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Unformatted text preview: October 3, 2010 1 1. Integration by substitution:
An integral often simliﬁes when we can denote a part of the function as a new variable
(y ). Suppose x = g (y ) where g is a strictly monotone function, then subsitute x
´b
using g (y ), a f (x)dx can be calculated by
ˆ ˆ b g −1 (b) f (x)dx = f (g (y ))d(g (y ))
g −1 (a) a (a) g must be a strictly monotone function here;
(b) The limits of integration change from “a to b” to “g −1 (a) to g −1 (b)”. Note that,
we don’t need b ≥ a or g −1 (b) ≥ g −1 (a) by using the convention that
ˆ ˆ b f (x)dx = −
a a f (x)dx
b (c) Consider the following integral
ˆ 2 xe−2x dx. 0 Let y = 2x, then x = y/2. By substitution (substitute x by y/2), and adjusting
the integration limites, the integral becomes
ˆ2
ˆ4
ˆ
1 4 −y
1
1 − 5e−4
−2x
−y
xe dx =
(y/2)e d(y/2) =
ye dy = (−ye−y −e−y )4 =
.
0
40
4
4
0
0
2. Exercises:
√
´∞
2
2
(a) We know that 0 e−x /2 dx = 2 π , use the integration by substitution technique, show that, for any σ > 0 and µ,
ˆ∞
(x−µ)2
1
√
e− 2σ2 dx = 1/2.
2πσ
µ (b) Evaluate ´4
0 2 /4 xex dx. ...
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This note was uploaded on 02/01/2012 for the course STAT 330B taught by Professor Zhou during the Spring '11 term at Iowa State.
 Spring '11
 Zhou

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