The First Fundamental Theorem of Calculus在线视频

Today we study the Calculus (I)

The topic is

The First Fundamental Theorem of Calculus

If f(x) is integrable on the

closed interval a and b

For any x belongs to

the closed interval

And then

how about this expression

f(t) dt

the lower limit is a

How about the upper limit

Pay attention to t

is changeable

t changeable means

the corresponding upper limit is x

x is changeable

the corresponding value

the definite value is changeable

We can write it in this form

a to x

f(x) dx

This equals φ(x)

capital φ(x)

Why

It is a definite integral

Lower limit is a

upper limit is x

x is changeable means

the corresponding value is changeable

The value is a function of x

We give a new name about capital φ(x)

means accumulation function

Now look at

we must distinguish

the independent variable x

This means independent variable x

and the integral variable t

Here it is our integral variable t

x is our independent variable

Of course we ask us some questions

How about the relationship

between f(x) and capital φ(x)

If f less than or equal to g

How to compare the values of a b f(x)

and a b g(x)

The corresponding definite values

If the definite integrals f(x) dx

bounded or not

So now we try to answer the questions

Look at the accumulation function

φ(x) equals a x f(t) dt

t is our integral variable

The geometrical meaning of this expression means

the area function of the blue part

as shown in the following figure

Look at the figure

y equals f(x)

x equals a

x equals b and the x-axis

This makes up

the corresponding curve trapezoid

How about the corresponding

area of the blue part

The blue part means

the lower limit is a

the upper limit is x

How about the area of the blue part

Now look at

if f(x) is a continuous function

on the closed interval a, b

the accumulation function φ(x) equals a x f(t) dt

Pay attention to

t is our integral variable

x

some fixed number

is a derivative function

on the given closed domain

And the derivative of this function

capital φ(x) means

what is derivative

Yes

prime

So φ prime x

according to the definition

equals dφ over dx

Taking dφ

what is our φ

Yes

φ is this expression

so equals φ

So dφ dx equals what

Equals

pay attention to

this is our integral function

x substituted for t

so equals f(x)

Pay attention to the first term

and the end term

Capital φ prime x equals f(x)

it is our final answer

It is a theorem

We need to prove

How to prove

Now look at

how to get the φ prime

Do you remember 3 steps

for the derivative

Step 1

get the increment of the function

Yes

so we try to write the φ x plus Δx

In order to get the increment of φ(x)

Yes here

φ(x) equals a x

φ(6) equals a to 6

φ(8) equals a to 8

So now φ x plus Δx

equals from a to x plus Δx

Pay attention to

they are the same expression

So now we write it

Δφ means Step 1

Find the increment of the function

equals φ x plus Δx minus φ(x)

Go on

we can get

Δφ equals the difference of the two expressions equals

look at the figure

how about φ x plus Δx

Yes

φ x plus Δx means this area

of the curve trapezoid

φ(x) means the blue corresponding area

so equals

equals a x plus Δx

minus from a to x

According to our expression of capital φ

equals

do you remember

the definite integral property

interval addition

So we can write it this form

Look at the figure

The first term from a to x plus Δx

the second term from a to x

The difference means x to x plus Δx

Of course you can write it

in the interval addition

According to the Mean Value Theorem

for the Integral

we can write it

equals some value of function f(x)

at one fixed point ξ time

bracket b minus a

In the special case

upper limit minus lower limit equals Δx

So f(ξ) time Δx

Pay attention to x

f(x)

ξ is between a and b

means ξ is between x and x plus Δx

So we have

do the step 2

Do the ratio Δ over Δx

Δ means this term

over Δx

delete this term

get the f(ξ)

ξ is some fixed number between

left end point x and the right end point x plus Δx

f is continuous on the interval a and b

Do you remember

the definition of continuous

This means if Δx tends to 0

Δx tends to 0

at this time

ξ tends to x

This means ok

It is our x

This point is x

and this point is x plus Δx

Any point ξ

Δx tends to 0

ξ tends to the fixed point x

So we have ξ tends to x

And then

we have the limit φ prime x

equals limit Δφ over Δx

Condition Δx tends to 0

This means

equivalently

ξ tends to x

f(ξ)

Why

The condition variable is changeable

because

this means Δφ over Δx

We talk something about Δx

So the condition is Δx tends to 0

Now the limit function has become f(ξ)

