Calculus-Sec-5-6-Solution

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,

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

(By the FTC 1)

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