If $f\left( y \right) = {e^y},\,\,\;g\left( y \right) = y;\;\;\,\;y > 0$ and $F\left( t \right) = \mathop \smallint \nolimits_0^t \,\,f\left( {t - y} \right)\,\,g\,\,\left( y \right)\;\,\,dy$, then

$F\,\,\left( t \right) = 1 - {e^{ - t}}\,\,\left( {1 + t} \right)$

$F\,\,\left( t \right) = t{e^t}$

$F\,\,\left( t \right) = t{e^{ - t}}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

f\left( y \right) = {e^y},\,\,\;g\left( y \right) = y;\;\;\,\;y > 0

and

F\left( t \right) = \int \nolimits_0^t \,\,f\left( {t - y} \right)\,\,g\,\,\left( y \right)\;\,\,dy

, then

View answer and explanation

Correct option: B

F\,\,\left( t \right) = {e^t} - \,\,\left( {1 + t} \right)

Option B (

F\,\left( t \right) = {e^t} - \,\left( {1 + t} \right)

) is correct because it is the result of evaluating the convolution integral using integration by parts.