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Unit 4: Definite Integral




                                  2 n (n   1)(2n  1)                                          Notes
                2
                       2
                    2
          9.   1   2   3   (n   1)       ................................
                                         6
                                            2
                                     ( n n 
                                 3 
                                         1)
                       3
                    3
                3
          10.  1   2   3    (n  1)       ................................
                                      2  
          4.2 Fundamental Theorem of Integral Calculus
          Fundamental theorem of integral calculus states that if f(x) is a continuous function defined on
          closed interval [a,b] and F(x) is integral of f(x) i.e., f  ( )dx   F ( ),then    f  ( )dx   F ( ),then  a is
                                                                      x
                                                             x
                                                   
                                                                             x
                                                      x
          called lower limit, b is called upper limit and F(b) – F(a) is called the value of the definite integral
          and is always unique.
             Notes  This theorem is  very useful  as it gives us a method of  calculating the  definite
             integral more easily, without calculating the limit of a sum.

             Did u know?  The crucial operation in evaluating a definite integral is that of finding a
             function whose derivative is equal to the integrand.
          This toughens the relationship among differentiation and integration.


                 Example: Evaluate the following integrals:
               3
                      3
          (i)   (2x   1) dx
               1
                2  1
          (ii)      dx
                1 3x   2
                /2
                    2
          (iii)    sin xdx
                0
                 /4
          (iv)    tanxdx
                0
                 /4
          (v)     secxdx
                0
               1   1
          (vi)       dx
               0 1  x 2
               1  dx
          (vii)    2
                  
               0  1 x
               3  x
          (viii)   2  dx
               2  x  1
          Solution:
                  3          (2x   1)  3  1  4  3  1   4 1    
                                   4
                                                       4
                         3
          (i)  I    (2x   1) dx   4 2      8 2x    1      8   (7)  (3)  8 (2320)   290
                                                                  
                                
                  1                  1         1                
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