Consider the function ƒ : R → R defined by 1 if x #0 if x = 0. f(x) Is f Riemann integrable on [-1, 1]? Justify your answer rigorously using the definition of the Riemann integral.

Let C = sin dx. As f(x) = sin(1/x) is continuous except at 0 and bounded, ƒ is integrable on [-1, 1]. Since the integrand is positive (never negative), C > 0. Put u = varibles to u so that f(x) = sin(u)² and dx = -u-2du. When x = u = -1. Therefore, 1/x and make the change of = 1, u = 1 and when x = -1, ´sin(u) = = - sin(u) (u-2)du C = du < 0. и Where does this argument go wrong?