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 
.
http://matrix.skku.ac.kr/cal-lab/cal-RiemannSum.html
(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.
