602 Essential Mathematical Methods 1 & 2 CAS

23

Integration

Objectives

Toantidifferentiate polynomial functions.

Toidentify families of curveswith the same gradient function. Tofind the definite integralof an algebraic function.

Todetermine the areabetween a curve and thex-axis. To usenumerical methodsfor finding areas under a graph.

23.1

Antidifferentiation of polynomial functions

The derivative ofx2_{with respect toxis 2x. Conversely, given that an unknown expression has}

derivative 2x, it is clear that the unknown expression could bex2. The process of finding a function from its derivative is calledantidifferentiation.

Consider the polynomial functions f(x)=x2+1 andg(x)=x2−7.

We have f(x)=2x andg(x)=2x, i.e. the two different functions have the same derived function.

Bothx2+1 and x2−7 are said to be

antiderivativesof 2x. If two functions have the same derivative on an interval, then they differ by a constant. If two functions have the same derived function then the graph of one function is obtained by a translation parallel to they-axis of the other. The following illustrates several antiderivatives of 2x.

y=x2_{+ 1}

y=x2_{– 1}

y=x2_{– 7}

y=x2

distance

7 units _distance

7 units –1

0 1

x

–7

y

Each of the graphs is a translation ofy=x2

parallel to they-axis.

The general antiderivative of 2xisx2 +c, wherecis an arbitrary real number. We use the notation of Leibniz to state this with symbols:

2xd x =x2+c

This is read as ‘the general antiderivative of 2xwith respect toxis equal tox2 plusc’ or ‘the

indefiniteintegral of 2xwith respect toxisx2_+c’.

To be more precise, the indefinite integral is the set of all antiderivatives and to emphasise this we write:

2xd x = {f(x) : f(x)=2x} = {x2+c:c∈R}

The set notation is not commonly used, but it should be clearly understood that there is not a unique antiderivative for a given function. We do not use the set notation but it is advisable to keep it in mind when considering further results.

In general:

If F(x)= f(x),

f(x)d x = F(x)+c,wherec is an arbitrary real number. We know that:

f(x)=x3 implies f(x)=3x2 f(x)=x8 _{implies f(x)=8x}7 f(x)=x20 _{implies f(x)=20x}19 f(x)=x implies f(x)=1 Reversing this process we have:

3x2_{d x =x}3_{+c,wherecis an arbitrary constant}

8x7_{d x =x}8_{+c,wherecis an arbitrary constant}

20x19_{d x =x}20_{+c,wherecis an arbitrary constant}

1d x =x+c,wherecis an arbitrary constant and from this:

x2d x = 1

3x

3_+c

x7d x = 1

8x

8_+c

x19d x = 1

20x

20_+c

1d x =x+c

From this we see that:

xn_{d x =} x n+1

We also record the following results which follow immediately from the corresponding results for differentiation in Chapter 19.

f(x)+g(x)d x=

f(x)d x+

g(x)d x

and

k f(x)d x=k

f(x)d x,wherekis a real number

Example 1

Find the general antiderivative (indefinite integral) of each of the following:

a 3x5 _{b 3x}2_+4x3₊₃

Solution

a

3x5d x =3

x5d x =3× x

6

6 +c

= x6

2 +c

b

3x2+4x3+3d x

=3

x2d x+4

x3d x+3

1d x =3

x3

3

+4x4

4 + 3x

1 +c

=x3+x4+3x+c

In many situations we are given extra information that helps us to find a unique antiderivative.

Example 2

It is known that f(x)=x3_+4x2_{andf(0)=0. Findf(x).}

Solution

x3+4x2d x = x 4

4 + 4x3

3 +c

∴ f(x)= x

4

4 + 4x3

3 +c and asf(0)=0,c=0

∴ f(x)= x

4

4 + 4x3

3

Using a CAS calculator

If the family of antiderivatives is required, enter

Example 3

If the gradient at a point (x,y) on a curve is given by 5xand the curve passes through (0, 6) find the equation of the curve.

Solution

Let the curve have equationy= f(x). Then f(x)=5x

5xd x = 5x 2

2 +c

∴ f(x)= 5x

2

2 +c

This describes the family of curves for which f(x)=5x. Here we are given the additional information that the curve passes through (0, 6), i.e.f(0)=6. Hence 6=c and f(x)= 5x

2

2 +6

Example 4

Findyin terms ofxif:

a d y

d x =x

2_{+2xandy=1 whenx=1 b} d y

d x =3−xandy=2 whenx=4

Solution

a

x2+2xd x = x3

3 +x

2_+c

∴ y= x

3

3 +x

2_+c

and, as y=1 when x =1, 1= 1

3+1+c and c= − 1 3

∴ y= x

3

3 +x

2₋ 1

3

b

3−xd x =3x− x

2

2 +c and y=2 whenx =4

∴ 2=3×4− 4

2

2 +c i.e. c= −2

∴ y=3x− x

2

2 −2

Exercise

23A

1 Find:

Example 1

a

1₂x3d x b

3x2−2d x c

5x3−2xd x

d

4₅x3_−2x2_{d x e}

_(x−1)2

d x f

xx+ 1_xd x,x =0

2 Findyin terms ofxin each of the following:

Example4

a d y

d x =2x−1 and y=0 when x=1 b d y

d x =3−x and y=1 whenx =0

c d y

d x =x

2_{+2x andy=2 whenx=0}

d d y

d x =3−x

2 _{andy =2 when x=3}

e d y

d x =2x

4_{+x and y=0 whenx =0}

3 d V

dt =t

2_{−t whent >1 andV =9 whent =3.}

a FindVin terms oft. b Calculate the value ofVwhent=10.

