HL Math 1 – Calculus - Santowski Lesson 45 - Antiderivatives and the Rules of Integration 1/6/2016...

HL Math 1 – Calculus - Santowski

Lesson 45 - Antiderivatives and the Rules of Integration

04/21/23

1

Calculus - Santowski

A function F is an antiderivative of f on an interval I if F’(x)=f(x) for all x in I.

Let’s use an example to figure out what this statement means…

Suppose we know f(x) = 2x and we want to find its antiderivative, F.

If f(x) = 2x, then F’(x) = 2x. So we know the derivative of F.

Think backwards, what function has a derivative equal to 2x ?

F(x) = x² !!!

To find the antiderivative, do the reverse of finding the derivative.

Is F(x) = x² the only function whose derivative is 2x ? Or in otherwords, is F(x) = x² the only antiderivative of 2x ? 04/21/232Calculus - Santowski

Theorem 1: Let G be an antiderivative of a function f. Then, every antiderivative F of f must be of the form F(x) = G(x) + C, where C is a constant.

Example 1: Let F(x) = x² + 4 and let G(x) = x²- 1 Then F’(x) = 2x and G’(x)= 2x

Thus both F and G are antiderivatives of f(x) = 2x.Note two functions which have the same derivative will only differ by the constant.

This means the antiderivative of a function is a family of functions as pictured on the next slide.

So the answer to the question is F(x) = x²the only antiderivative of2x is NO!!

04/21/233Calculus - Santowski

x

y

F(x)= x2 - 3

F(x)= x2 - 2

F(x)= x2 - 1

F(x)= x2

F(x)= x2 +1

When you integrate (which is another name for finding the antiderivative) you get a family of curves. Each of these is an antiderivative of f(x)=2x.


The Indefinite Integral

The process of finding all antiderivatives of a function is called antidifferentiation, or integration.

is the symbol used for integration and is called the integral symbol.

We write

€

f(x)dx=F(x)+C

This is called an indefinite integral, f(x) is called the integrand and C is called the constant of integration.

dx is the differential of x which denotes the variable of integration


Basic Integration Rules

€

kdx=kx+CRule 1: (k, a constant)

Example 2:

€

2dx=2x+C

Example 3:

€

π2dx=π2x+C

Keep in mind that integration is the reverse of differentiation. What function has a derivative k?

kx + C, where C is any constant.

Another way to check the rule is to differentiate the result and see if it matches the integrand. Let’s practice.


Before we list Rule 2, let’s go back and think about derivatives.

When we used the power rule to take the derivative of a power, we multiplied by the power and subtracted one from the exponent.

Example:

€

ddx

x3( ) =3x2

Since the opposite of multiplying is dividing and the opposite of subtracting is adding, to integrate we’d do the opposite. So, let’s try adding 1 to the exponent and dividing by the new exponent.

Check by differentiating the result:

€

ddx

14

x4 ⎛

⎝ ⎜

⎞

⎠ ⎟=44

x3 =x3

Since we get the integrand we know it works.

€

x3 dx=x3+1

3+1=14

x4Integrating:


Rule 2: The Power Rule C1n

xx

1nn

n 1

Example 4: Find the indefinite integral

€

t3 dt

Solution:

€

t3 dt=t4

4+C

Example 5: Find the indefinite integral

€

x32 dx

€

x32 dx=

x32+1

32+1

+C=X

52

52

+C=25

X52 +CSolution:



Example 6: Find the indefinite integral dxx13

Solution:

€

1x3 dx= x−3 =

x−3+1

−3+1+C=

x−2

−2+C=

−12x2

+C

Example 7: Find the indefinite integral dx1

€

1dx=x+CSolution:

Example 8: Find the indefinite integral dxx3 2

Solution:

€

−3x−2 dx=−3 x−2 =−3x−2+1

−2+1+C=

−3x−1

−1+C=

3x+C

Here are more examples of Rule 1 and Rule 2.


