If `f(y)=e^y,g(y)=y,y>0, and F(t)=int_0^t f(t-y)g(y) dy`, then

Options

A. `F(t)=e^(t)-(1+t)`
B. `F(t)=te^(t)`
C. `F(t)=te^(-t)`
D. `F(t)=1-e^(t)(1+t)`

Explanation

To solve the problem, we need to compute the function \( F(t) \) defined as:

\[ F(t) = \int_0^t f(t-y) g(y) \, dy \]

Given that \( f(y) = e^y \) and \( g(y) = y \), we can substitute these into the integral:

\[ F(t) = \int_0^t f(t-y) g(y) \, dy = \int_0^t f(t-y) \cdot y \, dy \]

Now substituting \( f(t-y) = e^{t-y} \):

\[ F(t) = \int_0^t e^{t-y} \cdot y \, dy \]

Next, we can factor out \( e^t \) since it is a constant with respect to \( y \):

\[ F(t) = e^t \int_0^t e^{-y} \cdot y \, dy \]

Now we need to evaluate the integral \( \int_0^t y e^{-y} \, dy \). We will use integration by parts for this integral. Let:

- \( u = y \) (thus \( du = dy \))
- \( dv = e^{-y} dy \) (thus \( v = -e^{-y} \))

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

We have:

\[ \int_0^t y e^{-y} \, dy = \left[ -y e^{-y} \right]_0^t + \int_0^t e^{-y} \, dy \]

Calculating the first term:

\[ \left[ -y e^{-y} \right]_0^t = -t e^{-t} - (0) = -t e^{-t} \]

Now we compute the second integral:

\[ \int_0^t e^{-y} \, dy = \left[ -e^{-y} \right]_0^t = -e^{-t} + 1 = 1 - e^{-t} \]

Putting it all together:

\[ \int_0^t y e^{-y} \, dy = -t e^{-t} + (1 - e^{-t}) = 1 - (t + 1)e^{-t} \]

Now substituting this back into our expression for \( F(t) \):

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

Thus, we simplify:

\[ F(t) = e^t - t - 1 \]

So the final answer is:

\[ F(t) = e^t - t - 1 \]