4 The gradient at any point (x, f(x)), on the curve with equationy=f(x), is given by

Example3

3x2−1. Findf(x), if the curve passes through the point (1, 2), i.e.f(1)=2.

5 a Which one of the following graphs represents dw

dt =2000−20t,t >0?

0 100 t

dw dt

0 100 t 0 ₁₀₀ t 0 100 t

dw dt

dw dt

dw dt

A B C D

b Findwin terms oft, if whenf=0,w=100 000.

6 The graph shows d y

d x againstxfor a certain curve

with equationy=f(x).

Findf(x), given that the point (0, 4) lies on the curve.

x

5

5 0

dw dt

7 Find the equation of the curvey=f(x) which passes through the point (2,−6) and for which f(x)=x2(x−3).

8 The curvey=f(x) for which f(x)=4x+k, wherekis a constant, has a turning point at (−2,−1).

a Find the value ofk.

b Find the coordinates of the point at which the curve meets they-axis.

9 Given that d y

d x =ax 2₊

1 and that whenx=1,d y

d x =3 andy=3, find the value ofy

whenx=2.

10 The curve for which d y

d x =2x+k,wherekis a constant, is such that the tangent

at (3, 6) passes through the origin. Find the gradient of this tangent and hence determine:

11 The curvey=f(x) for which f(x)=16x+k, wherekis a constant, has a stationary point at (2, 1).

a Find the value ofk. b Find the value off(x) whenx=7.

12 Suppose that a point moves along some unknown curvey=f(x) in such a way that at each point (x,y) on the curve the tangent line has slopex2_{. Find an equation for the curve,}

given that it passes through (2, 1).

23.2

Antidifferentiation of algebraic expressions

with rational exponents

(This section is not listed in the VCE study design for Mathematical Methods (CAS) Units 1 & 2 but can be used as a preparation for Mathematical Methods (CAS) Units 3 & 4.)

In Chapter 22 it was shown that if y =x

p

q _{where p},_q∈ _Z \ {₀}

then d y

d x = p qx

p q−1

For example, we know that

f(x)=x−2 implies f(x)= −2x−3 f(x)=x

1

2 _{implies f(x)}= 1

2x

−1₂

f(x)=x 3

2 _{implies f(x)}= 3

2x

1 2

Reversing this process we have

−2x−3d x =x−2+c,wherecis an arbitrary constant

1 2x

−1_{2d x =x}1_{2 +c,wherecis an arbitrary constant}

3

2x

1

2d x =x32 +c,wherecis an arbitrary constant and from this

x−3d x = −1

2x

−2_+c

x−12d x =2x12 +c

x12d x = 2

3x

3

2 +c

From this we see that

xrd x = x

r+1

r + 1+c, r ∈Q\ {−1}

Note:This definition can only be applied for suitable values ofxfor a given value ofr. For example ifr= 1

2 ,x∈R

+_{is a suitable restriction.}

Example 5

Findyin terms ofxif:

a d y

d x = 1

x2 andy=1 wherex=1 b

d y d x =3

√

x andy=2 whenx=4

c d y

d x =x

3 4 +_x−

3

4 _andy=_{0 whenx}=₀

Solution

a

1

x2d x =

x−2d x = x−1

−1 +c

∴ y = −1

x +c

and,

as y =1 whenx =1, we havec=2

∴ y = −1

x +2

b

3√xd x =3

x12d x =3× x

3 2

3 2

+c

∴y =2x32 +_{c and},

as y =2 whenx =4, 2=2×(4)32 +c

∴2=2×8+c

andc= −14 Thus y =2x

3

2 −₁₄

c

x

3 4 +_x−

3 4_{d x}

∴y= 4

7x

7

4 +4x14 +c

and,as y=0 whenx =0,c=0

∴y= 4

7x

7 4 +_4x

1 4

Exercise

23B

1 Find:

a

3x−2d x b

2x−4+6xd x c

√x(2+x)d x

d

3x13 −5x54d x e

_3z4 _{+ 2z}

z3 d z f

3x34 −7x12d x

2 Findyin terms ofxfor each of the following:

Example5

a d y

d x =x 1

2 +_{x andy}=_{6 whenx}=_{4 b} d y d x =

1

x2 andy=1 whenx=1

c d y

d x =3x+

1

x2 andy=5 whenx=1

3 A curve with equationy=f(x) passes through the point (2, 0) and f(x)=3x2− 1 x2 .

Findf(x).

4 Findsin terms oft, if ds

dt =3t−

8

t2 ands =1

1

5 Given that d y

d x = a

x2 +1 and whenx=1 d y

d x =3 andy=3, find the value ofywhen x=2.

6 The curve for which d y

d x =ax, whereais a constant, is such that the tangent at (1, 2)

passes through the origin.

Find the gradient of this tangent and hence determine:

a the value ofa b the equation of the curve.

7 Find the equation of the curve which passes through the point (−1, 2) and has the property that for each point (x,y) on the curve, the gradient equals the square of the distance between the point and they-axis.

23.3

Applications to kinematics

In Chapter 20 we considered examples in which the equation of motion defined the position of the particle in terms of time, and from it we derived equations for the velocity and the

acceleration by differentiation.

