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5
3
3
5
1
7
b−a
7
= f a+ b +f a+ b +f a+ b +f a+ b
8
8
8
8
8
8
8
8
4 b−a
4 f a+ (2)
a We’re now going to write out the midpoint Riemann sum approximation to b f (x) dx with
4 subintervals. Note that b is now the lower limit on the integral and a is now the upper
limit on the integral. This is likely to cause confusion when we write out the Riemann sum,
3 January 13, 2014 so we’ll temporarily rename b to A and a to B . The midpoint Riemann sum approximation
B
to A f (x) dx with 4 subintervals is
3B −A
5B −A
1B −A
+f A+
+f A+
+f
24
24
24
1
3
5
5
3
1
7
= f A+ B +f A+ B +f A+ B +f A+
8
8
8
8
8
8
8
f A+ A+
7
B
8 7B −A
24
B−A
4 B−A
4 Now setting A = b and B = a, we have that the midpoint Riemann sum approximation to
a
f (x) dx with 4 subintervals is
b
f 7
1
5
3
3
5
1
7
b+ a +f b+ a +f b+ a +f b+ a
8
8
8
8
8
8
8
8 a−b
4 (3) The curly brackets in (2) and (3) are equal to each other — the terms are just in the reverse
order. So (3)= −(2). The same computation with n subintervals shows that the midpoint
a
b
Riemann sum approximations to b f (x) dx and a f (x) dx with n subintervals are negatives
a
b
of each other. Taking the limit n → ∞ gives b f (x) dx = − a f (x) dx.
Example 4
Recall that
if x ≥ 0 x x = −x if x ≤ 0 So
1
−1 0 f x dx = −1
0 =
−1 1 f x dx + f x dx 0
1 f (−x) dx + f (x) dx
0 Example 4 4 January 13, 2014 Theorem 5 (Inequalities for Integrals).
Let a ≤ b be real numbers and let the functions f (x) and g (x) be integrable on the
interval a ≤ x ≤ b.
(a) If f (x) ≥ 0 for all a ≤ x ≤ b, then
a
b f (x) dx ≥ 0 (b) If there are constants m and M such that m ≤ f (x) ≤ M for all a ≤ x ≤ b,
then
a
m(b − a) ≤
f (x) dx ≤ M (b − a)
b (c) If f (x) ≤ g (x) for all a ≤ x ≤ b, then
a f (x) dx ≤ b (d) We have b g (x) dx
a a
b a f (x) dx ≤ f (x) dx b Proof. (a) Just says that if the curve y = f (x) lies above...
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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 The University of British Columbia.
 Spring '08
 Broughton
 Calculus, Derivative

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