So we will talk something about ξ

ξ tends to x

and f(x) is continuous function

So the final result equals f(x)

means the value of function f

at the point x

Try to understand the theorem

Pay attention to

Δx tends to 0

ξ tends to x

Now look at the corollary of the Theorem A

If f(x) belongs to

the continuous function family

means f(x) is a continuous function

on the closed interval a and b

And g(x) is a derivative function

on a and b

At this time

for every x

we have the first property

d this another accumulation function

over dx

Do you remember

if the order of

the lower limit and the upper limit

is changeable

We can get some sign

so equals minus f(x)

Try to remember the sign minus f(x)

Now look at 2

d a g(x)

the upper limit is g(x)

means d some expression over dx

d function of g over dx

Do you remember

the Chain Rule

So we can write it

d this expression

over dg time dg

and over dx

The final result equals f g

Doesn't finish

Why

Now the upper limit is

a compound function according to basic f g

time g prime x

So remember this

f g means d some accumulation function

over d g time dg over dx

So our final result means

f g time g prime x

Don't forget this term g prime x

Look at 3

d another accumulation function

over dx

Now look at

what is our integral variable

t is our integral variable

How about g(x)

g(x) can be looked as some values for t

means fixed values

So g(x)

we can take out

Write it in this form

g(x)

x is a variable

doesn't have any relationship about t

So we can write it

this expression prime

d this expression over dx

Now look at

using the product rules of functions

f time g

f prime g plus g prime f

So equals the first term g prime time

this expression plus g(x)

copy

this prime equals f(x)

So it is our final result

Try to understand every step

Now look at Theorem B

If f and g are integrable

on the given closed interval a and b

and f(x) less than or equal to g(x)

for all x in the a and b

And then we have

the integral a b f(x) dx

ess than or equal to a b g(x) dx

Two conditions

the first means a the left point

b the right point means

a less than or equal to b

Second

f less than or equal to g

At this time

we have this expression

So now prove this

A given condition

f less than or equal to g

we can get

f minus g

non-positive means

less than or equal to 0

So we can get

the corresponding

a b f minus g dx

less than or equal to 0

condition a less than or equal to b

Why

The geometrical meaning means the corresponding

area of the curve trapezoid

f minus g

So less than or equal to 0

We can get

the first term

corresponding integral

less than or equal to the second term definite integral

Try to remember the result

Now look at Theorem C

If f is integrable

on the given closed interval

and f between little m and capital M

for every x in the closed interval

and then

this corresponding definite integral

between the two numbers

The left little m time b minus a

the right capital M time b minus a

What is our b minus a

The difference of a and b

so now

because f(x) between little m and capital M

we can do the same computation

Get the definite integral

So every term

do the same computation

Look at the left

m taken out

we can look at

now integral function is 1

So little m time b minus a

in the similar meaning

the right integral capital M taken out

At this time

integral function become 1

so 1 time b minus a

So together

the middle term definite integral

large than or equals m bracket b minus a

less than or equals

capital M bracket b minus a

So try to understand this result

Look at this Theorem D

If special case

a equals b

at this time

a b f(x) dx means

at this time

the area of some segment

not some big region

So at this time equals 0

Case 2

if a larger than b

at this time

a b f(x) dx equals minus b a f(x) dx

means there are some relationship

between the orders of a and b

and the values

Change the order

we can write it

the sign minus

so equals minus from b to a f(x) dx

Case 3

a b f(x) plus or minus g(x) equals

the definite integrals plus or minus

Case 4

k f(x) dx

k is any constant

So we can take out

k times from a to b f(x) dx

k is some constant

According to the theorems

we have studied

We will do some examples

Now look at Example 1

If f(x) equals from 0 to x

t over t square plus t plus 1 dt

try to find f prime x

How to do

Yes

by Theorem A

It is a traditional

corresponding accumulation function

So now

f prime x equals

this is our special f(t)

This is our special φ(x)

So φ prime means f prime equals

every t becomes x

So the final answer is

x over x square plus x plus 1

by the theorem A

Now look at Example 2

If f(x) equals x cube to x square e t dt

try to find f prime x

It is another accumulation function

by the theorem A

according to the interval addition

We write it

x cube

Write it 0

0 means a equals 0

Now write it this term

Change the order

because we are familiar with from a to g

So change the order

Write it minus

Plus becomes minus

from 0 to x cube e t dt

Now we will do the same computation

f prime x equals

the first term prime

means d expression dx

minus d expression dx

How to do this

Yes this is our one theorem

From a to g

do you remember

equals f g time g prime

So we can write it

f g means e x square time g prime

g prime means x square prime equals double x

minus

the similar meaning

e g means e x cube times g prime

3 x square

Together final step equals

double x time e x square

minus 3 x square time e x cube

Try to understand step

according to interval addition

Try to understand this

f g time g prime

So now look at Example 3

Please compare the two values of

from 0 to minus 2

and 0 to minus 2 between e x and x

How to do

Compare the different values

by the theorem B

Now we can let

f(x) equals e x minus x

We try to get the signs of f(x)