We may be given a rule for acceleration at timetand, by the use of antidifferentiation with respect totand some additional information, we can deduce rules for both velocity and position.

Example 6

A body starts fromOand moves in a straight line. Aftertseconds (t ≥0) its velocity (vcm/s) is given byv=2t−4.

a Find its positionxin terms oft.

b Find its position after 3 seconds.

c What is the distance travelled in the first 3 seconds?

d Find its average velocity in the first 3 seconds.

e Find its average speed in the first 3 seconds.

Solution

a Antidifferentiate with respect totto find the expression for positionxm at time

tseconds.

x=t2−4t+c

Whent =0,x =0 and thereforec=0.

x=t2−4t

c v =0 when 2t−4=0, i.e. whent =2. Whent =2,x= −4.

Therefore the body goes fromx =0 tox= −4 in the first 2 seconds and then back to –3 in the next second. It has travelled 5 m in the first 3 seconds.

d Average velocity= −3−0

3 = −1 m/s

e From part c, distance travelled=5 m.

∴average speed is 5 3 m/s.

Example 7

A particle starts from rest 3 metres from a fixed point and moves in a straight line with an acceleration ofa=6t +8. Find its position and velocity at any timetseconds.

Solution

a= dv

dt =6t+8

By antidifferentiating v=3t2+8t+c

Att =0, v =0 and soc=0

∴ velocity v=3t2+8t

By antidifferentiating again x =t3+4t2+d

Att =0, x=3 and sod=3

∴position x=t3+4t2+3

Example 8

A stone is projected vertically upward from the top of a 20-m high building with an initial velocity of 15 m/s.

a Find the time taken for the stone to reach its maximum height.

b Find the maximum height reached by the stone.

c What is the time taken for the stone to reach the ground?

d What is the velocity of the stone as it hits the ground?

In this case we only consider the stone’s motion in a vertical direction, so we can consider it as rectilinear motion. Also we will assume that the acceleration due to gravity is approximately

Solution

Given thata= −10

v= −10t+c

Att =0, v=15,soc=15

∴ v= −10t+15

and x = −5t2_+15t+d

Att =0, x =20, sod =20

∴ x = −5t2+15t+20

a The stone will reach its maximum height whenv=0.

∴ −10t+15=0

which implies t =1.5

The stone reaches its maximum height whent =1.5 seconds.

b Att =1.5, x = −5 (1.5)2+15 (1.5)+20

=31.25

The maximum height reached by the stone is 31.25 metres.

c The stone reaches the ground whenx =0.

−5t2+15t+20=0

−5(t2−3t−4)=0

−5(t−4)(t +1)=0

∴ t =4 (The solution oft = −1 is rejected sincet ≥0.) The stone takes 4 seconds to reach the ground.

d Att =4,v = −10 (4)+15

= −25

Thus its velocity on impact is−25 m/s.

Exercise

23C

1 A body starts fromOand moves in a straight line. Aftertseconds (t ≥0) its velocity

Example 6

(vcm/s) is given byv =4t−6.

a Find its positionxin terms oft.

b Find its position after 3 seconds.

c Find the distance travelled in the first 3 seconds.

d What is its average velocity in the first 3 seconds?

2 The velocity (vm/s) at timetseconds (t ≥0) of a particle is given byv =3t2−8t +5. It is initially 4 m to the right of a pointO.

a Find its displacement and acceleration at any time.

b Find its displacement when the velocity is zero.

c What is its acceleration when the velocity is zero?

3 A body moves in a straight line with an acceleration of 10 m/s2_{. If after 2 seconds it passes} Example7

throughOand after 3 seconds is 25 metres fromO, find its initial displacement relative toO.

4 A body moves in a straight line so that its acceleration (am/s2_{) after timetseconds (t ≥0)}

is given bya=2t −3.

If the initial position of the body is 2 m to the right of a pointOand its velocity is 3 m/s, find the particle’s position and velocity after 10 seconds.

5 A body is projected vertically upwards with a velocity of 25 m/s. (Its acceleration due to gravity is−10 m/s2.)

a Find the particle’s velocity at any time.

b Find its height above the point of projection at any time.

c Find the time it takes to reach its maximum height.

d What is the maximum height reached?

e Find the time taken to return to the point of projection.

6 The lift in a tall building passes the 50th floor with a velocity of−8 m/s and an acceleration of 1₉(t −5) m/s2_.

If each floor spans a height of 6 metres, find at which floor the lift will stop.

23.4

Area — the definite integral

This section can serve as a preparation for Mathematical Methods (CAS) Units 3 & 4. It is not required for Mathematical Methods (CAS) Units 1 & 2.

Letf: [a,b]→Rbe a continuous function such thatf(x)≥0 for allx∈[a,b].

y

y=f(x)

x b x a

0 We define the functionF(x) geometrically by

saying that it is the measure of the area under the curve betweenaandx. We thus haveF(a)=0. It will be shown thatF(x)=f(x).

Consider the quotient F(x+h)−F(x)

h ,h>0.

By our definition ofF(x),F(x+h)−F(x) is the area betweenxandx+h.

Letcbe the point in [x,x+h] such that

f(c)≥f(z) for allz∈[x,x+h] and letd

be the point in the same interval such that

f(d)≤f(z) for allz∈[x,x+h].

y

x c d x+h

x

Thusf(d)≤f(z)≤f(c) for allz∈[x,x+h]. Thereforehf(d)≤F(x+h)−F(x)≤hf(c).

