Chapter 12 Mm34cas

8/12/2019 Chapter 12 Mm34cas

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AntidifferentiationIn year 11 we saw that antidifferentiation is the reverse ofdifferentiation. Antidifferentiation allows us to find the function for aparticular gradient and as we will see allows us to calculate the area under

the curve.

Eg1. Find the antiderivative of 523 34 x x

In general, if 1,)( nax x f n then

antiderivative is cnax n

1

1

where c is a constant.

If )()(' x f x F then c x F dx x f )()( where c is an arbitrary real

number.

NotationAs stated we often use a capital letter associated with the originalfunction symbol. Eg. the antiderivative of f(x) is given by F(x). Some

texts use the large capital A, Af(x). The antiderivative has also beencalled an indefinite integral with Leibnitz notation dx x f )( .

Linearity of the antiderivativeFor any two functions f(x) and g(x) and constants a and b.

dx x f k dx xkf

and

dx x g bdx x f adx xbg xaf

)()(

)()()()(

Eg2. Find dx x x )2( 3

Eg3. The gradient function for a particular curve is given by x x x f 621)( 2 . Find the equation of the curve given that the curve

passes through (1,9).


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1,0\,)()('1,,)()(,:

1 r Qr baxar x f

r Qr bax x f R R f r

r

c x x F then

x x f

0

1

01

)(

)(

For a functionusing chain rule.

So for a suitable domain 1,)()1(

1)( 1 r cbax

r a

dxbax r r

Eg4. Find the antiderivative of the following.a) 4)12( x b) 3)23( x

Roots and ReciprocalsRoots and reciprocals are easily handled by the rules governing powers except in one special case.

Eg5. Find F(x) for the following,

a) x x f )( b) 21

)( x

x f

Eg6. Find F(x) if x

x f 1)( .

This is using our rule but this is not possible. So our rule does not workwhen n=-1.

The rule for antidifferentiation of the reciprocal function is;

0,log1 x for c xdx x e

So

ab

x for cbaxa

dxbax e

,log11

Note: a and b are constants where a 0


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Eg7. Find

a) dx x 34 b) dx

x 123

Antiderivatives of other component functionsAntidifferentiation sometimes requires us to have the function inexpanded form.

Eg8. Find dx x x 124

In general dx x g dx x f dx x g x f )()()()(

There is no equivalent product rule, quotient rule or chain rule inantidifferentiation. We need to use a number of tricks t o find theantiderivative of compound expressions where this is possible.

Eg9. Find dx x

x1

Exponential FunctionsThe exponential function has its own derivative,

i.e. x x eedxd so cedxe x x .

The exponential function is also its own antiderivative (to within aconstant)The rule for antidefferentiation of exponential functions is;

cek dxe kxkx 1


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Eg10. Find the antiderivative of 22 4)( xe x f x given the curve passes

through the point ).21

,0(

Using CAS

To use Integral template: b42

Ex 12B: Q1-6

Ex 12C: Q1-3(1/2), 4-7

Ex 12D: Q1-4

Area The Definite Integral

Sums and Intergrals used to find the Area under a Curve

To calculate the area between a curve and the x-axis a limiting processcan be considered. The process of finding a limit of a sum of areas is thebasis of Integral Calculus. We need to find the area of the region of theplane surface bounded by the graph of a continuous function, the x-axisand the lines x=a and x=b where b>a.-We know how to find the areaof a rectangle.-The area of the curve boundedby the x-axis, x=a and x=b is lessthan the area of ABCF and morethan the area of ABDE.

-This is a rough estimate.

y

xA B

C

D E

F

a b

y=f (x)

0


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-The interval [a,b] can be dividedinto n equal parts by the pointsx0, x1, ,xn where a=x 0 and b=xn.-Each strip has a width of x .

-The area of each strip is givenby x x f )( .-So the area of the rectangularapproximation between a and bis the sum of the areas of all thestrips between a and b.

That is,n

inn x x f x x f x x f x x f

121 )()(...)()(

Area of the region enclosed by )( x f y isn

in x x x f

10 )(lim

Here the Greek S , is used to show sum. The limit replaces with an elongated S, , called the Integral

Symbol.

So enclosed area= b

a

n

in x

dx x f x x f )()(lim1

0

The terminals or values at the top and bottom of the integral

symbol revert back to the original boundaries of the area, a and bin this case.

The area of a region enclosed by the graph of a function and the x-axis

for the interval [a,b] is symbolised by the definite integralb

a

dx x f )( .

The Fundamental theorem Of Calculus

ba

b

a

x F

a F b F

dx x f Area

)(

)()(

)(

This is a very important Theorem because it indicates a much

method of evaluating definite integrals and calculating areas.

y

xa b

y=f(x)

0 x x 0 x 1 x n


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So the area under the curve y=f(x) between a and b is G(b)-G(a) whereG(x) is the antiderivative of f(x).

Definite and Indefinite Integrals

We use the same symbol for integration and antidifferentiation.

)()()()( a F b F x F dx x f bab

a

The terminals a and b are usually real numbers. If this is the case thenthe definite integral is a real number, NOT a function.

* dx x f )( is called the indefinite integral (no terminals). It is

represented by a family of antiderivatives. c x F x f A )()(

Eg11. Evaluate the following;

a)

4

3

2 dx x b)

2

0

3 dx x

Eg12. Find the area under the curve 2 x y between x=0 and x=5.


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Eg13. 3

1

2 123 dx x x Eg14. dx x

5

22

1

Using CAS obtain the Integral template and insert the lower and upper

terminals.

Signed and Unsigned AreaA For a curve which lies above the x-axis in the interval b xa , theb

a

dx x f )( is positive.

