)=te^{-t}\; \;

  • Option 4)

    \; F(t)=1-e^{t}(1+t).

 

Answers (1)

As we learnt in

Integration By PARTS -

Let u and v are two functions then 

\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx

- wherein

Where u is the Ist function v is he IInd function

 

 f (t-y) = e (t-y) ; g(y) = y


F(t)=\int_{0}^{t}e^{(t-y)}\times ydy

           =e^t\int_{0}^{t} ye^{-y}dy                    [Use ILATE]

           =e^t\left [ -ye^{-y}+\int e^{-y}dy \right ]_{0}^{t}

          =e^t\left [ -ye^{-y}- e^{-y} \right ]_{0}^{t}

           =e^t\left [ -te^{-t}-e^{-t}+1 \right ]

          =-t -1+e^t

          =e^t-(1+t)

                    


Option 1)

F(t)=e^{t}-(1+t)\;

This is correct option

Option 2)

\; \; F(t)=t\; e\, ^{t}\;

This is incorrect option

Option 3)

\; \; \; F(t)=te^{-t}\; \;

This is incorrect option

Option 4)

\; F(t)=1-e^{t}(1+t).

This is incorrect option

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