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

Options

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

Answer

To solve the problem, we need to evaluate the integral \( F(t) = \int_0^t f(t-y) g(y) \, dy \) where \( f(y) = e^y \) and \( g(y) = y \).

### Step-by-Step Solution:

1. **Substituting the Functions**:
We know that \( f(t-y) = e^{t-y} \) and \( g(y) = y \). Therefore, we can rewrite the integral:
\[ F(t) = \int_0^t e^{t-y} y \, dy \]

2. **Factoring Out the Constant**:
Since \( e^t \) is a constant with respect to \( y \), we can factor it out of the integral:
\[ F(t) = e^t \int_0^t e^{-y} y \, dy \]

3. **Integration by Parts**:
We need to evaluate the integral \( \int_0^t e^{-y} y \, dy \). We will use integration by parts, where we let:
- \( u = y \) (thus \( du = dy \))
- \( dv = e^{-y} dy \) (thus \( v = -e^{-y} \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int y e^{-y} \, dy = -y e^{-y} \bigg|_0^t + \int e^{-y} \, dy \]

4. **Evaluating the Boundary Terms**:
Now we evaluate the boundary terms:
\[ -y e^{-y} \bigg|_0^t = -t e^{-t} - (0) = -t e^{-t} \]

5. **Evaluating the Remaining Integral**:
The remaining integral \( \int e^{-y} \, dy = -e^{-y} \bigg|_0^t = -e^{-t} + 1 \):
\[ \int_0^t e^{-y} \, dy = 1 - e^{-t} \]

6. **Combining Results**:
Now substituting back into the integration by parts result:
\[ \int_0^t e^{-y} y \, dy = -t e^{-t} + (1 - e^{-t}) = 1 - (t + 1)e^{-t} \]

7. **Final Expression for \( F(t) \)**:
Now substituting this back into our expression for \( F(t) \):
\[ F(t) = e^t \left( 1 - (t + 1)e^{-t} \right) = e^t - (t + 1) \]

### Final Result:
Thus, we have:
\[ F(t) = e^t - (t + 1) \]