Calculus Preliminaries - Clemson CECAScecas.clemson.edu/~petersj/Courses/M208/Lectures/L... · 2 Fundamental Theorem Of Calculus 3 The Cauchy Fundamental Theorem Of Calculus 4 Some

Introduction to ODE

Calculus Preliminaries

James K. Peterson

Department of Biological Sciences and Department of Mathematical SciencesClemson University

May 16, 2017

Introduction to ODE

Outline

1 Properties of The Riemann Integral

2 Fundamental Theorem Of Calculus

3 The Cauchy Fundamental Theorem Of Calculus

4 Some more about the FToC

5 The Natural Logarithm Function

6 The Exponential Function

7 Adding Logarithms: both logarithms are bigger than 18 Adding Logarithms: one logarithm less than 1 and one bigger

than 1

9 General Results

10 The Exponential Function Rules

11 Graphing Logarithm and Exponential FunctionsLogarithm FunctionsExponential Functions

12 Half Lifes and Doubling Times

Introduction to ODE

Abstract

This lecture explains some mathematical preliminaries.

Introduction to ODE

Properties of The Riemann Integral

If you think about how the Riemann integral is defined in terms oflimits of Riemann sums, it is pretty easy to figure out some basicproperties.∫ a

a f (t)dt = 0 as the partitions are all one point and all the

changes in t are 0. Thus,∫ 11 t2dt = 0 as the interval is just

one point.∫ ba f (t)dt =

∫ ca f (t)dt +

∫ bc f (t)dt for any c between a and

b. For example:∫ 5

1t2dt =

∫ 3

1t2dt +

∫ 5

3t2dt.∫ 5

1t2dt =

∫ 2.2

1t2dt +

∫ 5

2.2t2dt

Having the order backwards just changes the sign of theRiemann integral value. So∫ 1

5t2dt = −

∫ 5

1t2dt.∫ −1

8t3dt = −

∫ 8

−1t3dt

Introduction to ODE

Fundamental Theorem Of Calculus

There is a big connection between the idea of the antiderivativeof a function f and its Riemann integral.

For a positive function f on the finite interval [a, b], we canconstruct the area under the curve functionF (x) =

∫ xa f (t) dt.

Let’s look at the difference in these areas: we assume h ispositive.

F (x + h) − F (x) =

∫ x+h

af (t) dt −

∫ x

af (t) dt

=

∫ x

af (t) dt +

∫ x+h

xf (t) dt

−∫ x

af (t) dt

where we have used standard properties of the Riemannintegral to write the first integral as two pieces.

Introduction to ODE


Now subtract to get

F (x + h) − F (x) =

∫ x+h

xf (t) dt

Now divide this difference by the change in x which is h. Wefind

F (x + h) − F (x)

h=

1

h

∫ x+h

xf (t) dt

We show F (x) and F (x + h) for a small positive h in thefigure which follows.

Introduction to ODE


F (x) is the area under this curve from a to x .

(a, f (a))(b, f (b))

a bx x + h

F (x) F (x + h)

Figure: The Function F (x)

Introduction to ODE


The difference in area,∫ x+hx f (t) dt, is the second shaded area in

the figure you just looked at. We see

If t is any number between x and x + h, the area of therectangle with base h and height f (t) is f (t) × h which isclosely related to the area difference.

Note the difference between this area and F (x + h) − F (x)is really small when h is small.

We know that f is bounded on [a, b] You can easily see that fhas a maximum value for the particular f we draw. Of course,this graph is not what all such bounded functions f look like,but you should be able to get the idea that there is a numberB so that 0 < f (t) ≤ B for all t in [a, b].

Thus, we see

F (x + h) − F (x) ≤∫ x+h

xB dt = B h (1)

Introduction to ODE


From this, it follows that

We see our difference lives between 0 and B.

0 ≤ (F (x + h) − F (x)) ≤ B h

And so taking the limit as h gets small, we find

0 ≤ limh→ 0

(F (x + h) − F (x))

≤ limh→ 0

B h = 0.

We conclude that F is continuous at each x in [a, b] as

limh→ 0

(F (x + h) − F (x)) = 0.

It seems that the new function F we construct by integratingthe function f in this manner always builds a new functionthat is continuous.

