Unformatted text preview: the x–axis and a ≤ b, then the
signed area of (x, y ) a ≤ x ≤ b, 0 ≤ y ≤ f (x) is at least zero, which is obvious.
(b) Since f (x) ≤ M , we have M − f (x) ≥ 0 so that
b 0≤ a b M − f (x) dx = a b M dx − b f (x) dx = M (b − a) − a a a
b which implies that b f (x) dx ≤ M (b − a). The argument showing
similar.
(c) Since f (x) ≤ g (x), we have g (x) − f (x) ≥ 0 so that
b 0≤ a b g (x) − f (x) dx = a b g (x) dx − a f (x) dx
a f (x) dx ≥ m(b − a) is a f (x) dx =⇒ b f (x) dx ≤ b g (x) dx
a (d) For any x, f (x) is either f (x) or −f (x) (depending on whether f (x) is positive or
negative), so we certainly have
f (x) ≤ f (x) − f (x) ≤ f (x) Applying part (c) to each of those inequalities gives
a
b a f (x) dx ≤ b a f (x) dx −
5 b a f (x) dx ≤ b f (x) dx
January 13, 2014 a a a a Since − b f (x) dx ≤ b f (x) dx is equivalent to − b f (x) dx ≤ b f (x) dx (move the
a
a
− b f (x) dx to the right hand side of the inequality and the b f (x) dx to the left hand
side of the inequality), we have
a − b a f (x) dx ≤ a f (x) ≤ b b f (x) dx That’s just another way to say
a
b Example 6 ( π/3 0 √ a f (x) dx ≤ b f (x) dx cos x dx) For x between 0 and π , the function cos x takes values between 1 and
3
√
1
cos x takes values between 1 and √2 . That is
0≤x≤ 1
2 and so the function √
π
1
=⇒ √ ≤ cos x ≤ 1
3
2 Consequently, by Theorem 5.b with a = 0, b = π , m =
3
π
√≤
32 π/3 √ 0 1
√
2 cos x dx ≤ and M = 1,
π
3
Example 6 6 January 13, 2014...
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This note was uploaded on 02/12/2014 for the course MATH 101 taught by Professor Broughton during the Spring '08 term at UBC.
 Spring '08
 Broughton
 Calculus, Derivative

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