given f(x) is continuous
integral 0 to 1 f(x) f'(x)dx=0
integral 0 to 1 [f(x)]^2 f'(x) dx = 18 , find
integral 0 to 1 [f(x)]^4 f'(x) dx
∫[0,1] f(x) f'(x) dx = 1/2 f^2(x)[0,1]
= 1/2 (f^2(1) - f^2(0)) = 0
so f^2(1) = f^2(0)
∫[0,1] f^2 f' dx = 1/3 f^3 [0,1] = 18
f^3(1) - f^3(0) = 54
(f(1)-f(0))(f^2(1) + f(1)f(0) + f^2(0)) = 54
since this is nonzero, f(1)≠f(0), but by part 1 above, that means that f(1) = -f(0). That gives us
2f(1)(f^2(1) - f^2(1) + f^2(1))
= 2 f(1) * f^2(1) = 2f^3(1) = 54
f^3(1) = 27
f(1) = 3
f(0) = -3
∫f^4 f' dx = 1/5 f^5 [0,1]
= 1/5 [f^5(1) - f^5(0)]
...