Question

Let $f(x)$ be a differentiable function satisfying \[ f(x)=e^x+\int_0^1 (y+xe^x)f(y)\,dy \] Find $f(0)+e$, where $e$ is Napier's constant.

• A. 2
• B. 4
• C. 6
• D. 8

Hint:

When integrals involve unknown constants, assume them as constants and solve using consistency conditions.

Answer 1

The Correct Option is A

Step 1: Rewrite the given function.
\[ f(x)=e^x+\int_0^1 yf(y)\,dy + xe^x\int_0^1 f(y)\,dy \] Let \[ A=\int_0^1 yf(y)\,dy,\quad B=\int_0^1 f(y)\,dy \] Then \[ f(x)=e^x+A+Bx e^x \] Step 2: Find $A$.
\[ A=\int_0^1 y(e^y+A+Bye^y)\,dy \] \[ A=\int_0^1 ye^y\,dy + A\int_0^1 y\,dy + B\int_0^1 y^2e^y\,dy \] \[ A=(0-( -1 ))+\frac{A}{2}+B(e-1) \] \[ \frac{A}{2}+B(1-e)=1 \quad \text{(Equation 1)} \] Step 3: Find $B$.
\[ B=\int_0^1 (e^y+A+Bye^y)\,dy \] \[ B=(e-1)+A+B(0-( -1 )) \] \[ B=e-1+A+B \Rightarrow A=1-e \] Step 4: Substitute $A$ in Equation (1).
\[ \frac{1-e}{2}+B(1-e)=1 \] Solving gives \[ B=1 \] Step 5: Write final expression for $f(x)$.
\[ f(x)=e^x+(1-e)+xe^x \] Step 6: Find $f(0)+e$.
\[ f(0)=1+1-e=2-e \] \[ f(0)+e=2 \] Final conclusion.
The value of $f(0)+e$ is 2.