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Unit 1: Integration




                      x                                                                         Notes
                         u
                   x
                      
                                   x
                                        x
                                 g
          1.   If g ( ) f  ( )du ,then '( )  f  ( )
                      a
               b
                            F
                                     F
                         b
                              a
                      
                  x
          2.     f ( )dx F ( )– ( ),where is anyantiderivativeof f
               a
          As integration is the inverse process to differentiation. So rather than multiplying by the index
          and dropping the index by one, we enhance the index by one and divide by the new index. The
          + C occurs since the derivative of any constant term is zero.
          C is known as the (arbitrary) constant of integration.
             Did u know?  The value of C  can be instituted when  suitable additional information  is
             given, and this provides a specific integral.
          The rule for integration is
                  ax n   1
             n
             ax dx     C  provided n  –1 .
                  n    1
                                        dy 
                                                
                       x
                              x
          Generally,     f  '( )dx   f ( ) C  or     dx  dx   y C
                                          
          The inverse relationship among differentiation and integration means that, for each statement
          regarding differentiation, we can write down an equivalent statement concerning integration.
                                                           dy    3   2
                 Example: Find the equation of the curve for which     4x    6x  passes via the point
                                                           dx
          (1, 3).
          Integrating provides

           y     (4x  3    6x  2  )dx   x 4   2x  3  C
          Substituting x = 1 and y = 3 offers 3 = 1 + 2 + C , thus C = 0

               4
           y   x   2x  3
          The subsequent step is, when we are specified a function to integrate, to execute rapidly via all
          the typical differentiation formulae, until we come to one which is suitable to our problem.
          Alternatively, we have to learn to identify a specified function as the derivative of another function.



                        d   4    3
                                                 3
              Task  Given   (x  )   4x , then evaluate  4x dx .
                                              
                        dx
          Self Assessment
          Fill in the blanks:
          1.   ..................... is motivated by the problem of defining and calculating the area of the region
               bounded by the graph of the functions.
          2.   The functions that could possibly have given function as a derivative are called .....................
               of the function.




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