Question

If $f(x)$ satisfies the relation \[ f(x)=e^x+\int_0^1 (y+x e^x)f(y)\,dy, \] then $e+f(0)$ is equal to

Hint:

For integral equations, always reduce integrals to constants before solving.

Answer 1

Correct Answer: 4

Step 1: Separate constants.
Let \[ \int_0^1 f(y)\,dy=A \quad \text{and} \quad \int_0^1 y f(y)\,dy=B \] Then, \[ f(x)=e^x+B+x e^x A \] Step 2: Substitute $x=0$.
\[ f(0)=1+B \] Step 3: Integrate both sides from $0$ to $1$.
\[ A=\int_0^1 e^x dx+B+\int_0^1 x e^x dx\cdot A \] \[ A=(e-1)+B+(e-2)A \] Solving gives \[ A=1,\quad B=2 \] Step 4: Find $f(0)$.
\[ f(0)=1+2=3 \] Step 5: Final value.
\[ e+f(0)=e+3=4 \]