That is, the shaded region has an area less than the area of the rectangle with basehand heightf(c) and an area greater than the area of the rectangle with basehand heightf(d).

Dividing byh: f(d)≤ F(x+h)−F(x)

h ≤ f(c)

Ash→0 bothf(c) andf(d) approachf(x). Thus we have shownF(x)=f(x).

We know that ifG(x) is an antiderivative off(x) thenF(x)=G(x)+k, wherek, is a constant. Letx=a. We then have

0= F(a)=G(a)+k, i.e.k = −G(a) Thus F(x)=G(x)−G(a) and letting x=byields

F(b)=G(b)−G(a) (equation 1)

The area under the curve y = f(x) between aandb isG(b)−G(a),whereG(x) is an antiderivative of f(x).

A similar argument could be used iff(x)<0 for allx∈[a,b], but in this case we must take

Fto be the negative of the area under the curve.

Signed area

For f(x)=x+1:

A1 =

1

2×3×3=4 1

2 (area of a triangle)

A2 =

1

2×1×1= 1 2

y

x A2

A1

–1 –1

1

1

1 2 3

2

y=x +

0 –2

The total area=A1+A2=5

Thesigned area=A1−A2=4

Regions below thex-axis havenegative signed areas. Regions above thex-axis havepositive signed areas. The total area of the shaded region

is A1+A2+A3 +A4.

The signed area of the shaded region isA1−A2+A3−A4.

y

x

0 A2

A3

A4 A1

We will denote the signed area between the curvey=f(x), the linesx=aandx=band the

x-axis by

b

Iff(x)≥0 forx∈[a,b] then

_abf(x)d x is the area enclosed between thex-axis and the curvey=f(x) and the linesx=aandx=b. We thus have (by equation 1)

b

a f(x)d x =G(b)−G(a), whereGis an antiderivative off.

This result holds generally and is known as the Fundamental Theorem of Integral Calculus. Let f be a continuous function on the interval [a,b]. Then

b

a f(x)d x =G(b)−G(a), whereGis any antiderivative off.

To facilitate setting out we writeG(b)−G(a)=

G(x)

b

a

b

a f(x)d x is called thedefinite integralfromatob.

The numberais called thelower limit of integrationandbis called theupper limit of integration. The functionfis called theintegrand.

We note that iff(x)≥0 for allx∈[a,b] then the area betweenbandais given by

b

a f(x)d x, and iff(x)<0 forx∈[a,b] then the area betweenbandais given by

−

_abf(x)d x.

Example 9

Find the area of the shaded region. y

x y=x+ 1

1 2 4

0 1 –1

Solution

Area =

4

2x+1d x

An antiderivative of x+1 is x

2

2 +x We write

4

2x + 1dx=

x2

2 +x

4

2

=

42 2 +4

−

22 2 +2

=12−4

=8

The shaded region has area 8 square units.

Example 10

Find the area of the shaded region.

(Note:The negative is introduced as the area is the negative of the integral from−4 to−2.)

– 4 –2 –11

y

x

0

Solution

Area= −

₋₄−2x+1d x = − x2

2 +x

−2

−4 = −(0−4)

The area of the shaded region is 4 square units.

Now consider

₋₄0 f(x)d x,where f(x)=x+1

=G(0)−G(−4)

To explain what this represents we use the fact thatG(−1)−G(−l)=0 to write

0

−4 f(x)d x =G(0)−G(−1)+G(−1)−G(−4) =

₋₁0 f(x)d x +

₋₄−1 f(x)d x A1 =

0

−1 f(x)d x,A2= −

−1

−4 f(x)d x

Thus

₋0₄ f(x)d x = A1+(−A2) = 1

2+

−41

2

= −4

This is the signed area.

A2

A₁

y=x+ 1

x y

0 – 4

1 –1

The actual areaA1+A2 =

1 2+4

1 2 =5

Note:The areas of examples 9 and 10 may also be determined geometrically.

In summary

Iff(x)≥0 for allx∈[a,b] the area of the region contained between the curve, thex-axis and the linesx=aandx=b

is given by

_abf(x)d x.

Iff(x)≤0 for allx∈[c,d] the area of the region contained between the curve, thex-axis and the linesx=candx=d

is given by−

d

c f(x)d x.

Ifc∈(a,b),f(c)=0 andf(x)≥0 forx∈(c,b] andf(x)≤0 forx∈[a,c) then the area of the shaded region is given by

b

c f(x)d x+ −

c

a f(x)d x.

y

x a

c b

0

Example 11

Find the area of the shaded region.

y

x y = x2–4

(0, – 4) 1

2 4

0

Solution

Area ₌

4

2x

2_{−4d x+ −}

2 1x

2_{−4d x}

= x3

3 −4x

4

2 − x3

3 −4x

3

1

= 64

3 −16−

8 3−8

− 8

3−8

−

1 3−4

= 56

3 −8− 7 3−4

= 37

3

The area is 37

3 square units.

Using a CAS calculator

Enter definite integrals as shown in the screen.

