The Second Fundamental Theorem of Calculus and the Method of Substitution在线视频

Today we study the Calculus (I)

The topic is The Second Fundamental Theorem of Calculus

and the Method of Substitution

So now look at

we have known some knowledge

in the physics

Now look at

some distance can be defined as

from the definite integral

time T1 to time T2 v(t) dt

v(t) is our integral function

It is our velocity function

means the distance equals

the definite integral

from T1 to T2

How to write it

Yes

it can be expressed as another form

Another form means

the distance at the Time 2

minus the distance at Time 1

This means

the difference of the two distances

In theory

the two expressions are equal

Now

maybe you ask yourself

is there any relationship between

the expression and the difference of the distances

How about the relationship

Yes

look at the distance

The distance at Time 2

minus the distance at Time 1

How about this

This means some distance

about velocity from T1 to T2

So in theory

yes they are equal

means the distance

from Time 1 to Time 2

How to find it

the relationship

Maybe you ask yourself

why they are equal

if you can find some relationship

between v(t) and s(t)

Yes

v is our velocity function

How about s

Yes

s is our distance function

we can write it

s prime t equals v(t)

s prime t equals v(t)

this means

s is an anti-derivative function of v

It's very good

If for every definite integral

always find the anti-derivative function

is very good

This means we can show

the kind of derivative integral

So now

we look at the theorems

If little f(x) be continuous

on the closed interval from a to b

and let the capital F(x)

be any anti-derivative of f(x)

on the corresponding a and b

the corresponding definite integral

from a to b f(x) dx

equals the difference of the two values

anti-derivative functions from b to a

How to understand

Yes

now we will give the concrete proof

tell you it is always true

If it is always true

we can find

the value of definite integral

So now let φ(x) equals a, x f(t) dt

Look at

where x

From a x

how about

It is a definite integral

In theory

it is a function of x and a

a is fixed

So this expression means

some function of x

Give it a new name

means capital φ(x)

So it is an anti-derivative of f(x) too

Because both capital F(x) and capital φ(x)

are anti-derivatives of little f(x)

why

Yes

capital F(x) is an anti-derivative function

It's the given condition

How about φ(x)

φ(x) we know

φ(x) is anti-derivative of f(x)

So there is some fixed C

satisfying

capital F(x) minus φ(x) equals C

C means some fixed real numbers

How about the real number

How about the variable

Yes

the variable between the given domain

means the closed interval from a to b

It is always true for every x

So now pay attention to case 1

if x equals a

x equals a

the formula is true

Taking the value of x

this means

capital F(a) minus φ(a) equals C

It's true

Yes for every x

it's true

Now look at

how about φ(a)

φ(a) equals 0

why

Look at the given formula

φ(a) means equals from a to a

From a to a

this means the area of the segment

a equals b

special case

So φ(a) equals 0

If φ(a) equals 0

we can get C

So C equals capital F bracket a

C is some fixed number

So we have

capital F(x) minus capital φ(x)

equals capital F(a)

Now look at the case 2

if x equals b

it is always true

we taking x equals b

At this time

capital F(b) write it

minus φ(b)

write it equals C

How about C

Yes here

C equals F(a)

So we can write it this form

F(b) minus φ(b) equals F(a)

We write it

φ(b) equals F(b) minus F(a)

So we can get this formula

How about φ(b)

So namely φ(b)

how to find it

Yes

the first line

φ(b) equals from a to b

So φ(b) means the definite integral

So φ(b)

this term we write it

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

So we have proved this

Try to understand the formula

The definite integral equals

the anti-derivative function

capital F(b) minus capital F(a)

Now look at the Theorem B

We have known

we have from a to b f(x) dx

equals capital F(x)

given

from b minus a

Capital F is our anti-derivative function

so equals

capital F(b) minus capital F(a)

Try to remember

So the problem of finding the integral problem

can be translated into

finding the anti-derivative function

So how to compute this integral

become how to find capital F

So we try to find the capital F

Now

Note one

if a less than or equal to b

at this time

the definite integral equals

capital F(b) minus capital F(a)