Positive or negative or zero

So now x between a and b

Do you remember our Theorem B

Condition means from a to b

So we write it

x belongs to the closed interval

minus 2 to 0

So now

because x changeable in this interval

f(x) is positive

At this time

we can get

corresponding from a to b f(x) dx

The definite integral is positive

Two conditions about Theorem B

The first

a less than or equal to b

a is minus 2

b equals 0

At this time

f larger than g

this integral large than 0

So we can get

the corresponding properties

Minus 2 to 0

larger than this expression

But compare our answer

and our question

Our question asks us from 0 to minus 2

So we need to change the orders of the

upper limit and lower limit

So now we change the order

So we can get

0 to minus 2 e x dx

Change the order time minus 1

change the order time minus 1

so less than

corresponding

less than this result

So the question is ok

So now compare

remember the condition of Theorem B

So now look at Example 4

Try to evaluate this expression

Evaluate means

what is our biggest value

maximum value

minimum value

How to do

Try to find little m and big M

capital M

So

do you remember the Theorem C

Little m and capital M

So we will talk something

about the function f(x)

In this question

f(x) equals 1 over 3 plus sinx cube

x between a and b

means x between 0 and π

In the closed interval

sinx cube

in high school

we have known corresponding range

between 0 and 1

Of course

cube between 0 cube and 1 cube

sinx cube plus 1

so we can get

this term plus 3

this term less than or equal to 1 over 3

larger than or equal to 1 over 4

And then we can get

the corresponding definite integral

larger than or equal to 1 over 4

is our little m

1 over 3 is our capital M

So now look at this

1 over 4 times bracket b minus a

means π minus 0

Similar meaning

this term 1 over 3

times π minus 0

So together we get this result

The definite integral

between the two values

Now we can give you some summary

about our theorems

Remember Theorem A

d some accumulation function

over dx equals f(x)

means t become the upper limit f(x)

Theorem B

f(x)

g(x)

the relationship is given

less than or equal to

Remember the condition

a is less than or equal to b

At this time

the definite integrals satisfying this expression

Look at Theorem C

f(x) between little m and capital M

the corresponding definite integral

between little m time

the distance of limits

means b minus a

less than or equal to

capital M time b minus a

Now we will use the theorems to do some questions

Look at Question 1

If f(x) equals sinx to 2

1 over 1 plus t square dt

try to find f prime x

How to find

We know some theorems about

from a to g(x)

