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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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 Fall '11
 StuartChalk

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