Introduction to ODE


Is F differentiable at x? Let’s do an estimate. We have a lowerand upper bound on the area of the middle slice in our figure.

minx ≤t ≤x+h

f (t) h ≤∫ x+h

xf (t)dt ≤ max

x ≤t ≤x+hf (t) h

Thus, we have the estimate

minx ≤t ≤x+h

f (t) ≤ F (x + h) − F (x)

h≤ max

x ≤t ≤x+hf (t)

Introduction to ODE


If f was continuous at x , then we must have

limh→ 0

minx ≤t ≤x+h

f (t) = f (x)

and

limh→ 0

maxx ≤t ≤x+h

f (t) = f (x)

Note the f we draw in our figure is continuous all the time,but the argument we use here only needs continuity at thepoint x! At any rate, we can infer for positive h,

limh→ 0+

F (x + h) − F (x)

h= f (x)

Introduction to ODE


You should be able to believe that a similar argument wouldwork for negative values of h: i.e.,

limh→ 0−

F (x + h) − F (x)

h= lim

h→ 0−f (t) = f (x)

This tells us that F ′(p) exists and equals f (x) as long as f iscontinuous at x as

F ′(x+) = limh→ 0+

F (x + h) − F (x)

h= f (x)

F ′(x−) = limh→ 0−

F (x + h) − F (x)

h= f (x)

Introduction to ODE


This relationship is called the Fundamental Theorem ofCalculus.

Our argument works for x equals a or b but we only need tolook at the derivative from one side. So the discussion is a bitsimpler.

Our argument used a positive f but it works just fine if f haspositive and negative spots. Just divide f into it’s postive andnegative pieces and apply these ideas to each piece and thenglue the result together.

We can actually prove this using fairly relaxed assumptions onf for the interval [a, b]. In general, f need only be RiemannIntegrable on [a, b] which allows for jumps in the function.But those arguments are more advanced!

Introduction to ODE


Theorem

The Fundamental Theorem of CalculusLet f be Riemann Integrable on [a, b]. Then the function Fdefined on [a, b] by F (x) =

∫ xa f (t) dt satisfies

1 F is continuous on all of [a, b]

2 F is differentiable at each point x in [a, b] where f iscontinuous and F ′(x) = f (x).

Introduction to ODE


We can do more!!!

Using the same f as before, suppose G was defined on [a, b]as follows

G (x) =

∫ b

xf (t) dt.

then

F (x) + G (x) =

∫ x

af (t) dt +

∫ b

xf (t) dt

=

∫ b

af (t) dt.

Introduction to ODE


So

G (x) =

∫ b

af (t) dt − F (x)

Since the Fundamental Theorem of Calculus tells us F isdifferentiable, we see G (x) must also be differentiable. Itfollows that since the derivative of a constant is 0, we have

G ′(x) = − F ′(x) = −f (x).

Introduction to ODE


Let’s state this as a variant of the Fundamental Theorem ofCalculus, the Reversed Fundamental Theorem of Calculus soto speak.

Theorem

Reversed Fundamental Theorem of CalculusLet f be Riemann Integrable on [a, b]. Then the function G

defined on [a, b] by G (x) =∫ bx f (t) dt satisfies

1 G is continuous on all of [a, b]

2 G is differentiable at each point x in [a, b] where f iscontinuous and G ′(x) = −f (x).

Introduction to ODE

The Cauchy Fundamental Theorem Of Calculus

We can use the Fundamental Theorem of Calculus to learn how toevaluate many Riemann integrals. This is how it works.

If f is continuous, the FToC tells us that F (x) =∫ xa f (t)dt

satisfies F ′(x) = f (x). So F is an antiderivative of f !!

If G is another antiderivative of f , then it also satisfiesG ′(x) = f (x).

We can use this to figure out a way to evaluate Riemannintegrals!

Introduction to ODE


Here’s the argument (very cool one too I might add!)

For our f which is continuous on [a, b], let

F (x) =

∫ x

af (t) dt,

So F ′ = f and note F (a) = 0.

Let G be an antiderivative of the function f on [a, b]. Then,by definition, G ′(x) = f (x) and also G is continuous since itis differentiable.

Let H = F − G . Then

H ′(x) =

(F (x)− G (x)

)′= f (x) − f (x) = 0.

The only function whose derivative is 0 is a constant. So forsome constant C ,

H(x) = F (x)− G (x) = C

Introduction to ODE


Almost done!

Thus H(a) = H(b) = C as H has the same value everywhere.

But H(a) = F (a)− G (a) and H(b) = F (b)− G (b).

