You Let f : [0, 1] → R, where f(x) := (1 if x ∈ {1/n: n ∈ N},0 otherwiseNote that f is Riemann integrable on [0, 1] by a previous assignment. Let F : [0, 1] → R, F(x) := integral 0 to x (f).(a) Prove that F is differentiable on [0, 1].(b) Find a c ∈ [0, 1] such that f is discontinuous at c and F'(c) = f(c).(c) Find a d ∈ [0, 1] such that f is discontinuous at d and F'(d) 6= f(d).Remark: This shows that if the indefinite integral F of a Riemann integrable function f isdifferentiable at a point c where f is discontinuous, then F'(c) may or may not equal f(c).

(a) To prove that F is differentiable on [0, 1], we need to show that the limit of the difference quotient exists for all x in [0, 1]. The difference quotient is (F(x+h) - F(x))/h as h approaches 0.

Since F(x) is the integral from 0 to x of f, we have F(x+h) - F(x) = integral from x to x+h of f. By the properties of the Riemann integral, this is less than or equal to Mh, where M is the maximum value of f on [x, x+h]. Since f takes only the values 0 and 1, we have M <= 1.

Therefore, the difference quotient is less than or equal to h/h = 1 as h approaches 0, so the limit exists and F is differentiable.

(b) Let c = 1/2. Then f is discontinuous at c because f(c) = 1 but f(x) = 0 for x close to c but not equal to c.

Since F is the integral of f, we have F'(c) = f(c) by the Fundamental Theorem of Calculus.

(c) Let d = 1/3. Then f is discontinuous at d because f(d) = 1 but f(x) = 0 for x close to d but not equal to d.

However, F'(d) = 0 because F is constant on any interval that does not include a point of the form 1/n. Since f(d) = 1, we have F'(d) != f(d).

This shows that if the indefinite integral F of a Riemann integrable function f is differentiable at a point c where f is discontinuous, then F'(c) may or may not equal f(c).