Properties of the definite integral

1

a

a f(x)d x =0

2

_abf(x)d x =

_cbf(x)d x+

_acf(x)d x,a<c<b

3

_abk f(x)d x =k

b

a f(x)d x

4

b

a f(x)±g(x)d x =

b

a f(x)d x±

b a g(x)d x 5

_abf(x)d x = −

a

Example 12

Evaluate each of the following definite integrals:

a

₂3x2d x b

₃2x2d x c

₀1x12 +x32d x

Solution a

3 2x 2 d x = x3

3 3 2 = 27 3 − 8 3

=9−22 3

=61 3

b

2 3x

2_{d x} = x3

3 2 3 = 8 3− 27 3

= −61 3

c

1 0x

1

2 +_x

3 2_{d x}

= 2

3x

3

2 + 2

5x 5 2 1 0 = 2 3+ 2 5 = 16 15 Example 13

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

y

x

0 2 3

Solution

y =x(−x2+5x−6)

= −x3+5x2−6x

Note:There is no need to find the coordinates of stationary points.

Area=

3

2(−x

3_+5x2_{−6x)d x+ −}

2 0(−x

3_+5x2_{−6x)d x}

= −x4

4 + 5x3

3 − 6x2

2

3

2

− −x4

4 + 5x3

3 − 6x2

2 2 0 = −81

4 +45−27

−

−4+ 40 3 −12

−

−4+ 40 3 −12

= −81

4 +18+32− 80

3

=50− 243+320 12

=3 1 12 Area is 3 1

Exercise

23D

1 Evaluate each of the following:

Example12

a

₁2x2d x b

₋₁3 x3d x c

₀1x3−xd x

d

₋2₁(x+1)2d x e

₁2x3d x f

₁4x+2x2d x

g

₀2x3+2x2+x+2d x h

4

12x+5d x

2 Evaluate each of the following:

a

₁2(2−x)(2+x)

x2 d x b

4 12x−3

√

xd x c

3

1

4x2+9

x2 d x

d

4

16x−3 √

xd x e

4

1 x2−1

x2 d x f

4 1

2x−3√x

x d x

3 Given that

₁5h(x)d x =4,evaluate:

a

₁52h(x)d x b

5

1h(x)+3d x c

1 5h(x)d x

4 Given that

5

2 f(x)d x =12,evaluate:

a

₅2f(x)d x b

₂53f(x)d x c

₂4f(x)+4d x+

5 4 f(x)d x

5 On a graph ofy=x2_{, shade the region corresponding to}

3 1x

2_{d x and calculate its value.} Example9

6 On a graph of the liney=2t+3 shade the trapezium between the line and thet-axis bounded byt=1 andt=5. Use geometry to find the area of this trapezium and verify your result by integration.

7 The figure shows part of the graph of the curve with equationy=x(x−1)(3−x).

0 1 3 x

y

Calculate the area of the shaded region.

Example11

8 Calculate the values of

3

1 f(x)d x,

4

3 f(x)d x,

4

1 f(x)d xfor:

a f(x)=6x b f(x)=6−2x

What is the relationship between your three answers in each case?

9 Sketch the graph ofy=5x−x2_{−4 and find the area enclosed by thex-axis and the}

portion of the curve above thex-axis.

10 Sketch the graph ofy=x(10−x) and hence find the area enclosed between thex-axis and the portion of the curve above thex-axis.

11 Sketch the graph ofy=x(x−2)(x+1) and find the area of the region contained between the graph and thex-axis. (Do not attempt to find the coordinates of the turning points.)

12 Find the area bounded by thex-axis and the graph of each of the following functions:

a x2_{−2x b (4−x)(3−x) c (x+2)(7−x)}

23.5

Numerical methods for finding areas

This section is not required for Mathematical Methods (CAS) Units 1 & 2 but can be used as a preparation for Mathematical Methods (CAS) Units 3 & 4.

In this section we consider three methods for determining the area under a graph.

The left-endpoint estimate

We first find an approximation for the area of the shaded region by dividing the region into rectangles as illustrated. The width of each rectangle is 0.5.

8

y

8

7 9 x

f(x) = 9 – 0.1x2

6

6 5 4

4 3 2

2 1 0

R₁R₂R₃R₄R₅R₆

The area of rectangleR1=0.5f(2.0)=0.5×8.60=4.30 square units.

The area of rectangleR2=0.5f(2.5)=0.5×8.38=4.19 square units.

The area of rectangleR3=0.5f(3.0)=0.5×8.10=4.05 square units.

The area of rectangleR4=0.5f(3.5)=0.5×7.78=3.89 square units.

The area of rectangleR5=0.5f(4.0)=0.5×7.40=3.70 square units.

The area of rectangleR6=0.5f(4.5)=0.5×6.98=3.49 square units.

The sum of the areas of the rectangles is 23.62 square units.

The left-endpoint estimate will be larger than the actual area for a graph that is decreasing over the interval and smaller than the actual area for a graph that is increasing.

The right-endpoint estimate

y

x

8 6 4 2

1 2 3 4 5 6 7 8 9

f(x) = 9 – 0.1x2

R₁R₂R₃R₄R₅R₆

Again the rectangles have width 0.5 and we find:

The area of rectangleR1=0.5f(2.5)=0.5×8.38=4.19 square units.

The area of rectangleR2=0.5f(3.0)=0.5×8.10=4.05 square units.

The area of rectangleR3=0.5f(3.5)=0.5×7.78=3.89 square units.

The area of rectangleR6=0.5f(5.0)=0.5×6.5=3.25 square units.