These values are the same, so

F (a)− G (a) = F (b)− G (b).

Rearranging, we have

F (b) =

∫ b

af (t)dt = G (b) − G (a).

This result is huge! It says we can evaluate any Riemannintegral if we can guess an antiderivative.

Introduction to ODE


Let’s formalize this as a theorem called the Cauchy FundamentalTheorem of Calculus. All we really need to prove this result isthat f is Riemann integrable on [a, b], which for us is usually trueas our functions f are continuous in general.

Theorem

Cauchy Fundamental Theorem of CalculusLet f be Riemann integral on [a, b] and let G be any

antiderivative of f . Then G (b) − G (a) =∫ ba f (t) dt.

We usually write this as∫ b

af (t)dt = G (t)

∣∣∣∣ba

Introduction to ODE


In the problems that follow it doesn’t matter which antiderivativewe choose as our result above doesn’t care. So we just chooseC = 0 always.

∫ 3

1t3 dt =

t4

4

∣∣∣∣31

=34

4− 14

4=

80

4

∫ 4

−2t3 dt =

t4

4

∣∣∣∣4−2

=44

4− (−2)4

4

=256

4− 16

4=

240

4

Introduction to ODE

Some more about the FToC

Consider the function f defined on [−2, 5] by

f (t) =

2t −2 ≤ t < 01 t = 0(1/5)t2 0 < t ≤ 5

Now as an exercise, draw this function on your paper. Drawthe straight line 2t starting at −2 and draw an open circle forthe point t = 0.

Draw the ordered pair (0, 1) as a filled in circle.

Draw the parabola 1/5 t2 starting with an open circle at 0 andfinish with the last ordered pair (5, 25/5 = 5).

Introduction to ODE


Here is quick sketch of this function. Note the right and left handlimits of f match at 0 but they don’t match f (0). So this functionhas a removeable discontinuity.

So if we build F as usual from the FToC what should happen? Weknow f is not continuous at 0, so we suspect either F ′(0) does notexist or it does exist with F ′(0) 6= f (0).

Introduction to ODE


Let’s calculate F (t) =∫ t−2 f (s) ds. This will have to be done in

several parts because of the way f is defined.

On the interval [−2, 0], note that f is continuous except atone point, t = 0. Hence, f is Riemann integrable.

Also, the function 2t is continuous on this interval and so isRiemann integrable. Since f on [−2, 0] and 2t match at allbut one point on [−2, 0], their Riemann integrals match.

Hence, if t is in [−2, 0], we compute F as follows:

F (t) =

∫ t

−2f (s) ds =

∫ t

−22s ds

= s2∣∣∣∣t−2

= t2 − (−2)2 = t2 − 4

Introduction to ODE


On the interval [0, 5], note that f is continuous except at onepoint, t = 0. Hence, f is Riemann integrable.

Also, the function (1/5)t2 is continuous on this interval andso is also Riemann integrable.

Then since f on [0, 5] and (1/5)t2 match at all but one pointon [0, 5], their Riemann integrals must match.

So if t is in [0, 5], we compute F as follows:

F (t) =

∫ t

−2f (s) ds =

∫ 0

−2f (s) ds +

∫ t

0f (s) ds

=

∫ 0

−22s ds +

∫ t

0(1/5)s2 ds

= s2∣∣∣∣0−2

+ (1/15)s3∣∣∣∣t0

= −4 +1

15t3

Introduction to ODE


Thus, we have found that

F (t) =

{t2 − 4, −2 ≤ t ≤ 0t3/15 − 4, 0 ≤ t ≤ 5

We see F is indeed continuous. What about the differentiability ofF? The FToC guarantees that F has a derivative at each pointwhere f is continuous and at those points F ′(t) = f (t). Hence,we know this is true at all t except perhaps at 0.

Introduction to ODE


Note at those t, we find

F ′(t) =

{2t, −2 ≤ t < 0(1/5)t2, 0 < t ≤ 5

which is exactly what we expect.

Also, note F ′(0−) = 0 and F ′(0+) = 0 as well.

Hence, since the right and left hand derivatives match, we seeF ′(0) does exist and has the value 0.

But this is not the same as f (0) = 1.

Note, F is not the antiderivative of f on [−2, 5] because ofthis mismatch.

Introduction to ODE


Next consider the function f defined on [−2, 5] by

f (t) =

2t −2 ≤ t < 01 t = 02 + (1/5)t2 0 < t ≤ 5

Now as an exercise, draw this function on your paper. Drawthe straight line 2t starting at −2 and draw an open circle forthe point t = 0.

