5.6   The Logarithm Defined as an Integral     

                                                                 by SGLee - HSKim- SWSun



 1. (a) By comparing areas, show that .

            (b) Use the Midpoint Rule with  to estimate .

     (a)  From above figure, we have


         Since ,   and ,

         we have .


     (b) Let . Then we get



2. By comparing areas, show that




    Note that . Let 

    If we use the left endpoint rule with  subintervals to estimate , then the th height is .

    If we use the right endpoint rule with  subinterval to estimate , the th  height is .

    Since  for both rules is 1 and , we obtain the desired expressions.

3. (a) By comparing areas, show that .

    (b) Deduce that .



    (a) By similar approach as Example 1,





       (b) Since  is an increasing function and ,

            we have .


4. Deduce the following laws of logarithms.





      Let  and .

      Then  and .







5. Show that  by using an integral.




      => .

6. Evaluate






 7. Evaluate .


      Let . Then



      Hence we have .

8. Evaluate 




     The above limit can be written as a definite integral, namely .

     To evaluate the integral, we first use the substitution . Then


      The last integral above is computed as follows:






9. Find 



     We know that

       as  and  as 

      while the second factor approaches

      , which is 0.

      The integral  has  as its upper limit of integration(Part 1 the FTC) suggesting that differentiation might be involved.

      Then the denominator  reminds us the definition of the derivative with .

      So this becomes .

        Thus define  then . Now




                                            (By the FTC 1)


Note  Alternatively you may use L’Hospital’s Rule.




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