# Integration by Substitution - PowerPoint by 9k9V48b

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```									Integration by Substitution

Antidifferentiation of a
Composite Function
Antidifferentiation of a Composite
Function
• Let g be a function whose range is an
interval I, and let f be a function that is
continuous on I. If g is differentiable on
its domain and F is an antiderivative of f
on I, then

 f  g  x  g '( x) dx  F ( g ( x)  C
Exploration
• Recognizing Patterns. Discover the rule
using the exploration
Recognizing the f’(g(x)) g’(x)
Pattern
• Evaluate

x        1 2 x dx
3
2

g '( x)  2 x and f ( g ( x))   g  x  
3
         


g ( x)    4
C 
 ( x 2  1
C
4                    4
Recognizing the f’(g(x)) g’(x)
Pattern
• Evaluate
1      1
  2 cos    d
 
1                    1
Letting g ( x )        and g '( x ) 
                    2
1          1  1 
f ( g ( x))  cos     cos    2  d
           
1
 sin    C
 
Multiplying and Dividing by a
Constant
• Evaluate

 x  4x            3 dx
2        3

Rewrite the integral as

  4x        3 x dx
2        3

   4 x  3 8 x dx
1       2    3

8
1  4 x  3       4 x  3  C
2        4        2   4

              C 
8       4              32
Evaluating an Integral
t  2t2

 t dt

We presently have no division rule for
integrals. As a matter of fact there will never be
a rule that involves division.

Suggestions?
Change of Variables
• We are allowed to multiply any integral
by a constant, but we can never multiply
by a variable. What can we do if there is
an extra variable in the given equation?
Integration by Substitution
• Evaluate

x   2 x  1 dx

• There is an extra x in this integrand. In
order to solve this equation, we will let
• u = 2x – 1.
Integration by Substitution
• Now solve this equation for x

u 1
x          Next find the derivative of this function
2
du
dx        Then substitute the variables into the given integral
2
 u  1  12 du
  2  u 2 Now symplify
       
Integration by Substitution
1
   u  1 u 2 du
1

4
1
   3

  u  u 2 du
4
2
1

 5   3 
1  u 2   u 2         1 52 1 32
            3   C  u  u  C
4  5                10        6
                
  2   2 
Now substitute (2 x  1) back in for u.
Integration by Substitution
1             1
  2 x  1   2 x  1 2  C
5            3
2
10             6

• This is a correct answer, but it can be
simplified a little further. If the problem
is not a multiple choice question on the
test or quiz, this answer is completed. If
it is a multiple choice question, it will be
given as
Integration by Substitution

  2 x  1 2  3  2 x  1  5 
1          3

30
1                                  1
  2 x  1  6 x  3  5    2 x  1 2  6 x  8 
3                            3
2
30                               30
Integrating Trig Functions
• Remember that you need the derivative
of the trig function and the function.
Evaluate

 sin 2 3x cos 3x dx
Guidelines for Making a Change
of Variable
1. Choose a substitution u = g(x). Usually, it is best
to choose the inner part of a composite function,
such as a quantity raised to a power.
2. Compute du = g’(x) dx.
3. Rewrite the integral in terms of the variable u.
4. Evaluate the resulting integral in terms of u.
5. Replace u by g(x) to obtain an antiderivative in
terms of x.
General Power Rule for
Integration
• If g is a differentiable function of x, then

 g ( x)
n 1

  g ( x)    g '( x) dx                  C , n  1
n

n 1
Equivalently, if u  g ( x) then
n1
u
 u du  n  1  C                n  1
n
Definite Integrals
3
       x( x  1) dx
2    3
Evaluate
0

   x  1 2 x dx
1 3 2      3

2 0
3
  x 2  1    x 2  1            3  1   02  1
3
4                   4
      2    4
1                                

2        4            8                  8         8
            0                  0
10, 000 1 9999
             
8        8     8
Integration of Even and Odd
Functions
• Let f be integrable on the closed
interval, [a, –a ],
• 1. If f is an even function, then
a                  a
   a
f ( x) dx  2  f ( x) dx
0

• 2. If f is an odd function, then
a
a
f ( x) dx  0.

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