A continuous function f:[0,1]→mathbb R satisfies f(x)=e^x+x∫0^1 t f(t) dt for all x∈[0,1]. Then ∫0^1 f(x) dx equals

Let $c=\int_0^1 t f(t)\,dt$, a constant, so $f(x)=e^x+cx$. Substituting back, $c=\int_0^1 t(e^t+ct)\,dt=\int_0^1 te^t\,dt+\dfrac c3=1+\dfrac c3$, because integration by parts gives $\int_0^1te^t\,dt=[te^t-e^t]_0^1=1$. Solving $c=1+c/3$ gives $c=\dfrac32$. Therefore $f(x)=e^x+\dfrac32x$ and $\displaystyle\int_0^1f(x)\,dx=(e-1)+\dfrac32\cdot\dfrac12=e-1+\dfrac34=e-\dfrac14$. Taking $c=1$ (forgetting the self-referential term) would give $e-\tfrac12$, which is not an option; the other options come from arithmetic slips. The structure is a Fredholm-type integral equation with a separable kernel, so the unknown reduces to a single constant that is found by a self-consistency equation.