Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{f}(\mathrm{y})=\mathrm{e}^{\mathrm{y}}, \mathrm{g}(\mathrm{y})=\mathrm{y} ; \mathrm{y}>0$ and $F(t)=\int_0^{\mathrm{t}} \mathrm{f}(\mathrm{t}-\mathrm{y}) \mathrm{g}(\mathrm{y})$, then
Options:
Solution:
1093 Upvotes
Verified Answer
The correct answer is:
$\mathrm{F}(\mathrm{t})=\mathrm{e}^{\mathrm{t}}-(1+\mathrm{t})$
$\mathrm{F}(\mathrm{t})=\mathrm{e}^{\mathrm{t}}-(1+\mathrm{t})$
$F(t)=\int_0^t f(t-y) g(y) d y$
$=\int_0^t e^{t-y} y d y=e^t \int_0^t e^{-y} y d y$
$=e^t\left[-y e^{-y}-e^{-y}\right]_0^t=-e^t\left[y e^{-y}+e^{-y}\right]_0^t$
$=-e^t\left[t e^{-t}+e^{-t}-0-1\right]=e^t\left[\frac{t+1-e^t}{e^t}\right]=e^t-(1+t)$
$=\int_0^t e^{t-y} y d y=e^t \int_0^t e^{-y} y d y$
$=e^t\left[-y e^{-y}-e^{-y}\right]_0^t=-e^t\left[y e^{-y}+e^{-y}\right]_0^t$
$=-e^t\left[t e^{-t}+e^{-t}-0-1\right]=e^t\left[\frac{t+1-e^t}{e^t}\right]=e^t-(1+t)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.