Draw the ordered pair (0, 1) as a filled in circle.

Draw the parabola 2 + 1/5 t2 starting with an open circle at 0and finish with the last ordered pair (5, 25/5 = 5).

Introduction to ODE


Here is quick sketch of this function. Note the right and left handlimits of f don’t match at 0 and they don’t match f (0). So thisfunction has a discontinuity which can’t be removed.

So if we build F as usual from the FToC what should happen? Weknow f is not continuous at 0, so we suspect either F ′(0) does notexist or it does exist with F ′(0) 6= f (0). Let’s see what happens.

Introduction to ODE


Let’s calculate F (t) =∫ t−2 f (s) ds. This will have to be done in

several parts because of the way f is defined.

On the interval [−2, 0], note that f is continuous except atone point, t = 0. Hence, f is Riemann integrable.

Also, the function 2t is continuous on this interval and so isRiemann integrable. since f on [−2, 0] and 2t match at allbut one point on [−2, 0], their Riemann integrals match.

Hence, if t is in [−2, 0], we compute F as follows:

F (t) =

∫ t

−2f (s) ds =

∫ t

−22s ds

= s2∣∣∣∣t−2

= t2 − (−2)2 = t2 − 4

Introduction to ODE


On the interval [0, 5], note that f is continuous except at onepoint, t = 0. Hence, f is Riemann integrable.

Also, the function 2 + (1/5)t2 is continuous on this intervaland so is also Riemann integrable.

Then since f on [0, 5] and 2 + (1/5)t2 match at all but onepoint on [0, 5], their Riemann integrals must match.

So if t is in [0, 5], we compute F as follows:

F (t) =

∫ t

−2f (s) ds =

∫ 0

−2f (s) ds +

∫ t

0f (s) ds

=

∫ 0

−22s ds +

∫ t

0(2 + (1/5)s2) ds

= s2∣∣∣∣0−2

+ (2s + (1/15)s3)

∣∣∣∣t0

= −4 + 2t +1

15t3

Introduction to ODE


Thus, we have found that

F (t) =

{t2 − 4, −2 ≤ t ≤ 0t3/15 + 2t − 4, 0 ≤ t ≤ 5

We see F is indeed continuous. What about the differentiability ofF? The FToC guarantees that F has a derivative at each pointwhere f is continuous and at those points F ′(t) = f (t). Hence,we know this is true at all t except perhaps at 0.

Introduction to ODE


Note at those t, we find

F ′(t) =

{2t, −2 ≤ t < 0(1/5)t2 + 2, 0 < t ≤ 5

which is exactly what we expect.

Also, note F ′(0−) = 0 and F ′(0+) = 2.

Hence, since the right and left hand derivatives don’t match,we see F ′(0) does not exist.

Note, F is not the antiderivative of f on [−2, 5] F ′(0) doesnot exist and so can’t match f (0).

Introduction to ODE


Homework 1

For the functions f

1 Graph f carefully labeling all interesting points.2 Verify that F is continuous and differentiable at all points

except one.

1.1 Compute F (t) =∫ t−3 f (s) ds for

f (t) =

3t −3 ≤ t < 06 t = 0(1/6)t2 0 < t ≤ 6

1.2 Compute F (t) =∫ t2 f (s) ds for

f (t) =

−2t 2 ≤ t < 512 t = 53t − 25 5 < t ≤ 10

Introduction to ODE


Homework 9 Continued

For the functions f

1 Graph f carefully labeling all interesting points.2 Verify that F is continuous and differentiable at all points

except one.

1.3 Compute∫ t−3 f (s) ds for

f (t) =

3t −3 ≤ t < 06 t = 0(1/6)t2 + 2 0 < t ≤ 6

1.4 Compute∫ t2 f (s) ds for

f (t) =

−2t 2 ≤ t < 512 t = 53t 5 < t ≤ 10

Introduction to ODE

The Natural Logarithm Function

Now let’s do something useful with all this complicatedmathematics. Look at the continuous function f (t) = 1/t for allt ≥ 1.

Pick any x ≥ 1.

Pick any L > x .

Let F (x) =∫ x1 1/t dt on the interval [1, L].

The FToC applies and we know F is continuous on [1, L] – inparticular F is continuous at x .

