46. Bound on an integral Let f be a continuously differentiable function on [a, b] satisfying " f(x) dr = 0. a. If c = (a + b)/2, show that (x - c)f(x)dx. xf(x)dx = (x - c)f(x)dx + = |x – c| and e = (b – a)/2. Show that (f(c + t) - f(c - 1)) dt. xf(x)dx = c. Apply the Mean Value Theorem from Section 4.2 to part (b) to prove that (b – a) -M, 12 f(x) dx s where M is the absolute maximum of f' on [a, b].

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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46. Bound on an integral Let f be a continuously differentiable
function on [a, b] satisfying " f(x) dr = 0.
a. If c = (a + b)/2, show that
(x - c)f(x)dx.
xf(x)dx =
(x - c)f(x)dx +
= |x – c| and e = (b – a)/2. Show that
(f(c + t) - f(c - 1)) dt.
xf(x)dx =
c. Apply the Mean Value Theorem from Section 4.2 to part (b)
to prove that
(b – a)
-M,
12
f(x) dx s
where M is the absolute maximum of f' on [a, b].
Transcribed Image Text:46. Bound on an integral Let f be a continuously differentiable function on [a, b] satisfying " f(x) dr = 0. a. If c = (a + b)/2, show that (x - c)f(x)dx. xf(x)dx = (x - c)f(x)dx + = |x – c| and e = (b – a)/2. Show that (f(c + t) - f(c - 1)) dt. xf(x)dx = c. Apply the Mean Value Theorem from Section 4.2 to part (b) to prove that (b – a) -M, 12 f(x) dx s where M is the absolute maximum of f' on [a, b].
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