We have proved

yes

this formula is true

If a larger than b

this definite integral

equals F(b) minus F(a)

still holds

So we don't care a and b

which one is bigger

the formula is always true

Now look at another theorem

Substitution idea means

Substitution Rule for the Definite Integral

If f(x) is a continuous function

on the closed interval from a to b

f is a continuous function

and the function

x equals φ(t)

satisfying the following two conditions

x is φ(t) means

x is a function of t

satisfying Condition 1

φ(α) equals a

φ(β) equals b

this means x equals a

equals φ(α)

a

corresponding t equals α

In the similar meaning

x equals b

corresponding

t equals β

Look at Condition 2

φ(t) has a continuous derivative

on the given domain

from a to b

and the corresponding range of φ means

function range is a subset of a and b

means we write it

have some mathematical meaning

And then we have

the definite integral from a to b f(x) dx

How to write it

Yes

change the variable

means substitution

How to substitute

f

f

what

Yes

f(x)

x equals φ(t)

So we can write it φ(t)

d what

dx

We can do the same computation

dx equals dφ

dφ equals what

Yes

φ prime t dt

This is our dx

This is our f(x)

Now pay attention to

the integral variable

Look at the left

the left

x is our integral variable

Look at the right

what

Yes

t

t is a new variable

our new integral variable

So x from a to b

corresponding

t from what

Yes

a

corresponding α

b

corresponding β

So now t from α to β

Pay attention to the order

ok

a corresponding α

b corresponding β

Yes

the original integral variable is x

substitution becomes t

So try to remember the formula

So if α larger than β

the corresponding theorem still holds

means we don't care α and β

the values

We care the order

don't care the values

a corresponding α

b corresponding β

So now look at Example 1

Try to use the above theorem

finish the following examples

Try to find the definite integral

from 0 to half π

it is our integral function

dx

How to write it

Yes

try to find the corresponding

anti-derivative function capital F

So we write it

this integral equals

double cosx

the anti-derivative equals double sinx

sinx

pay attention to the signs ok

minus cosx

minus 1 means minus x

So it is our anti-derivative function

from half π to 0

Taking the values half π

taking the value 0

Final value equals 3 minus half π

So now try to use the formula

It's very useful

Using the idea we now look at

Example 2

Try to

now

define A to be the area

under the curve of y equals sinx

above the x-axis

and between the vertical lines

x equals 0

and x equals π

Try to understand

Yes

this means our a

this means our b

This means our f(x)

Yes

the question means

evaluate the area of A

Evaluate the area of A

How to draw the figure

How to draw the region

Now look at the figure

Yes

the figure

The curve means y equals sinx

The curve is y equals sinx

It is our x-axis

x equals 0

x equals π

the area according to

geometrical meaning of definite integral

corresponding A

equals some definite integral

What

Yes

f(x) is sinx

a equals 0

b equals π

Yes

π is some real number

so equals integral function

from 0 to π

Now we will be very happy

Yes

the area equals

some definite integral values

How to do

Do you remember the formula

Try to find the anti-derivative

sinx

yes

equals minus cosx

So it is our

the capital F(x)

minus cosx

so equals minus sign

taken out means

cosπ minus cos0

cosπ equals minus 1

cos0 equals 1

Minus 1 minus 1

minus

equals positive 2

Of course you can check by yourself

if the answer is negative number

it's wrong

Why

Because the area must be non-negative

so your answer must be non-negative

If equals some negative number

means you are wrong

Try to understand the geometrical meaning

Now look at Example 3

Let f(x) equals double x

if x is between 0 and 1

and another between 1 and 2 equals 5

evaluate the definite integral

between 0 and 1

How to understand the f(x)

Yes

we can draw the corresponding figure

Look at the figure

Between this

this is our

yes

between 0 and 1

equals what

Yes

equals double x

Between 1 and 2

1 and 2 equals 5

5 means a constant ok

Include 2

doesn't include 1

So include this point

this point doesn't include

So we have corresponding this figure

Between 0 and 1 equals double x

between 1 and 2

this segment

ok

So it is some piecewise function

How to do

Yes

do you remember

the property of definite integral

Interval additivity

so we can write it

Evaluate the definite integral

how to write it f(x)

There are two cases

So you write it every case

So we can write it

from 0 to 1 plus what

Yes

0 to 1 plus 1 to 2

So plus 1 to 2 f(x) dx

It is always true

Why

According to our property

interval additivity

so 0 1 1 2

Between 0 and 1

yes

equals double x

So we can write it double x

Between 1 and 2 equals what

Equals 5

we can write it 5

So write it double x plus 5

So now we can compute the final result equals 6

Try to understand the question

using what idea

Yes

x to the piecewise function

using the interval additivity property

So now look at Example 4

Try to find

the definite integral

It is another special case

Why

a is given

minus 2

b is given

positive 2

Now

our integral function not f(x)

