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.
6. Check your answer by differentiating.
          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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