The sum of the areas of the rectangles is 22.67 square units. This is called the right-endpoint estimate for the area. It is an approximation for

₂5(9−0.1x2)d x

Forf decreasing over [a,b]:

left-endpoint estimate≥true area≥right-endpoint estimate

Forf increasing over [a,b]:

left-endpoint estimate≤true area≤right-endpoint estimate

From this it can be seen that a further estimate for the area may be achieved by the average of the two estimates:

Average area= left-endpoint estimate+right-endpoint estimate 2

For the example discussed above, average area=23.15 square units.

It is clear that if narrower strips are chosen we obtain an estimate that is closer to the true value. This is time-consuming to do by hand, but a computer program or spreadsheet makes the process quite manageable. In the spreadsheet shown below the right-endpoint estimate is calculated using rectangles of width 0.1 units.

x y=f(x) f(x)×0.1 x y=f(x) f(x)×0.1 2.1 8.559 0.8559 3.6 7.704 0.7704 2.2 8.516 0.8516 3.7 7.631 0.7631 2.3 8.471 0.8471 3.8 7.556 0.7556 2.4 8.424 0.8424 3.9 7.479 0.7479 2.5 8.375 0.8375 4 7.4 0.74 2.6 8.324 0.8324 4.1 7.319 0.7319 2.7 8.271 0.8271 4.2 7.236 0.7236 2.8 8.216 0.8216 4.3 7.151 0.7151 2.9 8.159 0.8159 4.4 7.064 0.7064 3 8.1 0.81 4.5 6.975 0.6975 3.1 8.039 0.8039 4.6 6.884 0.6884 3.2 7.976 0.7976 4.7 6.791 0.6791 3.3 7.911 0.7911 4.8 6.696 0.6696 3.4 7.844 0.7844 4.9 6.599 0.6599 3.5 7.775 0.7775 5 6.5 0.65

Sum of areas of rectangles=22.9945, therefore

5

2(9−0.1x

The trapezoidal estimate

For this estimate we work with trapeziums instead of rectangles. The area of a trapezium= 1

2(a+b)h,whereaandbare lengths of the parallel sides andh their distance apart.

y

x

8 6 4 2

1 2 3 4 5 6 7 8 9

f(x) = 9 – 0.1x2

T1T2T3T4T5T6

0

The area of trapeziumT1 =

1

2(f(2.0)+ f(2.5))0.5=4.24375 square units. The area ofT2=

1

2(f(2.5)+ f(3.0))0.5=4.11875 square units. The area ofT3=

1

2(f(3.0)+ f(3.5))0.5=3.96875 square units. The area ofT4=

1

2(f(3.5)+ f(4.0))0.5=3.79375 square units. The area ofT5=

1

2(f(4.0)+ f(4.5))0.5=3.59375 square units. The area ofT6=

1

2(f(4.5)+ f(5.00))0.5=3.36875 square units. The sum of the areas=23.0875 square units.

This method is called thetrapezoidal rule. We see from adding that:

the required area= 1

2(f(2)+ f(5)+2[f(2.5)+ f(3)+ f(3.5)+ f(4)+ f(4.5)]) Once again we have found an approximation

for

₂5(9−0.1x2)d x.

y

x

0 x0 =a x1 x2x3 b = xn

If the interval [a,b] on thex-axis is divided intonequal sub-intervals [a,x1], [x1,x2],

[x2,x3], . . . [xn−1,xn] as illustrated, the three

methods may be summarised as follows.

The left-endpoint estimate

Ln =

b−a

n [f(x0)+ f(x1)+ · · · + f(xn−1)] The right-endpoint estimate

Rn =

b−a

The trapezoidal estimate

Tn =

b−a

2n [f(x0)+2f(x1)+2f(x2)+ · · ·2f(xn−1)+ f(xn)]

These methods are not limited to situations in which the graph is either increasing or decreasing for all the interval but may be used to determine the area under any continuous curve. They may be applied to any continuous function on an interval [a,b] to determine an approximate value of

_abf(x)d x.

Exercise

23E

1 To find the area of the region shaded in the figure, calculate:

a the left-endpoint estimate

b the right-endpoint estimate

c the trapezoidal estimate. (Use sub-intervals as shown.)

y

x

0

(1, 5) (2, 3.5)

(3, 2.5) (4, 2.2)

(5, 2)

1 2 3 4 5

2 Calculate an approximation to an area under the graph of

y=x(3−x) betweenx=0 andx=4 using strips of width:

i 0.5 ii 0.2

3 The graph is that ofy = 1

1+x2. It is known that

the area of the shaded region is ␲

4. Apply the trapezoidal rule with strips of width 0.25 and hence find an approximate value for␲.

y

x

0 1

1

4 A table of values is given for the ruley=f(x).

x 0 1 2 3 4 5 6 7 8 9 10

y 3 3.5 3.7 3.8 3.9 3.9 4.0 4.0 3.7 3.3 2.9

Find the area enclosed by the graph ofy=f(x), the linesx=0 andx=10, and thex-axis by using:

a the left-endpoint estimate b the trapezoidal estimate.

5 Use the trapezium rule to find approximate values for:

a

₀22xd x (Use intervals of width 0.5.)

b

₀0.9√ 1

6 A man takes soundings at intervals of 3 metres across a river 30 metres wide. Using the trapezium rule find an estimate for the area of the cross-section of the river’s channel.

Review

Chapter summary

To find thegeneral antiderivativeF: If F(x)= f(x)

then

f(x)d x = F(x)+c,wherecis an arbitrary real number. This gives the following general results:

xrd x = x

r+1

r +1+c, r ∈Q\ {−1}

f(x)+g(x)d x =

f(x)d x+

g(x)d x

k f(x)d x =k

f(x)d x, wherek is a real number

Thefundamental theorem of integral calculusstates that

b

a f(x)d x =G(b)−G(a),

whereGis any antiderivative off, and

b

a f(x)d xis called the definite integral fromatob.