Since the integrand 1/t is continuous here, FToC tells usF ′(x) = f (x) = 1/x .

We can do this argument for any x ≥ 1. So we know our F iscontinous at x and F ′(x) = 1/x for all x ≥ 1.

Introduction to ODE


Look at the continuous function f (t) = 1/t for all 0 < t ≤ 1.

Pick any x with 0 < x ≥ 1.

Pick any L so that 0 < L < x ≤ 1.

Let G (x) =∫ 1x 1/t dt on the interval [L, 1]].

The FToC applies and we know G is continuous on [L, 1] – inparticular G is continuous at x .

Since the integrand 1/t is continuous here, Reversed FToCtells us G ′(x) = −f (x) = −1/x .

We can do this argument for any such x . So we know our Gis continous at x and G ′(x) = −1/x for all 0 < x ≥ 1.

Introduction to ODE


Now we define the natural logarithm of x for any x > 0.

ln(x) =

∫ x

11/t dt

We get two pieces:

ln(x) =

{F (x) =

∫ x1 1/t dt, x ≥ 1∫ x

1 1/t dt = −∫ 1x 1/t dt = −G (x), 0 < x ≤ 1.

Introduction to ODE


So we have ln(x) is continuous for 0 < x ≤ 1 and for x ≥ 1; soln(x) is continuous for all positive x . We also have

(ln(x)

)′=

{F ′(x) = 1/x , x ≥ 1−G ′(x) = −(−1/x) = 1/x , 0 < x ≤ 1.

We see (ln(x))′ = 1/x for all positive x .

Introduction to ODE


Here is a picture of what we have defined for the two cases:0 < x < 1 and x > 1

ln(x) always increases: just look at the graph. Also,(ln(x))′(x) = 1/x > 0 always here.

Introduction to ODE


ln(1) = 0 as∫ 11 1/tdt = 0.

Since ln(x) is always increasing, there is a unique x valuewhen the area under the curve first hits 1: call this e.

ln(x) is negative for 0 < x < 1 and positive for x > 1.

Introduction to ODE

The Exponential Function

Since ln(x) is always increasing, we know it has an inverse,(ln(x))−1 which we will call exp(x). A little thought tells us therange of ln(x) is all real numbers as for x > 1, ln(x) gets as largeas we want and for 0 < x < 1, as x gets closer to zero, thenegative area −

∫ 1x 1/tdt approaches −∞. By definition then

ln( exp(x) ) = x for −∞ < x <∞; ie for all x .

exp( ln(x) ) = x for all x > 0.

Introduction to ODE

The Exponential Function

We know ln( exp(x)) = x . Take the derivative of both sides:(ln( exp(x)

)′=

(x

)′= 1

Using the chain rule, for any function u(x),(ln( u(x)

)′=

1

u(x)u′(x).

So (ln( exp(x)

)′=

1

exp(x)

(exp(x)

)′.

Using this, we see 1exp(x)

(exp(x)

)′= 1 and so

(exp(x)

)′= exp(x).

Introduction to ODE

Adding Logarithms: both logarithms are bigger than 1

Let’s look at ln(3) + ln(5).

First, rewrite using the definition.

ln(3) + ln(5) =

∫ 3

11/t dt +

∫ 5

11/t dt.

Now rewrite the second integral∫ 51 1/t dt so it starts at 3.

This requires a change of variable.

Let u = 3t. Then when t = 1, u = 3 and when t = 5,u = 3× 5 = 15.So u = 3t and du = 3dt or 1/3 du = dtFurther, u = 3t means 1/3 u = tMake the substitutions in the second integral.∫ 5

1

1/t dt =

∫ t=5

t=1

1

1/3 u

1

3du =

∫ t=5

t=1

3

u

1

3du

=

∫ t=5

t=1

1

udu

Introduction to ODE


Now switch the lower and upper limits of integration to uvalues. We didn’t do this before although we could have. Thisgives ∫ 5

11/t dt =

∫ t=5

t=1

1

udu =

∫ u=15

u=3

1

udu.

Now, as we have said, the choice of letter for the name ofvariable in the Riemann integral does not matter. The integralabove is the area under the curve f (u) = 1/u between 3and 15 which is exactly the same as area under the curvef (y) = 1/y between 3 and 15 and indeed the same as areaunder the curve f (t) = 1/t between 3 and 15. So we have∫ 5

11/t dt =

∫ 15

3

1

udu =

∫ 15

3

1

tdt.