We write it special sign

maximum x and x square

means if x is changeable

the corresponding integral function

taking the maximum value of x and x square

How to write it

Thinking in your mind in Example 3

What

Yes

piecewise function

So try to delete the three letters m a x

How to delete

Write it piecewise function

So now look at

as shown in the following figure

look at the figure

y equals x square

our curve

How about y equals x

Yes

y equals x is our straight line

Together

which is bigger

which is smaller

Yes

in left

this is bigger

ok

In this part

this is bigger

And this part it is bigger

So different x values

different expression

According to the given domain

minus 2 to positive 2

We write it three cases

Look at x equals minus 2

between x equals 1

and x equals 2

There are three cases

Different cases

different expressions

Write it

f(x) equals maximum value

Maximum value

delete m a x

How to delete

Write it three cases

Look at case 1

x between minus 2 and 0

At this time

which one is above

which one is maximum

At this time equals the expression

so equals x square

Similar meaning

between 0 and 1

Between 0 and 1

which is bigger

Yes

the right line

so equals x

The last interval between 1 and 2

Between 1 and 2

which one is bigger

Yes

the function

function means x square

So now the question integral function f(x)

become our piecewise function

In the similar idea in Example 3 is ok

So the original integral equals

minus 2 to 0 plus 0 to 1 plus 1 to 2

How about the property

The interval additivity

1 2 3

so 1 2 3 intervals

Compute every definite integral

Submission the values

the final result equals 11 over 2

So if you

have the special minimum or maximum

try to write it into the corresponding

piecewise function

and do the definite integral

So now look at Example 5

Try to find

between 0 and half π sinx cube dx

sinx cube dx

how to do

Do you remember

the substitution idea

Yes

the Method 1

we will use the substitution method

Write it sinx cube

pay attention to cube

Cube is an odd number

How to write the odd number

Write sinx square time sinx

One doesn't change

Another becomes differential variable

So we write it

Copy the definite signs

from 0 to half π

Look at the first term

sinx square equals

1 minus cosx square

How about sinx dx

Yes

equals dcosx

Pay attention to the minus

must write the minus

Otherwise

your answer is wrong

dcosx equals minus sinx

so we can write it

substitution

let t equals cosx

Why

In the expression

we only have the expression cosx

So let t equals cosx

The original integral becomes what

t equals cosx

Pay attention to

in this expression

x is our integral variable

So x equals 0

x equals 0 yes

x equals 0

t equals cos0

cos0 equals what

Yes equals 1

So t equals 1

Another

x equals half π

means upper limit

x equals half π

t equals cos half π

cos half π equals 0

Yes

so t equals 0

We have the idea

what idea

Substitution

x from 0 to half π

t from 1 to 0

Pay attention to the order

We don’t care the value

We only care the order

So equals 1 minus t square dt

What is our variable

t

so taking the value of t

t from 1 to 0

ok

Pay attention to the order

so from 1 to 0

How about this definite integral

Become very easy ok

it's a polynomial integral

So the final result

equals 2 over 3

It is our final result

Pay attention to the new variable t

If you still write it

from 0 to half π

It’s wrong

Pay attention to new variable t

the corresponding new upper limit

What

Yes

corresponding

new upper limit is 0

The new lower limit is 1

So pay attention to

the new variable limits are changeable

Now look at Method 2

Maybe we don't use the substitution

We can finish Question 2

Look at Method 2

Write it

the first step they are equal

sinx square time sinx

Write it

yes

this term doesn't change

Now we don’t use the substitution idea

We write it

what

1 dcosx

cosx square dcosx

We write it the anti-derivative

using cosx directly

so equals minus

minus copy

1 dcosx equals cosx

minus u square du

1 over 3 u cube

equals this

Similarly

we can get

the final answer 2 over 3

Compare the two different methods

The final answers are equal

always equal 2 over 3

Why

Because it is our definite integral

In fact it is some limit

The limit means some real numbers

They are equal

Different idea may be different methods

Every method is ok

But you have to

try to compute the correct answer

So now

look at our summary

We have the first formula

the integral from a to b f(x) dx

equals capital F(x) from b to a

Capital F(x) is our anti-derivative function

means the distance of F(b) and F(a)