The numberais called thelower limit of integrationandbis called theupper limit of integration.

The functionfis called theintegrand.

Iff(x)≥0 for allx∈[a,b] the area of the region contained between the curve, thex-axis and the linesx=aandx=bis given by

b

a f(x)d x.

Iff(x)≤0 for allx∈[c,d] the area of the region contained between the curve, thex-axis and the linesx=candx=dis given by –

_cdf(x)d x.

Ifc∈[a,b], f(c)=0 and f(x)≥0 for

x∈(c,b] andf(x)≤0 forx∈[a,c), then the area of the shaded region is given by

b

c f(x)d x + −

c

a f(x)d x c _b x

a y

0

Three numerical methods are used for finding areas.

x0= a x1x2x3

0

y

x b = xn

The interval [a,b] on thex-axis is divided intonequal sub-intervals [a,x1], [x1,x2], [x2,x3], . . . , [xn– 1,xn].

r The left-endpoint estimate Ln =

b−a

n [f(x0)+ f(x1)+ · · · f(xn−1)]

r The right-endpoint estimate Rn =

b−a

n [f(x1)+ f(x2)+ · · · f(xn)]

r The trapezoidal estimate Tn =

b−a

Review

Multiple-choice questions

1 An antiderivative ofx3_+3xis

A x(x2+3) B 3(x2 +1) C x

4

4 + 3x2

2 +c

D x3+3+c E 3x3+x2+c

2

√x +xdxequals

A

√ x2+_x2

2 +c B x

3

2 +x2_{+c C} 1

2x+1

D 2x

3 2

3 + 1 2x

2_{+c E 2}

3 3

x−4 dxequals

A −1

x3 +c B −x

3_{+c C −12x}−3_+c

D −12x−5+c E −3x

−4

5 +c

4 An expression fory, if d y

d x = 2x+5 andy=1 whenx=0, is

A y=x2_{+5x B y=2 C y=x}2_+5x−1

D y=x2+5x+1 E y=x2 +5x−5

5 Given that f(x)=5x4_{– 9x}2_{andf(1)=2, thenf(x) is equal to}

A x5– 3x2 +2 B x5 – 3x3+4 C x5– 9x2+5

D x5_{– 3x}3 _{E x}5 _{– 3x}2₊₂

6 An expression fory, if d y

d x =

4

x3 andy=0 whenx=1, is

A y=−2

x2 B y=

−2

x2 +2 C y=

−2 3x2 −

2 3

D y=−4

x2 −2 E y=2x

7 IfF(x)=f(x), then

₃5f(x)d x is

A f(5)−f(3) B f(5)+c C f(5)−f(3)+c

D F(5)−F(3) E F(5)+F(3)

8

2

0(3x

2_{−2x)d x =}

A 12 B 4 C 8 D 32 E 2

9 An equivalent expression for

2

03f(x)+2dxis

A 3

2

0f(x)d x+2x B 3

2

0 f(x)d x+2x C 3

2

0 f(x)d x +4

Review

10 The graph with the equationy=k(x−3)2_{is shown.}

If the area of the shaded region is 36 square units, the value ofkis

A 4 B −4 C 9

D 4

7 E 32

x (3, 0)

y

0

Short-answer questions (technology-free)

1 Find:

a

1

2d x b

1 2x

2

d x c

(x2+3x)d x d

(2x+3)2d x

e

atdt f

1

3t

3_dt

g

(t +1)(t−2)dt h

(2−t)(t+1)dt

2 The curve with equationy=f(x) passes through the point (3, –1) and f(x)=2x+5. Find f(x).

3 The curve with equationy=f(x) passes through the origin. Let f(x)=3x2_{– 8x+3.}

a Findf(x). b Find the intercepts of the curve with thex-axis.

4 Find:

a

2x−3d x b

√x(x−2)d x c

3x

2_+2x

x d x

d

3x+1

x3 d x e

5x−2√xd x f

5x 3 4 −_2x

1 3_{d x}

g

2−√xd x h

3x

2_+2x x4 d x

5 Findsin terms oft, if ds

dt =t +3−

1

t2 ands=6 whent=1.

6 Evaluate each of the following definite integrals:

a

₁22xd x _b

5

22d x c

5 3(3x

2_{+2x)d x}

d

5

1(x

2_{+2x)d x}

e

₋₃−25d x

7 Evaluate each of the following definite integrals:

a

₁4√xd x _b

4

1x

3_{−2xd x}

c

2

1

1

x2d x

d

4

1

2

x3d x e

1 0

√

x(x+1)d x

8 On a graph ofy=x3_{, shade the region corresponding to}

2 1x

Review

9 The figure shows the graph ofy=(1 –x)(2+x). Find the area of the shaded region.

y= (1 –x)(2 + x)

–2 1

2

x y

0

10 The figure shows the graph of the curve with equation

y=x(x−3)(x+2).

Calculate the area of the shaded region.