Introduction to ODE


Now plug this new version of the second integral back into theoriginal sum.

ln(3) + ln(5) =

∫ 3

11/t dt +

∫ 5

11/t dt

=

∫ 3

11/t dt +

∫ 15

31/u du

=

∫ 3

11/t dt +

∫ 15

31/t dt

But∫ 151 1/t dt = ln(15) so we have shown

ln(3) + ln(5) = ln(3× 5) = ln(15).

Now let’s do it again.

Introduction to ODE


Let’s look at ln(4) + ln(12).


ln(4) + ln(12) =

∫ 4

11/t dt +

∫ 12

11/t dt.

Now rewrite the second integral∫ 121 1/t dt so it starts at 4.


Let u = 4t. Then when t = 1, u = 4 and when t = 12,u = 4× 12 = 48.So u = 4t and du = 4dt or 1/4 du = dtFurther, u = 4t means 1/4 u = tMake the substitutions in the second integral.∫ 12

1

1/t dt =

∫ t=12

t=1

1

1/4 u

1

4du =

∫ t=12

t=1

4

u

1

4du

=

∫ t=12

t=1

1

udu

Introduction to ODE



11/t dt =

∫ t=12

t=1

1

udu =

∫ u=48

u=4

1

udu.

Now, as we have said, the choice of letter for the name ofvariable in the Riemann integral does not matter. The integralabove is the area under the curve f (u) = 1/u between 4and 48 which is exactly the same as area under the curvef (y) = 1/y between 4 and 48 and indeed the same as areaunder the curve f (t) = 1/t between 4 and 48. So we have∫ 12

11/t dt =

∫ 48

4

1

udu =

∫ 48

4

1

tdt.

Introduction to ODE



ln(4) + ln(12) =

∫ 4

11/t dt +

∫ 12

11/t dt

=

∫ 4

11/t dt +

∫ 48

41/u du

=

∫ 4

11/t dt +

∫ 48

41/t dt

But∫ 481 1/t dt = ln(48) so we have shown

ln(4) + ln(12) = ln(4× 12) = ln(48).

Time for some homework.

Introduction to ODE


Homework 2

Follow the steps above to show

2.1 ln(2) + ln(7) = ln(14)

2.2 ln(3) + ln(9) = ln(27)

2.3 ln(8) + ln(10) = ln(80)

Introduction to ODE

Adding Logarithms: one logarithm less than 1 and one bigger than 1

Let’s look at ln(1/4) + ln(5).


ln(1/4) + ln(5) =

∫ 1/4

11/t dt +

∫ 5

11/t dt.

Now rewrite the second integral∫ 51 1/t dt so it starts at 1/4.


Let u = 1/4 t. Then when t = 1, u = 1/4 and when t = 5,u = 1/4× 5 = 5/4.So u = 1/4t and du = 1/4dt or 4 du = dtFurther, u = 1/4 t means 4 u = tMake the substitutions in the second integral.∫ 5

1

1/t dt =

∫ t=5

t=1

1

4 u4 du

=

∫ t=5

t=1

1

udu

Introduction to ODE



11/t dt =

∫ t=5

t=1

1

udu =

∫ u=5/4

u=1/4

1

udu.

Now, as we have said, the choice of letter for the name ofvariable in the Riemann integral does not matter. The integralabove is the area under the curve f (u) = 1/u between 1/4and 5/4 which is exactly the same as area under the curvef (y) = 1/y between 1/4 and 5/4 and indeed the same asarea under the curve f (t) = 1/t between 1/4 and 5/4. Sowe have ∫ 5

11/t dt =

∫ 5/4

1/4

1

udu =

∫ 5/4

1/4

1

tdt.

Introduction to ODE



ln(1/4) + ln(5) =

∫ 1/4

11/t dt +

∫ 5

11/t dt

=

∫ 1/4

11/t dt +

∫ 5/4

1/41/u du

= −∫ 1

1/41/t dt +

∫ 5/4

1/41/t dt

= −∫ 1

1/41/t dt +

∫ 1

1/41/t dt +

∫ 5/4

11/t dt

=

∫ 5/4

11/t dt

But∫ 5/41 1/t dt = ln(5/4) so we have shown

ln(1/4) + ln(5) = ln(1/4× 5) = ln(5/4).

