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.