So now it gives you idea

pay attention to

now

lower limit is sinx

So we write it

f(x) equals

change the order of the limits

equals minus 2 to sinx

1 over 1 plus t square dt

At this time

gives substitution

Let u equals sinx

So this expression become

minus from 2 to u

sinx equals u

1 over 1 plus t square dt

As to upper limit

it is our u

And f prime x means

do the same operations

f prime x

f prime this expression

so equals d this expression

You should write it dx

Why

You write du

Over du time du

So we can write time du dx

Original d this expression over dx

but this expression

there are some relationship about u

so we over du time du

Now we write in this form

the question becomes very easy

Look at this expression

this means from a to u d du

this expression means du over dx

so equals

minus copy

1 over 1 plus u square

u prime

what is our u

Here

substitution u prime

means sinx prime

means cosx prime

So u taking equals this expression

It's our f prime for x

Now look at

Question 2

Try to find the limit

x tends to 0

1

cosx

e minus t square dt

over x square

How to understand this question

If x tends to 0

this term means cos0

cos0 tends to 1 means 1 1

So this expression

limit is 0

x tends to 0

Still 0 means the limit of the type

0 over 0

So

through the analysis

we know

this is an indeterminate form

of the type 0 over 0

We don't know the limit

So try to use the L' Hospital Rule

How to use this

According to

d this expression over dx

we can get

this means

minus from 1 to cosx

Why

Because change the order

cosx upper limit

1 becomes lower limit

lower limit becomes upper limit

So we can write it 1 minus

At this time equals

minus copy

e minus t square equals

e minus cosx square time

cosx prime

So together

cosx prime we know equals

minus sinx

Minus time minus

together become positive

so equals sinx time this expression

At this time we know ok

This prime equals this expression

Do you remember

in the L' Hospital Rule

we know f over g

the limit

ok

Tends to f prime and g prime

They are equal

They are the equal limits

So now we can write it

limit

original expression

the first term prime

equals this expression

x square prime equals

double x

x tends to 0

They are the same variables

So sinx over x

the limit equals 1

And cosx

the limit equals 1

we can get

1 over double e

So our final result

is 1 over double e

Look at this question

If f(x) is a continuous function

on the closed interval a and b

and f is bounded

f less than 1

in this question

Prove double x minus this expression

equals 1

How about this expression

It is some accumulation function

Double x minus some accumulation function

this expression equals 1

It's the equation

The equation and the corresponding interval

from 0 to 1

there is only one solution

It is the equation

There is some solution and only one solution

How to prove

So now look at our answer

Let capital F equals

the left side minus the right side

Try to find

using the Zero Theorem

So left side minus right side

this equals this

How to compare

How to think of this question

Capital F prime equals

this term equals 2

this term equals f(x)

so capital F prime equals

2 minus f(x)

The given condition

f(x) less than 1

So 2 minus f(x) is a positive number

Positive number means capital F(x)

is a monotonically increasing function

means increasing function

on the corresponding closed interval 0, 1

is an increasing function

Now we consider

the values of the two end points

Capital F(0)

F(0) means x equals 0

0 0

double 0 minus 0 0

0 0

0 minus 0 minus 1

so equals minus 1

minus 1 negative so less than 0

Now consider another value

about right end point

Capital F(1)

capital F(1) double minus 0 to 1 minus 1

together equals 1 minus 0 to 1 f(t) dt

Do you remember

the Mean Value Theorem

So we can write it

copy y minus

equals f(ξ) time b minus a

using the Mean Value Theorem

f(ξ) time b minus a

means 1 minus 0

Now consider the given condition

f(x) less than 1

So this result positive

so large than 0

Ok now

Condition 1

capital F is an increasing function

means monotonically increasing function

Condition 2

the two points

end points values

less than 0

and another is large than 0

According the Zero Theorem

we know something

Of course you can

use another knowledge to get F(1)

From 0 to 1

1 dt

How to write it

Yes

1 taking out time 1 minus 0

So this definite integral equals

number 1 minus this

According to the property

what's property

Linear property

we can write it

0 1 0 1

The same upper limits and lower limits

So write it 1 minus f(t)

Do you remember f(t) values

Yes

f(t) less than 1

So 1 minus f(t) is a

positive between 0 and 1 positive number

So we know another idea

Anyway it is ok

we know the two values

F(0) less than 0

capital F(1) larger than 0

And then

according to the Zero Theorem

increasing function

and the two end points

one is positive

another is negative

So there is only one solution

on the given interval from 0 to 1

Look at the last question

Try to find the limit

Yes

it's very difficult

How to understand

Now we look at

how about this expression

x tends to 0

0 time 1 minus cos0

0 time 0 equals 0

So this is 0

The limit equals 0

Look at this term

If x tends to 0

0 to 0

We don't care this expression

Why

Anyway it's some definite integral

from 0 to 0 equals 0

Together

the limit becomes 0 over 0

It's indeterminate type

how to do

Using the theorem in this class

The original limit equals

double limit x tends to 0

How to do

From 0 to 0

0 time 1 minus cos0

means 0 over 0

using the L' Hospital Rule

So this term

we can get

1 minus cosx

equivalent to x square over 2

x tends to 0

Together this term

prime

we can write it

x time something

equals 2 x cube

And then we can write it

still 0 over 0

using the L' Hospital Rule

we can write it

x cube prime

equals n x n minus 1

means 3 x square

Look at this term

x prime

prime for x means

this term u together

original this expression is

a function of u

u is replaced by x

So this term becomes

from 0 to x square

And then

we compare the limits

This tends to 0

Of course using the L' Hospital Rule

one time again

We can write it

2 over 3

it is our constant taken out

At this time

x square prime equals double x

Now consider this expression

from 0 to x square

so we can write it

arctan 1 plus t

t using the x square

means arctan 1 plus g time g prime

g is our x square

so time double x

Now consider this form

double x

we can omit

And then

left arctan 1 plus x square

1 plus x square

x tends to 0

means arctan1

so 2 over 3 time π over 4

together equals π over 6

So the final result

the limit equals π over 6

Now consider the question

Pay attention to 2 points

Point 1

it is the 0 over 0 type limit

Another

using 2 times of L' Hospital Rule

So now class is over

See you next time