Introduction to ODE


Homework 3

Follow the steps above to show

3.1 ln(1/2) + ln(5) = ln(5/2)

3.2 ln(1/3) + ln(8) = ln(8/3)

3.3 ln(1/8) + ln(13) = ln(13/8)

Introduction to ODE

General Results

Consider ln(1/7× 7). Since 1/7 < 1 and 7 > 1, this is the caseabove.

ln(1/7) + ln(7) = ln(1/7× 7) = ln(1) = 0.

So ln(1/7) = − ln(7).

In general, if a > 0, then ln(1/a) = − ln(a).

So we can now handle the sum of two logarithms of numbersless than one. ln(1/5) + ln(1/7) = − ln(5)− ln(7) =−(ln(5) + ln(7)) = −ln(35) = ln(1/35)!

So in general, for any a > 0 and b > 0,

ln(a) + ln(b) = ln(a× b).

Introduction to ODE

General Results

Now we can do subtractions:

ln(7)− ln(5) = ln(7) + ln(1/5) = ln(7/5)

ln(1/9)− ln(1/2) = − ln(9)− ln(1/2) = − ln(9/2) = ln(2/9).

So in general, for any a > 0 and b > 0,

ln(a) − ln(b) = ln(a/b).

Introduction to ODE

General Results

What about powers? Start with positive integer powers:

ln(42) = ln(4× 4) = ln(4) + ln(4) = 2 ln(4).

ln(43) = ln(42 × 4) = ln(42) + ln(4) = 3 ln(4).

Continuing, we see if n is a positive integer, ln(4n) = n ln(4).

Now do negative integer powers:

ln(4−1) = ln(1/4) = −ln(4).

ln(4−2) = ln(1/42) = −ln(42) = −2 ln(4).

ln(4−3) = ln(1/43) = −ln(43) = −3 ln(4).

Continuing, we see if n is a negative integer, ln(4n) = n ln(4).

Introduction to ODE

General Results

What about fractional powers?

ln(4) = ln( (41/4)4 ) = 4 ln( 41/4 ) and so ln( 41/4 ) = 1/4 ln(4).

ln( 45/3 ) = ln( (41/3)5 ) = 5 ln( 41/3 ). Butln( 41/3 ) = 1/3 ln(4). So combining, ln(45/3) = 5/3 ln(4).

In general, we see if p and q are any positive integers, Thenfor any a > 0,

ln(ap/q) = (p/q)× ln(a).

Introduction to ODE

General Results

If r is any real number power, we know we can find a sequence ofrational numbers which converges to r . We can use this to seeln(ar ) = r ln(a).√

2 = 1.414214.... So the sequence of fractions that convergeto√

2 is {p1/q1 = 14/10, p2/q2 = 141/100, p3/q3 =1414/1000, p4/q4 = 14142/10000, p5/q5 =141421/100000, p6/q6 = 1414214/1000000, . . .}.So

lim ln(4pn/qn) = lim (pn/qn) ln(4) −→√

2 ln(4).

Define 4x = exp(ln(4) x). Since exp(x) is continuous, so is

exp(ln(4)x). So 4x is continuous and lim 4pn/qn −→ 4√2.

Thus, since ln is continuous at 4√2, we have

lim ln(4pn/qn) = ln( 4√2 ) −→ ln( 4

√2 ) =

√2 ln(4).

In general, for any power r , ln(ar ) = r × ln(a).

Introduction to ODE

General Results

The rules for the natural logarithm:

The sum rule: ln(x) + ln(y) = ln(x y) for all positive x andy .

The difference rule: ln(x) − ln(y) = ln(x/y) for all positivex and y .

The power rule: ln(x r ) = r ln(x) for all positive x and anypower r .

Introduction to ODE

The Exponential Function Rules

Let u = ln(x) and v = ln(y). Then

x = exp(u) and y = exp(v).

Let w = exp(u + v). Then by definition, ln(w) = u + v .

But u + v = ln(x) + ln(y) = ln(x y).

So ln(w) = ln(x y).

Then by definition, w = x y = exp(u) exp(v).

So in general exp(u + v) = exp(u) exp(v).

Introduction to ODE


Let u = ln(x) and v = ln(y). Then

x = exp(u) and y = exp(v).

Let w = exp(u − v). Then by definition, ln(w) = u − v .

But u − v = ln(x) − ln(y) = ln(x/y).

So ln(w) = ln(x/y).

Then by definition, w = x/y = exp(u)/exp(v).

