5.6 The Logarithm Defined as an Integral


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 way in Example 1,

           and

          

          Hence

      

       (b) Since is an increasing function and ,

          we have .

          Therefore 


4. Deduce the following laws of logarithms.

   (a)

   (b)

   (c)

 Let and .

      Then and .

      (a)

      (b)

      (c)


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:

      

                 

                 

      Hence

      


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

      

      

      

      

         (the FTC 1)

        


Note Alternatively use l’Hospital’s Rule.