Animated Solution for Mathematics - Definite Integration: If $f(y)=e^y,g(y)=y;y>0$ and $F(t)={\int }_0^{t}f(t-y)g(y)dy$, then

We are given the functions $f(y)=e^y$ and $g(y)=y$. We are tasked with evaluating the convolution integral defined as:

This is a classic JEE Advanced style problem where the setup is half the battle.

The Substitution

First, let us substitute the functions into our integral. We replace $f(t-y)$ with $e^{t-y}$ and $g(y)$ with $y$:

Here is the secret that separates the masters from the novices: $t$ is a constant relative to $y$. Because we are integrating with respect to $y$, the term $e^t$ does not change as $y$ changes.

Using the laws of exponents, we can rewrite $e^{t-y}$ as $e^t\cdot e^{-y}$. Since $e^t$ is independent of $y$, we can pull it out of the integral entirely:

This simple step transforms a potentially confusing expression into a clean, manageable integral.

The IBP Battle

Now, we face the integral $\int y{e}^{-y}dy$. We have a product of an algebraic function ($y$) and an exponential function ($e^{-y}$), which is the classic signature of Integration by Parts (IBP).

Following the ILATE rule, we choose $u=y$ and $dv=e^{-y}dy$. Differentiating $u$ gives us $du=dy$, and integrating $dv$ gives us $v=-e^{-y}$.

The IBP formula is $\int udv=uv-\int vdu$. Substituting our parts, we get:

Simplifying the signs, we get $-y{e}^{-y}+\int e^{-y}dy$. Since the integral of $e^{-y}$ is $-e^{-y}$, our result is:

The Final Reunion

Now, we apply the limits from $0$ to $t$. Evaluating at the upper limit $t$, we get $-e^{-t}(t+1)$. Evaluating at the lower limit $0$, we get $-e^0(0+1)=-1$.

Subtracting the lower limit from the upper limit gives us:

Finally, we must not forget the $e^t$ we pulled out earlier. We multiply it back:

Distributing $e^t$, we get $e^t(1)-e^t{e}^{-t}(t+1)$. Since $e^t\cdot e^{-t}=e^0=1$, the expression simplifies beautifully to:

This is the elegance of calculus. You have successfully navigated the trap, handled the integration, and arrived at the solution.