If f(x) satisfies the relation f(x) = e^{x} + | vedclass

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(B) Given $f(x) = e^{x} + \int_{0}^{1} yf(y) dy + xe^{x} \int_{0}^{1} f(y) dy$. Let $A = \int_{0}^{1} yf(y) dy$ and $B = \int_{0}^{1} f(y) dy$. Then $

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$2$

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(B) Given $f(x) = e^{x} + \int_{0}^{1} yf(y) dy + xe^{x} \int_{0}^{1} f(y) dy$. Let $A = \int_{0}^{1} yf(y) dy$ and $B = \int_{0}^{1} f(y) dy$. Then $f(x) = e^{x} + A + Bxe^{x}$. Now, $B = \int_{0}^{1} (e^{y} + A + Bye^{y}) dy = [e^{y} + Ay + B(ye^{y} - e^{y})]_{0}^{1} = (e + A + 0) - (1 + 0 - B) = e + A - 1 + B$. This simplifies to $A = 1 - e$. Next, $A = \int_{0}^{1} y(e^{y} + A + Bye^{y}) dy = \int_{0}^{1} (ye^{y} + Ay + Bye^{y}) dy$. Using integration by parts, $\int ye^{y} dy = (y-1)e^{y}$. So, $A = [(y-1)e^{y} + \frac{A}{2}y^{2} + B(y-1)e^{y}]_{0}^{1} = (0 + \frac{A}{2} + 0) - (-1 + 0 - B) = \frac{A}{2} + 1 + B$. Substituting $A = 1 - e$, we get $1 - e = \frac{1-e}{2} + 1 + B$, which gives $B = \frac{1-e}{2} - e = \frac{1-3e}{2}$. Finally, $f(0) = e^{0} + A + B(0)e^{0} = 1 + A = 1 + (1 - e) = 2 - e$. Therefore, $e + f(0) = e + 2 - e = 2$.

If $f(x)$ satisfies the relation $f(x) = e^{x} + \int_{0}^{1} (y + xe^{x}) f(y) dy$, then $e + f(0)$ is equal to . . . . . . .

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