Rule 3: The Indefinite Integral of a Constant Multiple of a Function

€

( )cf x dx=c ( ) ,f x dxcisaconstant

Rule 4: The Sum Rule (or difference)

€

f x( )+g x( )[ ] dx= ( )f x dx+ ( )g x dxf x( )−g x( )[ ] dx= ( )f x dx− ( )g x dx

Rule 5:

€

ex dx=ex+C

Rule 6:

€

1x

dx=lnx+C


To check these 2 rules, differentiatethe result and you’ll see that it matchesthe integrand. 04/21/2310Calculus - Santowski

Integrate and check your answer by taking the derivative.

€

( ) a 3+x4 +exdx

€

( ) b (2x−1x

+3x2

+ x+3ex)dx

€

( ) c 2x−4( )1x2

+3x

⎛

⎝ ⎜

⎞

⎠ ⎟dx


Here is the solution for Q(a)in detail.

€

( ) a 3+x4 +exdx = 3dx + x4dx + exdx

= 3x +x4+1

4+1+ ex +C

=3x+15

x5 +ex +C


Q(b) Integrate

€

(2x−1x

+3x2

+ x+3ex)dx

Using the sum rule we separate this into 5 problems.

€

2xdx−1x

dx+3x2 dx+ x dx+ 3 exdx

€

2xdx−1x

dx+3x2 dx+ x dx+ 3 exdx=x2 −lnx−

3x+23

x32 +3ex+C

You may be wondering why we didn’t use the C before now. Let’s say that we had five constants . Now we add all of them together and call them C. In essence that’s what’s going on above.

54321 CCCCC

ANSWER:


(c) Integrate

€

2x−4( )1x2

+3x

⎛

⎝ ⎜

⎞

⎠ ⎟dx

€

Multiplythetwofactorsintheradicandtogether

andcombinethelike .terms

2x+6−

4x2

−12x

dx= 6−10x−4x2

dx

€

NextuseRule4 tointegrateeachterm .separately

6dx−101xdx−4 x−2dx

6x−10lnx−4x−1

−1+C

6x−10lnx+4x+C


Differential Equations

A differential equation is one which has a derivative expression in it.

For example: f’(x)=2x+1 is a differential equation.When we integrate this type of equation we get the general solution which contains the constant, C .

To find a particular solution we need another piece of information. It could be a point that we know the function passes through. We call this piece of information the initial condition.

A typical problem might be:f’(x)= 3x2 - 4x + 8f(1)= 9


Let’s look at the solution to the problem:f’(x)= 3x2 - 4x + 8f(1)= 9

Solution: First integrate both sides:

Cx82x4

3x3

xf23

)(

Simplify: Cx8x2xxf 23 )(

Now find C by using the initial condition.Substitute 1 for x and 9 for f(x)

C2

C79

C8219

C181219 23

This gives the particular solution.

2x8x2xxf 23 )(04/21/2316Calculus - Santowski

Review - Basic Integration Rules

Rule 1: (k, a constant)

€

kdx=kx+CRule 2: The Power Rule C

1nx

x1n

n

Rule 3: The Indefinite Integral of a Constant Multiple of a Function

€

( )cf x dx=c ( ) ,f x dxcisaconstantRule 4: The Sum Rule (or difference)

€

f x( )+g x( )[ ] dx= ( )f x dx+ ( )g x dxf x( )−g x( )[ ] dx= ( )f x dx− ( )g x dx

Rule 5:

€

ex dx=ex+C

Rule 6:

€

1x

dx=lnx+C04/21/2317Calculus - Santowski

Integration Formulas for Trigonometric Functions

Cxdxxx

Cxdxxx

Cxdxx

Cxdxx

Cxdxx

Cxdxx

csccotcsc

sectansec

cotcsc

tansec

sincos

cossin

2

2

04/21/23

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Examples.

€

( ) a (3cscxcotx−7sec2 x)dx

€

( ) b3 tanθ −4cos2 θ

cosθdθ

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Solution to Example (a).

€

(3cscxcotx−7sec2 x)dx=3 cscxcotxdx−7 sec2 x dx=3 −cscx+C1[ ] −7 tanx+C2[ ]

=−3cscx−7 tanx+C

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Solution to Example (b)

€

3 tanθ −4cos2 θcosθ

dθ

=31

cosθtanθdθ −4 cosθdθ

=3 secθ tanθdθ −4 cosθdθ=3secθ −4sinθ +C

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Exercises:

θθθ d

dxx

x

dyy

yy

dxxx

dyyy

)tan3cot2(.5

cos

sin.4

12.3

)1(.2

)32(.1

22

2

24

23

04/21/23

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Applications of Indefinite Integrals

1. Graphing

Given the sketch of the graph of the function, together with some function values, we can sketch the graph of its antiderivative as long as the antiderivative is continuous.

