Final answer:
To prove that the indefinite integral function F(x) is continuous on the interval [a, b], we need to show that the limit of F(x) as x approaches any point c in the interval exists and is equal to F(c). By applying the mean value theorem for integrals and using the continuity of f, we can demonstrate that F(x) is continuous at every point within the interval [a, b].
Step-by-step explanation:
To show that the indefinite integral function F(x) is continuous on the interval [a, b], we need to prove that the limit of F(x) as x approaches any point c in the interval exists and is equal to F(c). Let's assume that c is a point in the interval [a, b].
Since f is integrable on [a, b], it means that for every ε > 0, there exists a partition P of [a, b] such that the difference between the upper and lower Riemann sums of f over P is less than ε. We can choose a partition P that contains the point c. By the continuity of f, for every ε > 0, there exists a δ > 0 such that |x - c| < δ implies |f(x) - f(c)| < ε. Let's choose δ such that the subinterval (c - δ, c + δ) is contained in P.
Now, for any x in the interval (c - δ, c + δ), we have |x - c| < δ, which implies |f(x) - f(c)| < ε. We can apply the Mean Value Theorem for Integrals to the integral of f(x) over the subinterval [c, x] to get:
|F(x) - F(c)| = |∫[c, x] f(t) dt - ∫[c, c] f(t) dt| = |∫[c, x] f(t) dt| ≤ ∫[c, x] |f(t)| dt ≤ M(x - c),
where M is the upper bound on f in the interval [a, b]. According to the partition P, the length of the subinterval (c - δ, c + δ) is less than ε/M. So, we have |F(x) - F(c)| ≤ M(x - c) < Mε/M = ε. Therefore, we can conclude that the limit of F(x) as x approaches c exists and is equal to F(c), which means that F(x) is continuous at c within the interval [a, b].