Integral Equation Analysis

We are given the functional relation:

$f(x) = e^x + \int_0^1 (y + xe^x)f(y)dy$

Our objective is to determine the value of $e + f(0)$.

Simplifying the Integral Term

The integral part of the equation can be split and rearranged:

$\int_0^1 (y + xe^x)f(y)dy = \int_0^1 yf(y)dy + \int_0^1 xe^x f(y)dy$

Since $x$ and $e^x$ are constants with respect to the integration variable $y$, we can factor them out:

$= \int_0^1 yf(y)dy + xe^x \int_0^1 f(y)dy$

Let's define two constants:

• $C_1 = \int_0^1 f(y)dy$
• $C_2 = \int_0^1 yf(y)dy$

Substituting these constants, the expression for $f(x)$ becomes:

$f(x) = e^x + C_1 x e^x + C_2$

Calculating the Constants $C_1$ and $C_2$

To find the values of $C_1$ and $C_2$, we substitute the expression for $f(x)$ back into their definitions. We'll need the results of some basic definite integrals:

• $ \int_0^1 e^y dy = [e^y]_0^1 = e^1 - e^0 = e - 1 $
• $ \int_0^1 y e^y dy = [y e^y - e^y]_0^1 = (1 \cdot e^1 - e^1) - (0 \cdot e^0 - e^0) = (e - e) - (0 - 1) = 1 $
• $ \int_0^1 y^2 e^y dy = [y^2 e^y - 2y e^y + 2e^y]_0^1 = ((1^2 - 2(1) + 2)e^1) - ((0^2 - 2(0) + 2)e^0) = (1 - 2 + 2)e - (2)(1) = e - 2 $
• $ \int_0^1 dy = [y]_0^1 = 1 - 0 = 1 $
• $ \int_0^1 y dy = \left[\frac{y^2}{2}\right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} $

Using the definition $C_1 = \int_0^1 f(y)dy$:

$ C_1 = \int_0^1 (e^y + C_1 y e^y + C_2) dy $

$ C_1 = \int_0^1 e^y dy + C_1 \int_0^1 y e^y dy + C_2 \int_0^1 dy $

$ C_1 = (e - 1) + C_1(1) + C_2(1) $

$ C_1 = e - 1 + C_1 + C_2 $

Simplifying this equation gives:

$ 0 = e - 1 + C_2 \implies C_2 = 1 - e $

Now, using the definition $C_2 = \int_0^1 y f(y) dy$:

$ C_2 = \int_0^1 y(e^y + C_1 y e^y + C_2) dy $

$ C_2 = \int_0^1 (y e^y + C_1 y^2 e^y + C_2 y) dy $

$ C_2 = \int_0^1 y e^y dy + C_1 \int_0^1 y^2 e^y dy + C_2 \int_0^1 y dy $

Substituting the known integral values:

$ C_2 = 1 + C_1 (e - 2) + C_2 \left(\frac{1}{2}\right) $

Substitute $C_2 = 1 - e$ into this equation:

$ 1 - e = 1 + C_1 (e - 2) + (1 - e)\left(\frac{1}{2}\right) $

$ 1 - e = 1 + C_1 (e - 2) + \frac{1}{2} - \frac{e}{2} $

Rearranging to solve for $C_1 (e - 2)$:

$ C_1 (e - 2) = 1 - e - 1 - \frac{1}{2} + \frac{e}{2} $

$ C_1 (e - 2) = -\frac{1}{2} - \frac{e}{2} $

$ C_1 = \frac{-(1 + e)}{2(e - 2)} $

We have found $C_2 = 1 - e$. The value of $C_1$ is determined, though not strictly necessary for the final step.

Determining $f(0)$ and Final Result

Recall the simplified form of $f(x)$:

$f(x) = e^x + C_1 x e^x + C_2$

To find $f(0)$, we substitute $x=0$:

$f(0) = e^0 + C_1 (0) e^0 + C_2$

$f(0) = 1 + 0 + C_2$

$f(0) = 1 + C_2$

Using the value $C_2 = 1 - e$:

$f(0) = 1 + (1 - e) = 2 - e$

The question asks for the value of $e + f(0)$:

$e + f(0) = e + (2 - e)$

$e + f(0) = 2$