Question:

Step 1: Rewrite the integral by factoring out the exponential term. We start with the given functional equation: $$f(x) = \int_0^x e^{x-y} f'(y)\,dy - (x^2-x+1)e^x$$ Factor out $e^x$ from the integral: $$f(x) = e^x \int_0^x e^{-y} f'(y)\,dy - (x^2-x+1)e^x$$ Step 2: Introduce a substitution to simplify the expression. Let us define: $$g(x) = \int_0^x e^{-y} f'(y)\,dy$$ Then we can rewrite $f(x)$ as: $$f(x) = e^x g(x) - (x^2-x+1)e^x = e^x\left[g(x) - (x^2-x+1)\right]$$ Step 3: Differentiate both sides with respect to $x$. Using the product rule on $f(x) = e^x[g(x) - (x^2-x+1)]$: $$f'(x) = e^x\left[g(x) - (x^2-x+1)\right] + e^x\left[g'(x) - (2x-1)\right]$$ Step 4: Use the relationship between $g'(x)$ and $f'(x)$. From the definition of $g(x)$, we have: $$g'(x) = e^{-x}f'(x)$$ Substituting this into the differentiated equation: $$f'(x) = e^x\left[g(x) - (x^2-x+1)\right] + e^x\left[e^{-x}f'(x) - (2x-1)\right]$$ $$f'(x) = e^x\left[g(x) - (x^2-x+1)\right] + f'(x) - (2x-1)e^x$$ Step 5: Simplify to find an explicit form for $f(x)$. Notice that $e^x[g(x) - (x^2-x+1)] = f(x)$ from Step 2. Substituting: $$f'(x) = f(x) + f'(x) - (2x-1)e^x$$ The $f'(x)$ terms cancel: $$0 = f(x) - (2x-1)e^x$$ Therefore: $$f(x) = (2x-1)e^x$$ Step 6: Solve for $x$ when $f(x) = 0$. Setting $f(x) = 0$: $$(2x-1)e^x = 0$$ Since $e^x > 0$ for all real $x$, we must have: $$2x - 1 = 0$$ $$x = \frac{1}{2}$$ **Final Answer:** The value of $x$ for which $f(x) = 0$ is $\boxed{\frac{1}{2}}$, which corresponds to **Option 1**.