James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A50
appendix G The Logarithm Defined as an Integral
Our treatment of exponential and logarithmic functions until now has relied on our intuition,
which is based on numerical and visual evidence. (See Sections 1.4, 1.5, and 3.1.)
Here we use the Fundamental Theorem of Calculus to give an alternative treatment that
provides a surer footing for these functions.
Instead of starting with b x and defining log b x as its inverse, this time we start by defining
ln x as an integral and then define the exponential function as its inverse. You should
bear in mind that we do not use any of our previous definitions and results concerning
exponential and logarithmic functions.
The Natural Logarithm
We first define ln x as an integral.
1 Definition The natural logarithmic function is the function defined by
ln x − y x
1
1
t dt x . 0
y
y= 1 t
0
1
area=ln x
x t
FIGURE 1
y
area=_ln x
y= 1 t
0
x 1
t
FIGURE 2
y
y= 1 t
A
E
D
B C
0 1 2 t
FIGURE 3
The existence of this function depends on the fact that the integral of a continuous
function always exists. If x . 1, then ln x can be interpreted geometrically as the area
under the hyperbola y − 1yt from t − 1 to t − x. (See Figure 1.) For x − 1, we have
For 0 , x , 1,
ln x − y x
ln 1 − y 1
1
1
1
t dt − 0
1
t dt − 2 y 1 x
1
t dt , 0
and so ln x is the negative of the area shown in Figure 2.
EXAMPLE 1
(a) By comparing areas, show that 1 2 , ln 2 , 3 4 .
(b) Use the Midpoint Rule with n − 10 to estimate the value of ln 2.
SOLUTION
(a) We can interpret ln 2 as the area under the curve y − 1yt from 1 to 2. From Figure
3 we see that this area is larger than the area of rectangle BCDE and smaller than
the area of trapezoid ABCD. Thus we have
1
2 ? 1 , ln 2 , 1 ? 1 2 s1 1 1 2 d
1
2 , ln 2 , 3 4
(b) If we use the Midpoint Rule with f std − 1yt, n − 10, and Dt − 0.1, we get
ln 2 − y 2 1
1 t dt < s0.1df f s1.05d 1 f s1.15d 1 ∙ ∙ ∙ 1 f s1.95dg
− s0.1dS 1
1.05 1 1
1.15 1 ∙ ∙ ∙ 1
1.95D 1 < 0.693 ■
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