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.