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Chem Differential Eq HW Solutions Fall 2011 5

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

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Section 2.1 Periodic Functions 5 25. We have | F ( a + h ) - F ( a ) | = a 0 f ( x ) dx - a + h 0 f ( x ) dx = a + h a f ( x ) dx M · h, 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 a f ( x ) dx ( b - a ) · M, 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 . (b) If f is continuous and F ( a ) = a 0 f ( x ) dx , the fundamental theorem of calculus implies that F ( a ) = f ( a ). If f is only piecewise continuous and a 0 is a point of continuity of f , let ( x j - 1 ,x j ) denote the subinterval on which f is continuous and
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