B - For a curve which lies below the x-axis in the interval b xa , theb

a

dx x f )( is negative.

Eg15.

A -41

041

4

1

0

41

0

3 xdx x

B -41

41

04

0

1

40

1

3 xdx x

NOTE - 041414

1

1

41

1

3 xdx x

This is telling us that there is zero area between -1 and 1. Why? From Aand B we see that there is an area. The integral gives the measure of thesigned area. To find the actual area, the absolute value of the signedarea is taken, or if the curve crosses the x-axis the area of the regionsabove and below the axis must be calculated separately. In the above eg,

the area of the shaded region is equal to21

41

41 square units.

A

B

y

x1 2 1 2

1

2

1

2

y = x 3


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Eg16. Find the area of the shaded region.

Eg17. Find the area enclosed by the graph of )3)(2( x x x y and the x-axis.

Using CASAlternative method to find definite Integral.

y

x

y = x 2 4

1 24


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Properties of Definite Integrals

1. a

a

dx x f 0)( * If terminals are the same there is no area*

2. b

a

c

a

b

c

bcawheredx x f dx x f dx x f )()()(

3. b

a

b

a

dx x f k dx xkf )()(

4. b

a

b

a

b

a

dx x g dx x f dx x g dx x f )()()()(

5. b

a

a

b

dx x f dx x f )()( * reversing the limits changes the sign of the

integral*

Ex 12E: Q1-3(1/2), 4, 5

Ex 12F: Q1, 2-3(1/2), 4, 6, 7, 9, 10, 11

Circular FunctionsThe circular functions sin x and cos x are related in antidifferentiation as

they are in differentiation.

The rule for antidefferentiation of circular functions is;

ckxk

dxkx cos1

sin

ckxk

dxkx sin1

cos

We dont have an antiderivative immediately obvious for x x f tan)( but

we did find that x xdxd 2sectan .

So ckxk dxkx tan

1sec 2

Eg18. Find the antiderivative of x x x

x f 3sin2cos42

)(

So if we have sin or cos functions of the form;


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cakxk

dxakx )cos(1

)sin(

cakxk

dxakx )sin(1

)cos(

Eg19. Find the antiderivatives of the following;

a) )3

4sin(

x b) )4

2cos(

x

Eg20. Evaluate

a) 2

0

2cos3

dx x b) 4

0

sin3

dx x

Ex 12G: Q1-3(1/2), 4ac(CAS), 5(CAS)

We often find antiderivatives by using a result found in differentiation.


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Eg21. Find the derivative of x x y elog and hence find the antiderivativeof xelog .

Eg22. Differentiate 32 )5( x x and hence integrate 22 )5)(12( x x x .

Eg23. Differentiate )72sin( x x y and hence integrate )72cos( x x .

Ex 12H: Q1(1/2), 2, 3-6(1/2), 8(1/2), 9, 11, 14(1/2), 16, 18(1/2)


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Area of a Region Between two CurvesThe area between the two curves is found by subtracting the areabetween the graph of g(x) and the x-axis from the area between thegraph of f(x) and the x-axis.

)()(

)()(

)()(

x g x f If

dx x g x f

dx x g dx x f Area

b

a

b

a

b

a

Eg24. Calculate the area of the region enclosed by the graphs withequations 34)( 2 x x x f and 23)( x x x g .

Eg25. Calculate the area of the segment of the curve 2 x y cut off bythe line x y .

Note: Graphs are essential to seewhich graph is greater.

Ex 12I: Q1, 2, 3(1/2), 5, 6, 8

y

x

f(x)

g(x)

a b

y

x

1 2 3 4

1

2

3

4

1

2

3

f(x)

g(x)

y

x

1

1

1

y=x

y = x 2


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Applications of Intergration

Average value of a functionThe average value of a function with rule for an interval [a, b]is defined as:

In terms of the graph the average is the height of a rectanglewhich has the same area under the graph for the interval [a, b].

Eg 26. Find the average value of for the interval [0, 2].Illustrate with a horizontal line determined by this value.

KinematicsRecall:Kinematics is the study of objects in straight line motion. We look at therelationship between position, velocity and acceleration of an object.

DisplacementWhen considering a particle in motion we need to relate its position to areference point or starting point. This point is taken as zero and anyposition from this point can be given a positive or negative value. Note:the fixed point may not always be 0 or O.

VelocityThe average velocity is the average rate of change in the position overa period of time.If a particle moves from d 1 at time t 1 to d 2 at time t 2, then its

12

12

t t d d

timeinChange positioninChange

velocityaverage

This can actually be found by finding the average value of the velocityfunction over the period of time.


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Speed is the magnitude of Velocity. (Speed is always positive)

timeinChangetravelled Distance

speed Average

This can actually be found by finding the average value of the speed function over the period of time.

The instantaneous velocity is the instantaneous rate of change , whichspecifies the rate of change at a given instant in time.

Instantaneous velocity )(' t xdt dx

v where x is a function of time.

Acceleration

Average acceleration for the time interval [t 1 , t 2] is defined by12

12

t t vv

where v 1 and v2 are the velocities at time t 1 and t 2 respectively.

Instantaneous acceleration = )(''22

t xdt

xd dt dx

dt d

dt dv

a

Eg 27. A body starts from O and moves in a straight line. After t seconds(t0) its velocity (v cm/s) is given by t t v 83 2 . Finda) its position x in terms of t.

b) its position after 3 seconds.

c) its average speed in the first 3 seconds.


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d) its average velocity in the first 3 seconds.

Ex 12J: 1ace, 2, 3b, 5, 6, 7,11, 12, 13, 15, 17

Review: ODDS

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Chapter 12 Mm34cas