Recitation Note 1 Some Integration Theorem

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Recitation Note
17.872 TA Jiyoon Kim
Sep. 30, 2003
1
Some Integration Theorem
1. A constant can be moved through an integral sign.
�
cf (x)dx = c
�
f (x)dx
2. An integration of a sum is the sum of the integrations.
�
[f (x) + g(x)]dx =
�
f (x)dx +
�
g(x)dx
3. An integration of a difference is the difference of the integrations.
�
2
[f (x) − g(x)]dx =
�
f (x)dx −
�
Integration Rules
1. Power Rule
�
2. Exponential Rule I
xn dx =
�
1
xn+1 + c
n+1
ex dx = ex + c
3. Exponential Rule II
�
4. Logarithm Rule I
f � (x)ef (x) dx = ef (x) + c
�
1
dx = lnx + c
x
5. Logarithm Rule II
�
f � (x)
dx = lnf (x) + c
f (x)
1
g(x)dx
Examples:
�
14x
dx
2+5
7x
�
4x
dx
2)
2+1
2x
�
x
3)
dx
2
3x + 5
1)
3
2e2x +
Substitution Rule
When you have integration with high power or with complicated form, you
can use substitution rule. First, you put the complicated part as u and find
du
du
dx . Then, substitute the original equation with u and dx to solve it in u term.
Example:
1)
2)
�
�
4
2x(x2 + 1)50 dx
6x2 (x3 + 2)99 dx
Integration by Parts
The integral of v with respect to u is equal to uv minus the integration of u
with respect to v.
�
�
vdu = uv −
udv
Here, you use the less complicated part (usually one term) as v and the rest
of the� integration including integral sign and dx as� u. For example, if you
1
1
have x(x + 1) 2 dx, v should be x and u should be (x + 1) 2 dx. Then, find
dv and du to plug them in the formula.
Example:
1)
2)
�
�
xex dx
ln x dx
2
5
Properties of Sigma Notation
The following properties of sigma notation will help to manipulate sums:
1.
n
�
cak = c
n
�
ak
k=1
k=1
2.
n
�
(ak + bk ) =
(ak − bk ) =
ak +
k=1
n
�
k=1
n
�
n
�
ak −
n
�
bk
n
�
bk
k=1
3.
k=1
k=1
k=1
Question Are these true?
1)
� ��
n
d
2)
dx
3)
4)
5)
n
�
�
fi (x) dx =
�i=1
n
�
�
fi (x) =
i=1
ai bi =
n
�
n ��
�
i=1
n �
�
i=1
n
�
fi (x)dx
�
�
d
[fi (x)]
dx
bi
ai
i=1
i=1
�n
ai
ai
= �in=1
b
i=1 bi
i=1 i
� n
�2
n
�
�
2
ai =
ai
i=1
i=1
i=1
n
�
Quick Question
Let x̄ denote the arithmetic average of the n numbers x1 , x2 , · · · , xn . Prove
3
that
n
�
i=1
¯ =0
(xi − x)
4
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