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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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 Spring '08
 Broughton
 Calculus, Derivative

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