F(\pi) + F(0) is the integral of f(x) sin(x) from 0 to \pi. f(x) sin(x) is the product of four factors (not necessarily integers):
f(x) sin(x) = (1/n!) (x^n) (a-bx)^n sin(x)
Examine the signs of these terms on the interval 0 to \pi. All four factors are non-negative everywhere, so the integral has to be non-negative.
Could the integral be zero? 1/n! is a nonzero constant. x^n = 0 only when x = 0. (a-bx)^n = 0 only when x = a/b = \pi. And sin(x) = 0 only when x = 0 or x = \pi.
The four factors are all positive inside the interval, and nice continuous functions, so there's definitely a rectangle of positive area underneath their product (and the product can never be negative so you can never get a negative area to get the integral back down to 0).
f(x) sin(x) = (1/n!) (x^n) (a-bx)^n sin(x)
Examine the signs of these terms on the interval 0 to \pi. All four factors are non-negative everywhere, so the integral has to be non-negative.
Could the integral be zero? 1/n! is a nonzero constant. x^n = 0 only when x = 0. (a-bx)^n = 0 only when x = a/b = \pi. And sin(x) = 0 only when x = 0 or x = \pi.
The four factors are all positive inside the interval, and nice continuous functions, so there's definitely a rectangle of positive area underneath their product (and the product can never be negative so you can never get a negative area to get the integral back down to 0).