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
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.
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.