Basic Properties of the Integral Notes

Taking the limit n gives b b b af x bg x dx a a

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Unformatted text preview: Rn (f, a, b) + BRn (g, a, b) So Rn (Af + Bg, a, b) = ARn (f, a, b) + BRn (g, a, b) for all n. Taking the limit n → ∞ gives b b b [Af (x) + Bg (x)] dx = A a f (x) dx + B a g (x) dx. a Geometrically, formula (e) just says that the area of the rectangle with x running from a to b and y running from 0 to 1 is b − a, which is obvious. It is also easy to see formula (e) algebraically since n b 1 dx = lim Rn (1, a, b) = lim a n→∞ n→∞ i=1 b−a = lim (b − a) = b − a n→∞ n Example 2 In Example 1 of the notes “Definition of the Integral”, we saw that 1 1 ex + 7 dx = 0 1x e 0 dx = e − 1. So 1 ex dx + 7 0 dx 0 (by Theorem 1.d with A = 1, f = ex , B = 7, g = 1) = (e − 1) + 7 × (1 − 0) (by Example 1 in the notes “Definition of the Integral” and Theorem 1.e) =e+6 Example 2 2 January 13, 2014 Theorem 3 (Arithmetic for the Domain of Integration). Let a, b, c be real numbers. Let the function f (x) be integrable on an interval that contains a, b and c. Then a (a) f (x) dx = 0 a a (b) b f (x) dx = − b b (c) f (x) dx a c f (x) dx = a b f (x) dx + a f (x) dx c Proof. For notational simplicity, let’s assume that a ≤ c ≤ b and f (x) ≥ 0 for all a ≤ x ≤ b. a The identities a f (x) dx = 0, that is Area (x, y ) a ≤ x ≤ a, 0 ≤ y ≤ f (x) and b a f (x) dx = c a f (x) dx + b c =0 f (x) dx, that is, Area (x, y ) a ≤ x ≤ b, 0 ≤ y ≤ f (x) = Area (x, y ) a ≤ x ≤ c, 0 ≤ y ≤ f (x) + Area (x, y ) c ≤ x ≤ a, 0 ≤ y ≤ f (x) are intuitively obvious. See the figures below. We won’t give a formal proof. y y y = f (x) a y = f (x) x a a So we concentrate on the formula b f (x) dx = − b sum approximation to a f (x) dx with 4 subintervals is c x b b f (x) a dx. The midpoint Riemann 3b−a 5b−a 7b−a 1b−a +f a+ +f a+ +f a+ 24 24 24 24...
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