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					Integration
Integration (Anti-derivative)
∫2x dx =» What is the function whose derivative is 2x
Example 1 : Find the indicated integral


     3x   2
                       
                 5x  6 dx
Solution

= x3 
       5x 2
             6x  C          PLUS A CONSTANT
        2
Example 2: Find the indicated integral

       1    

      x2
      
             dx
             
Solution


     x dx
       2
=


   1
=  C             PLUS A CONSTANT
   x
Example 3: Find the indicated integral


     x  1x  1dx
Solution


     x         
               1 dx
           2
=


  x3
=     xC             PLUS A CONSTANT
  3
How to integrate common funtions:

In general,

                   n 1

      
               x
        x dx 
          n
                     C ; n  1
               n 1
Example 4: Find the indicated integral

    x 4  3x 3  4 x 2  1
             2
                          dx
             x

Solution

                   1
=    x  3x  4  x 2 dx
       2




= x 3 3x 2       1
           4x   C
  3    2         x
Exponential Rule

In general,

                  1 kx
          e dx  k e  C
              kx
Examples 5,6,&7: Find the indicated integral
 ∫e dx = ex + C



  e x dx =           ex  C


                       e 2 x 5
    e
         2 x 5
                 dx =          C
                          2
The Logarithmic Rule

In general,


              1
             x
                dx =   ln | x | C
In general,


              f ' ( x)
             f ( x)
                       dx  ln | f ( x) | C
Example 8: Find the indicated integral

         2x

       x 1
         2
            dx


=       
    ln x 1  C
            2
                
Example 9: Find the indicated integral

           x 1
      x2  2x  5
Solution

 1  2x  1
=  2
 2 x  2x  5

  1
=   ln | x 2  2 x  5 | C
  2
Example 10
 If the slope of the tangent of f(x) is 3x 2  5x, find the
   equation of the curve of this function which passes
   through (1,2).
Solution
f ’(x)= 3x 2  5x
f(x)=∫ 3x 2  5x dx
         2
y= x  5 x  C
    3

         2
Since the curve passes through (1,2), then (1,2) satisfies
   its equation.
       5
2=1+ 2 +C                  5x 2 3
C=  3              yx 
                        3
                               
     2                      2     2
 The Definite Integral
The Fundamental Theorem of Calculus



If the function f(x) is continuous on the interval
                  b
a ≤ x ≤ b, then    f ( x)dx  F (b)  F (a)
                  a
                                               where F(x) is any

anti-derivative of f(x) on a ≤ x ≤ b.
Example 11: Evaluate the definite integral

  x 3  3x dx
    1


    2



Solution
                  1

=  x  3x    
        4     2

             
     4     2 2


    1 3
=       4  6 
    4 2

       1
=   8
       4
Example 12: Evaluate the definite integral


     3x               
    2
           2
                 2 x  1 dx
    1


Solution
    
= x3  x 2  x 2
               1        
= 14-13 = 11
Example 13: Find the area under the curve of the
 following function


      x             
     3
            2
                  1 dx
     1


Solution

          x             
         3
    A           2
                      1 dx
         1

                     3
       x3    
          x
       3     1

             1 
  9  3    1            12 
                                       4
                                            10.67
             3                       3
Example 13 (Graph)


                     f ( x)  x 2  1
Area Between two Curves

In general, Area between 2 curves f(x) and g(x)


               b
          A    f ( x)  g ( x)dx
               a
Example 14
 Find the area of the region R enclosed by the curves y  x 3
  and y  x 2

Solution
To find the points of intersection of the 2 curves, we solve the equations
simultaneously : x 3  x 2 => x 2  x  1  0 => x=1 or x=0
                                    1
 A   x  x dx => A 
      1
                          x 4 x3  1 1                    1    1
                            =  4  3             = 
         3   2
                                                               =
     0
                         4    3 0                       12   12
Example 15
 Find the area of the region enclosed by the line y  4 x and the curve
   y  x 3  3x 2



Solution
To find the point of intersection, put x 3  3x 2  4 x
  
 x x 2  3x  4  0 
                                     x                             x                   
                                     0                               1
                               A          3
                                                3 x  4 x dx 
                                                   2                        3
                                                                                 3 x 2  4 x dx
 x x  4 x  1  0              4                               0

 x  0, x  4, x  1              x4
                                                               0                                1
                                               2       x4              
                               A      x  2x 
                                          3
                                                           x3  2 x 2 
                                   4            4    4               0

                                       1                         1          
                               A  0    4  (4)3  2(4) 2     1  2 
                                              4

                                       4                         4          
                                                       3                        3
                               A   32                          A  32
                                                       4                        4

				
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posted:8/29/2012
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