0 3

–2

y

x

y = x(x– 3)(x + 2)

11 a Find the coordinates of pointsBandC.

b Find the area of rectangleABCD.

c Find the area of the shaded region.

y

x

1 3

y = x(4 –x)

A D B C

0

Extended-response questions

1 The slope of a children’s slide is given by d y

d x =

9 32(x

2_{−4x) forx ∈[0,4], where the}

origin is taken to be at ground level beneath the highest point of the slide, which is 3 m above the ground. All units are in metres.

a Find the equation of the curve which describes the slide.

b Sketch the curve of the slide, labelling stationary points.

c Does the slope of the slide ever exceed 45◦?

2 A swimming pool has a cross-sectional area as shown.

a Find the area of the rectangleOABC.

b Find the equation of the curve given that it is of the formy=k(x– 4)2.

c Find the total area of the region enclosed between the curve and thex-axis forx∈[0, 9].

d Find the area of the cross-section of the pool (i.e. the shaded region).

x y

(4, 0)

O

B (9, 3)

Review

3 a Water flows into a container at the constant rate (R) of 2 litres/second. A graph of the rate of flow is as shown.

i How much water has flowed into the container after 1 minute?

ii Illustrate this quantity as an area under the graph ofR=2. 0

R

2 (L/s)

t (s)

b Water flows into a container at a rateRL/s, where

R= t

2 andtis the time measured in seconds.

The graph is as shown.

i How much water has flowed into the container after 1 minute?

ii Illustrate this quantity as an area under the graph ofR= t

2.

iii How much water has flowed into the container afteraminutes?

0 _{t (s)}

R R= t

2 (L/s)

c Water flows into a container at a rateRL/s, where

R= t 2

10andtis the time measured in seconds.

i Find the area of the shaded region.

ii What does the area represent?

iii After how many seconds will 10 000 litres have flowed into the container?

0

R

R=

60 10

t2

(L/s)

t (s)

4 a i A car travels on a straight road at 60 km/h for 2 hours. Sketch the speed–time graph illustrating this.

ii Shade the region which indicates the total distance travelled by the car after 2 hours.

b i Sketch the speed–time graph of a car travelling for 5 minutes, if the car starts from rest and accelerates at a rate of 0.3 km/min2.

ii How far has the car travelled at the end of 5 minutes?

c A particle starts from a pointOand travels at a velocityVm/s given byV=20t−3t2, wheretis the time the particle has been travelling measured in seconds.

i Find the acceleration of the particle at timet.

ii Sketch the graph ofVagainsttfor 0≤t≤ 20

3.

iii How far has the particle travelled after 6 seconds? Illustrate this quantity by shading a suitable region under the graph.

5 A large mound of earth has a constant cross-sectional area.The cross-section is described by the rule

y= x 2

1000(50−x), whereydenotes the height of the mound in metres at a distance ofxmetres from the edge. (xis measured from 0.)

0 x

y

50

a Find the height of the mound when:

Review

b Find the gradient of the boundary curvey = x 2

1000(50−x) when:

i x=10 ii x=40

c i Find the value ofxfor which the height of the mound is a maximum.

ii Find the maximum height of the mound.

d Find the cross-sectional area of the mound. y

x

0 ₅₀

B(b, 12)

A(20, 12)

e It is decided to take the ‘top’ off the mound as shown.

i PointAhas coordinates (20, 12). Find the coordinates of pointB.

ii Find valuesp,qandRsuch that the area of the cross-section of the top is determined by

p q

x2

1000(50−x)d x−R.

6 a The curve with equationy=f(x) passes through the point with coordinates (1, 6) and

f(x)=6x+3.

i Find the gradient of the curve at the point with coordinates (1, 6).

ii Find the equation of the tangent to the curvey=f(x) at this point.

iii Find the equation of the curve.

b The curve with equationy=f(x) is such that the tangent to the curve at the point with coordinates (2, 10) passes through the origin andf(x)=6x+k.

i Find the gradient of this tangent in terms ofk.

ii Find the value ofk. iii Find the equation of the curve.

7 a An irrigation channel 2 metres deep is constructed. Its cross-section, trapezium

ABCD, is shown in the diagram.

AD=5 metres andBC=1 metre. Calculate the area enclosed by this cross-section.

A D

C B

2 metres

b The cross-section is placed symmetrically on coordinate axes as shown.

i Find the equation of the lineCD.

ii Calculate the area of the cross-section of the water (shaded) when the water isymetres deep.

Give your answer in terms ofx.

x x

y y

0

B A

C

Review

c In an attempt to improve water flow a metal chute is added. Its cross-section has the shape of a parabolaPQORS

and it just touches the original channel atQ,OandR.

i Given that thex-coordinate ofR

is 1, find the equation of the parabola and find the coordinates ofPandS.

ii Calculate the area of the cross-sectionPQORS.

T B C V

R

D S P

A

y

Q _O

0

(Hint: You could find the area of rectanglePTVSand subtract the area ‘under’ the parabola.)

8 A small hill has a cross-section as shown. The coordinates of four points are given. Measurements are in metres.

a Find the equation of the cubic.

b Find the maximum height of the hill correct to the nearest metre.

x y

0

(0, 60)

(300, 96)

(800, 0) (700, 15.2)

c i Using a CAS calculator, plot the graph of the gradient function.

ii State the coordinates of the point on the curve where the magnitude of the gradient is a maximum.

d Use a CAS calculator to find the cross-sectional area correct to the closest square metre.

9 a Use a CAS calculator to plot the graph of g(x)=

x

0 f(t)dt,where f(t)=2t for x ∈[0,5].

b Solve the equation

x

0 f(t)dt =10.

x y

y = 2x