Chem Differential Eq HW Solutions Fall 2011 5

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

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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. ...
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