So the definite integral

become

try to find capital F

Try to understand

Look at another substitution idea x

Yes

x equals φ(t) dx

φ prime t dt

t is a new variable from α to β

This is our

the two important theorems

Try to remember

So now using the two theorems

to do our questions and answers

Now look at Question 1

Question 1

try to find the definite integral

Pay attention to

the upper limit and lower limit

How to do

Maybe it is difficult

Why

Because our integral function become difficult

It is so compounded

How to do

Yes

try to find the anti-derivative

So now look at

the original integral equals

copy

we doesn't change

Yes

pay attention to y over x

change into differential

d lnx equals 1 over x dx

So the first step

x taking become d lnx

How to write it

Go on write it

Look at this term

Square root a time b

we write it

square root a time square root b

If a and b non-negative

So we write this form

This term write it

square root a time square root new b

ok

So now do you remember this property

How to take this term

into differentiable variable

Yes

remember this

How to do this

Differentiable square root of lnx equals

half 1 over square root of lnx d lnx

So this term we can take

time some constant

Go on write it

equals this step

You can check by yourself

d term equals half this

go back itself

So it's ok

Look at this

Do you remember substitution

du

square root 1 minus u square

We have the original anti-derivative function

equals what

Yes equals

arcsinu

So equals double

2 2 copy

du over square root 1 minus u square equals

arcsinu given from b minus a

We can write it

2 time π over 12

equals π over 6

Pay attention to this means lnx

ok

means 3 over 4

Square root 3 over 4 ok

Another

square root e

This means e half power

Try to write it

ok

So double π over 12

We can get the final result

π over 6

The question is difficult

Why

Because how to find the anti-derivative

We need to do the computation carefully

So now look at

this question

Evaluate the definite integral

from 0 to a

square roof of a square minus x square dx

a is a positive number

How to do

Using the idea substitution

how to use

Now this is our integral

Let x equals a sint

In order to delete the square root sign

Let x equals a sint

We can write it

square root a square minus a square sint square

equals a square cost square

means absolute value a cost

if t is between 0 and half π

It is positive

So we can write it a cost

ok

Now

integral function becomes a function of t

Look at dx

Where is our x

Yes here

So dx equals d a sint

equals a cost dt

Still the similar idea

x between 0 and a

means x from 0 to a

How about t

t from 0 to half π

Go on write it

From 0 to half π

a square cost square dt

How to do

a square is a constant

cost square

yes

double angle formula

We can write it

from 0 to half π

1 plus cos double t

Write it

a square over 2 bracket

1 corresponding

the anti-derivative equals t

cos double t

time 2 over 2

so equals sin double t over 2

This is our

the anti-derivative function

So we can write it

The final result equals

1 over 4 π a square

It is a positive number

Of course why is it positive

The function is positive

So the final result should be positive number

Do you remember

how to change the limit

So we can give you the concrete idea

Square root of a square minus x square

Go on

equals the absolute value of a cost

t in the first quadrant

So it is a positive value

dx

do you remember

d a sint equals a cost dt

x equals 0

here

x equals 0

t equals 0

t equals 0

x equals a

yes

x equals a

sint equals 1

what number sint equals 1

Yes

t equals half π

So we have this

from a to b

from α to β

The question

try to find

the upper limits and the lower limits

about the original function

and the new function

Can you have some idea about

the definite integral

Yes

thinking by the geometrical meaning

Geometrical meaning

this means

1 by 4 time

the area of some circle

Try to understand after class

So now look at the last question

The last question means

try to find the definite integral

between 0 and 1 absolute value x time

bracket double x minus 1 dx

Do you have some idea about

this integral function

Do you remember

in our example

Try to delete the max

Yes

try to delete the max

becomes the integral function into

piecewise function

Similar meaning

try to delete absolute signs

How to delete

Try to find

what is our equals 0

So let the integral function

x time bracket double x minus 1 equals 0

What value equals 0

Two values

one is 0

another is half

So there are two zero points

One is 0

Another is half

0 is the lower limit

Half between 0 and 1

So we try to write

between 0 and half

if x is changeable

between the interval

At this time

our integral function

less than or equal to 0

You can check ok

Between half and 1

at this time

the integral function

is non-negative

large than or equal to 0

We can write it

the integral equals

delete the absolute sign equals minus

0 to 1

What's the property

Yes

interval additivity

0 to half equals some expression

less than or equals 0

Absolute value

so write it minus

Minus

0 to half

half to 1

x time double x minus 1

our integral function

Pay attention to the signs

Between the interval

between the interval

it is an expression less than 0

So we write it minus

Between another interval positive

we write it plus directly

Finally the result equals 1 by 4

So in the last question

pay attention to

how to delete the absolute signs

Try to find the zero points

According to the zero points

our original integral

becomes some new interval

And finally

compute carefully

we can get the correct answer

So now class is over

See you next time