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Example 1. Given the sketch of the function f =F’(x) below, sketch the possible graph of F if it is continuous, F(-1) = 0 and

F(-3) = 4.

-1

4

5

-3 -2 -1 0 1 2 3 4 5

1

-4-5

3

2

-2

-3

-4

-5

F(x) F’(x)

F’’(x)

Conclusion

X<-3 + - Increasing, Concave down

X=-3 4 0 - Relative maximum

-3<x<-2

- - Decreasing, Concave down

X=-2 - 0 Decreasing, Point of inflection

-2<x<-1

- + DecreasingConcave up

X=-1 0 0 + Relative minimum

X>-1 + + Increasing,Concave up

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The graph of F(x)

-1

4

5

-3 -2 -1 0 1 2 3 4 5

1

-4-5

3

2

-2

-3

-4

-5

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1. Boundary/Initial Valued Problems

There are many applications of indefinite integrals in different fields such as physics, business, economics, biology, etc.

These applications usually desire to find particular antiderivatives that satisfies certain conditions called initial or boundary conditions, depending on whether they occur on one or more than one point.

Applications of Indefinite Integrals

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Example 2.

6 1dy

xdx

Suppose we wish to find a particular antiderivative satisfying the equation

and the initial condition y = 7 when x = 2.

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Sol’n of Example 2

2

2

(6 1)

(6 1)

3

2 7,

7 3(2) 3 7

dy x dx

dy x dx

y x x C

but x when y then

C C

23 7y x x Thus the particular antiderivative desired,

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Example 3.

The volume of water in a tank is V cubic meters when the depth of water is h meters. The rate of change of V with respect to h is π(4h2 +12h + 9), find the volume of water in the tank when the depth is 3m.

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Sol’n of Example 3

2

2

32

32

32

32 3

4h 12h 9

4h 12h 9

46h 9h

3

4(0 )0 6(0 ) 9(0)

3

0

46h 9h

3

4(3 )6(3 ) 9(3) 207

3

dV

dh

dV dh

hV C

C

C

hThus V

V m

π

π

π

π

π

π π

Volume V=0 if depth h =0

04/21/23

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The Differential Equations

Equation containing a function and its derivative or just its derivative is called differential equations.

Applications occur in many diverse fields such as physics, chemistry, biology, psychology, sociology, business, economics etc.

The order of a differential equation is the order of the derivative of highest order that appears in the equation.

The function f defined by y= f(x) is a solution of a differential equation if y and its derivatives satisfy the equation.

04/21/23

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If each side of the differential equations involves only one variable or can be reduced in this form, then, we say that these are separable differential equations.

Complete solution (or general solution)y = F(x) + C

Particular solution – an initial condition is given

The Differential Equations

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Example 4.

342

2

xdx

yd

Find the complete solution of the differential equation

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Solution to Example 4.

2

2

21

4 3

(4 3)

(4 3)

2 3

d y dyx

dx dxdy x dx

dy x dx

y x x C

21

21

3 2321 23 2

2 3

(2 3 )

dyx x C

dx

dy x x C dx

y x x C x C

2

2

d y dyddx dxdx

dydxlet y

342

2

xdx

ydFind the complete solution of the differential equation

04/21/23 34Calculus - Santowski

Example 5.

Find the particular solution of the differential equation below for which y=2 and y’=-3 when x=1.

342

2

xdx

yd

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Solution to Example 5.

21

21

1

2 3

3 2(1) 3

8

y x x C

x C

C

3 2321 23 2

3 23223 2

472 6

2 (1) (1) 8(1)

y x x C x C

C

C

3 23 4723 2 68y x x x

Find the particular solution of the differential equation in Ex. 14 for which y=2 and y’=-3 when x=1.

04/21/23 36Calculus - Santowski

Example 6.

A stone is thrown vertically upward from the ground with an initial velocity of 20ft/sec.