Further, exp(0) = 1 because ln(1) = 0, soexp(x − x) = exp(0) = 1 implying exp(x) exp(−x) = 1.Hence, exp(−x) = 1/exp(x).

So in general exp(u − v) = exp(u)/exp(v) = exp(u) exp(−v).

Introduction to ODE


Let u = ln(x) and r be any power. Then

x = exp(u).

Let w = (exp(u))r . Then by definition,ln(w) = r ln(exp(u)) = r u.

Or w = exp(r u).

So in general

(exp(u)

)r

= exp(ru).

Introduction to ODE


The rules for the exponential function:

The sum rule: exp(x + y) = exp(x) exp(y) for all x and y .

The difference rule: exp(x − y) = exp(x)/exp(y) for all xand y . Further, exp(−x) = 1/ exp(x).

The power rule:

(exp(x)

)r

= exp(rx) for all x and any

power r .

Introduction to ODE


Let’s graph exp(x) and ln(x) on the same graph.

Introduction to ODE

Graphing Logarithm and Exponential Functions

Logarithm Functions

Let’s graph ln(2x) and ln(x) on the same graph.

Introduction to ODE


Logarithm Functions

Let’s graph ln(3x) and ln(x) on the same graph.

Introduction to ODE


Logarithm Functions

Homework 4

4.1 Graph ln(4x) and ln(3x) on the same graph.

4.2 Graph ln(0.5x) and ln(1.5x) on the same graph.

4.3 Graph ln(x) and ln(3x) on the same graph.

Introduction to ODE


Exponential Functions

Let’s graph e−2t and e−t on the same graph.

Introduction to ODE



Let’s graph et and e1.5t on the same graph.

Introduction to ODE



Homework 5

5.1 Graph e−t and e−1.4t on the same graph.

5.2 Graph e−2.3t and e−1.5t on the same graph.

5.3 Graph e−0.8t and e−1.2t on the same graph.

5.4 Graph et and e1.4t on the same graph.

5.5 Graph e2.3t and e1.5t on the same graph.

5.6 Graph e0.8t and e1.2t on the same graph.

Introduction to ODE

Half Lifes and Doubling Times

Let’s examine the exponential decay/ growth problem from a newperspective. Consider the function x(t) = 100 e−2t . How longdoes it take for this function to decay to the value 50? Since it hasdropped to 50% of its original value, we call this the half - life ofthe function.

Let t 12

denote the time at which x has decayed to 50. Then,

we want 50 = 100 e−2 t 1

2

Solving, we have

1

2= e

−2 t 12

ln(1

2) = −2 t 1

2

− ln(2) = −2 t 12

t 12

=ln(2)

2

Introduction to ODE


Consider the function x(t) = 400 e−5.4t . How long does it takefor this function to decay to the value 200?

Then, we want 200 = 400 e−5.4 t 1

2

Solving, we have

1

2= e

−5.4 t 12

ln(1

2) = −5.4 t 1

2

− ln(2) = −5.4 t 12

t 12

=ln(2)

5.4

Introduction to ODE


Consider the function x(t) = 400 e−5.4t again How long does ittake for this function to decay to the value 100? Call this time t 1

4.

Then, we want 100 = 400 e−5.4 t 1

4

Solving, we have

1

4= e

−5.4 t 14

ln(1

4) = −5.4 t 1

4

−2 ln(2) = −5.4 t 14

t 14

= 2ln(2)

5.4= 2 t 1

2

Introduction to ODE


Summary: For x(t) = Ae−rt for a positive r , we have

At t 12

ln(2)r , x has decayed to 50% of its original value.

At 2 t 12

ln(2)r , x has decayed to 25% of its original value.

At 3 t 12

ln(2)r , x has decayed to 12.5% of its original value.

Introduction to ODE


For exponential growth, we have a similar concept: the doublingtime, td . Summary: For x(t) = Aert for a positive r , we have

At td , x has increased by a factor of to 2 times its originalvalue.

At 2 td , x has increased by a factor of to 4 times its originalvalue.

At 3 td , x has increased to 8 times its original value.

Introduction to ODE


Let’s graph e−.5t and e−t on the same graph using half lives as ourtime units.

Introduction to ODE


Let’s graph et and e2t on the same graph using doubling times asour time units.

Introduction to ODE


Homework 6

6.1 Graph e1.2t and e1.4t on the same graph using doubling timesas our time units.

6.2 Graph e−1.2t and e−1.4t on the same graph using half lifetimes as our time units.




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