5.6 The Logarithm Defined as an Integral

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

   (b) Use the Midpoint Rule with to estimate .


   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,





       (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





         (the FTC 1)


Note Alternatively use l’Hospital’s Rule.