Which condition on the function f(x)f(x)f(x) is sufficient to guarantee that ∫0πf(x)sinxdx=∫0πf(π−x)sinxdx\int_{0}^{\pi} f(x) \sin x dx = \int_{0}^{\pi} f(\pi - x) \sin x dx∫0πf(x)sinxdx=∫0πf(π−x)sinxdx?
f(x)f(x)f(x) is an odd function
f(x)f(x)f(x) is differentiable on [0,π][0, \pi][0,π]
No condition; it is always true for integrable functions
f(x)=f(π−x)f(x) = f(\pi - x)f(x)=f(π−x)