math 161 calculus i - millersville...

Natural Logarithm as an IntegralMATH 161 Calculus I

J. Robert Buchanan

Department of Mathematics

Summer 2019

Background

I Previously we had worked with functions ln x and ex

numerically and hypothesized the general properties ofthese two functions.

I Now that we have encountered the Fundamental Theoremof Calculus we can treat these two functions rigorously.

I We will see that the rigorous treatment will lead to thesame properties we had hypothesized earlier.

Definition of ln x

DefinitionFor x > 0 we define the natural logarithm function as

ln x =

∫ x

1

1t

dt .

Note:

ddx

[ln x ] =ddx

[∫ x

1

1t

dt]

=1x

by the FTC, Part II.

Properties of the Natural Logarithm

TheoremFor any a, b > 0 and any rational number r ,

1. ln 1 = 02. ln(a b) = ln a + ln b

3. ln(a

b

)= ln a− ln b

4. ln(ar ) = r ln a.

Proof (1 of 2)

1.

ln 1 =

∫ 1

1

1t

dt = 0

2.

ln(ab) =

∫ ab

1

1t

dt =

∫ a

1

1t

dt +

∫ ab

a

1t

dt

= ln a +

∫ ab

a

1t

dt

Let a u = t then a du = dt .

ln(ab) = ln a +

∫ b

1

1u

du = ln a + ln b

Proof (2 of 2)

4.

ddx

[ln(x r )] =1x r

ddx

[x r ] =1x r

[rx r−1

]=

rx

ddx

[r ln x ] =rx

The two functions have the same derivative and thus bythe MVT

ln(x r ) = r ln x + Cln(1r ) = r ln 1 + C

ln 1 = r ln 1 + C0 = r(0) + C0 = C

which implies ln(x r ) = r ln x .

Approximating Values of the Natural LogarithmUse a Riemann sum to approximate ln 3 =

∫ 3

1

1t

dt .

Let n = 100 and use midpoint evaluation points.

∆t =b − a

n=

3− 1100

=1

50

ti = a +

(i − 1

2

)∆t = 1 +

(i − 1

2

)1

50

Therefore

ln 3 =

∫ 3

1

1t

dt ≈n∑

i=1

f (ti)∆t

=100∑i=1

[1

1 +(i − 1

2

) 150

]150

≈ 1.0986.

The Transcendental Number e

DefinitionWe define e to be the real number such that ln e = 1.

Use Newton’s method to approximate e by solving the equation

ln x − 1 = 0.

Approximation

Let f (x) = ln x − 1 and x0 = 2 and use the Newton’s Methodformula:

xn = xn−1 −f (xn−1)

f ′(xn−1)= xn−1 −

ln xn−1 − 11

xn−1

.

Then

n xn0 2.01 2.613712 2.716243 2.718284 2.71828

The Exponential Function

Sinceddx

[ln x ] > 0 then the natural logarithm function isinvertible on (0,∞).

DefinitionFor any real number x we define y = ex to be the number forwhich

ln y = ln(ex ) = x .

Thus we have

ln(ex ) = x for −∞ < x <∞,eln x = x for x > 0.

Properties of the Exponential Function

TheoremFor any real numbers r and s and any rational number t,

1. er es = er+s

2.er

es = er−s

3. (er )t = ert

Proof

ln(er es) = ln(er ) + ln(es)

= r ln e + s ln e= (r + s) ln e= ln(er+s)

Since ln x is a one-to-one function, then er es = er+s.

Derivative of the Exponential Function

We have defined y = ex if and only if ln y = x .

ddx

[ln y ] =ddx

[x ]

1y

dydx

= 1

dydx

= y

ddx

[ex ] = ex

General Exponential Functions

If y = ax where a > 0 and a 6= 1 then

ln y = ln(ax )

= x ln aeln y = ex ln a

y = ax = ex ln a.

Thus

ddx

[ax ] = (ln a)ax∫ax dx =

ax

ln a+ C

General Logarithmic Functions

If a > 0 and a 6= 1 then x = ay if and only if loga x = y .

Question: how can we work with logarithms with generalbases?

ln x = ln(ay )

ln x = y ln aln xln a

= y

ln xln a

= loga x

Derivative of General Logarithmic Functions

ddx

[loga x ] =ddx

[ln xln a

]=

1ln a

ddx

[ln x ]

=1

ln a

[1x

]=

1x ln a

Homework

I Read Section 4.8I Exercises: 1–39 odd

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