Basic Properties of the Integral Notes

The argument showing similar c since f x g x we have

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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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