(a) How long will the ball be going up?

(b) How high will the ball go?

(c) With what velocity will the ball strike the ground?

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ANS to Example 6.

A stone is thrown vertically upward from the ground with an initial velocity of 20ft/sec.

(a) How long will the ball be going up?Ans. 0.625 sec

(b) How high will the ball go?Ans. 6.25 ft

(c) With what velocity will the ball strike the ground?

Ans. 20 ft/sec

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Other Properties of Indefinite Integrals


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And two other interesting “properties” need to be highlighted:

Interpret what the following 2 statement mean:

Let f(x) = x3 - 2x

What is the answer for f `(x)dx ….?

What is the answer for d/dx f(x)dx ….. ?

Examples with Motion


40

An object moves along a co-ordinate line with a velocity v(t) = 2 - 3t + t2 meters/sec. Its initial position is 2 m to the right of the origin.

(a) Determine the position of the object 4 seconds later

(b) Determine the total distance traveled in the first 4 seconds

Examples – “B” Levels


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Sometimes, the product rule for differentiation can be used to find an antiderivative that is not obvious by inspection

So, by differentiating y = xlnx, find an antiderivative for y = lnx

Repeat for y = xex and y = xsinx

Integration by Chain Rule/Substitution

For integrable functions f and g

where is an F antiderivative of f and C is an arbitrary constant.

( ( ))[ '( ) ] ( ( ))f g x g x dx F g x C

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Example 1.

14

54

54

2 34

3 24

3 2

3

54

385

36 6 5

2 6 5 (18 )

2 (6 5) (18 )

(6 5)2

(6 5)

x x dx

x x dx

x x dx

xC

x C

Let g(x) = 6x3+5

g’(x)=18x2

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Example 1. Take 2!

14

54

54

54

2 3 3 24 4

4

54

85

385

36 6 5 2 6 5 (18 )

2

2 ( )

2

(6 5)

x x dx x x dx

u du

u du

uC

u C

x C

Let u = 6x3 + 5

du = 18x2 dx

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Example 2.

3 3

7 74 4

74 3

64

64

2 1 1 2(2 1)

22 2

12 2(2 1)

2

21

2 61

12 2

t t dxdx

t t t t

t t t dx

t tC

Ct t

Let g(t) = t4 + 2t

g’(t) = 4t3 + 2

= 2(2t3 + 1)

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Example 3.

5 2 12

22 2 12

2 12

14 13 12

15 14 131 2 115 14 13

2 15 2 14 2 131 1 115 7 13

( 1) 2

( 1) 2

( 1)

( 2 )

( 1) ( 1) ( 1)

x x dx

x x xdx

u u du

u u u du

u u u C

x x x C

Let u = x2 -1

du = 2x dx

x2 = u+1

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Example 4.

Let u = 2 – cos2x

du = 0 – (-sin2x)(2dx)

=2sin2xdx

Cx

Cu

duu

duu

xdxx

dxxx

2/331

2/332

21

2/121

21

21

)2cos2(

)(

)(

)2sin2(2cos2

2cos22sin

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Example 5.

xdxx

dxxx

dxxx

xx

dxx

x

x

x

dxxx

2csc2sec

2sin2cos

1

2sin2cos

2cos2sin

2sin

2cos

2cos

2sin

)2cot2(tan

22

22

222

2

2

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Example 6

Cxx

Cxx

xdxdxxx

xdxdxxx

xdxxdxx

dxxx

xdxx

2tan2cot

2tan)2(tan

2sec22sec2)2(tan

2sec2sec)2(tan

2sec2cot2sec

)12(cot2sec

2csc2sec

21

21

21

21

22122

21

222

222

22

22

1

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Exercises:

drr

r

xx

dxxx

7

3 22

62

)1(

2.3

)49(5.2

)12(7.1

424

2

22

)123(

)13(.6

3sin21

3cos.5

3cot3csc.4

xx

dxxx

dxx

x

dyyyy

04/21/23

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HL Math 1 – Calculus - Santowski Lesson 45 - Antiderivatives and the Rules of Integration 1/6/2016...

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Transcript of HL Math 1 – Calculus - Santowski Lesson 45 - Antiderivatives and the Rules of Integration 1/6/2016...