Chem Differential Eq HW Solutions Fall 2011 5

Chem Differential Eq HW Solutions Fall 2011 5 - Section 2.1...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Section 2.1 Periodic Functions 5 25. We have a |F (a + h) − F (a)| = a+h f (x) dx − 0 f (x) dx 0 a+h = f (x) dx ≤ M · h, a where M is a bound for |f (x)|, which exists by the previous exercise. (In deriving the last inequality, we used the following property of integrals: b f (x) dx ≤ (b − a) · M, a which is clear if you interpret the integral as an area.) As h → 0, M · h → 0 and so |F (a + h) − F (a)| → 0, showing that F (a + h) → F (a), showing that F is continuous at a. a (b) If f is continuous and F (a) = 0 f (x) dx, the fundamental theorem of calculus implies that F (a) = f (a). If f is only piecewise continuous and a0 is a point of continuity of f , let (xj −1, xj ) denote the subinterval on which f is continuous and a0 is in (xj −1, xj ). Recall that f = fj on that subinterval, where fj is a continuous a component of f . For a in (xj −1, xj ), consider the functions F (a) = 0 f (x) dx and a xj −1 G(a) = xj−1 fj (x) dx. Note that F (a) = G(a) + 0 f (x) dx = G(a) + c. Since fj is continuous on (xj −1, xj ), the fundamental theorem of calculus implies that G (a) = fj (a) = f (a). Hence F (a) = f (a), since F diﬀers from G by a constant. ...
View Full Document

Ask a homework question - tutors are online