IBHM 373 402 .pdf



Nom original: IBHM_373-402.pdf
Titre: IBHM_Ch14v3.qxd
Auteur: Claire

Ce document au format PDF 1.6 a été généré par Adobe Acrobat 7.0 / Acrobat Distiller 7.0.5 for Macintosh, et a été envoyé sur fichier-pdf.fr le 07/06/2014 à 21:15, depuis l'adresse IP 87.66.x.x. La présente page de téléchargement du fichier a été vue 487 fois.
Taille du document: 514 Ko (30 pages).
Confidentialité: fichier public


Aperçu du document


14 Integration 1
Jakob Bernoulli was a Swiss
mathematician born in Basel,
Switzerland, on 27 December 1654.
Along with his brother, Johann, he is
considered to be one of the most
important researchers of calculus after
Newton and Leibniz. Jakob studied
theology at university, but during this
time he was studying mathematics and
astronomy on the side, much against
Jakob Bernoulli
the wishes of his parents. After
graduating in theology he travelled
around Europe and worked with a number of the great mathematicians of the time.
On return to Basel, it would have been natural for him to take an appointment in the
church, but he followed his first love of mathematics and theoretical physics and took
a job at the university. He was appointed professor of mathematics in 1687 and, along
with his brother, Johann, started studying Leibniz’s work on calculus. At this time
Leibniz’s theories were very new, and hence the work done by the two brothers was at
the cutting edge. Jakob worked on a variety of mathematical ideas, but in 1690 he
first used the term “integral” with the meaning it has today. Jakob held the chair of
mathematics at the university in Basel until his death in 1705. Jakob had always been
fascinated by the properties of the logarithmic spiral, and this was engraved on his
tombstone along with the words “Eadem Mutata Resurgo” which translates as “I shall
arise the same though changed.”

14.1 Undoing differentiation
In Chapters 8–10 we studied differential calculus and saw that by using the techniques
of differentiation the gradient of a function or the rate of change of a quantity can be
found. If the rate of change is known and the original function needs to be found, it is
necessary to “undo” differentiation. Integration is this “undoing”, the reverse process
to differentiation. Integration is also known as anti-differentiation, and this is often the
best way of looking at it.
dy
2x, what is the original function y?
dx
This is asking what we started with in order to finish with a derived function of 2x.
Remembering that we differentiate by multiplying by the power and then subtracting
one from the power, we must have started with x2.
dy
Similarly, if
4, then this must have started as 4x.
dx
If

Exercise 1

373

14 Integration 1

Exercise 1
Find the original function.
1

dy
5
dx

2

dy
10
dx

3

dy
2
dx

4

dy
4x
dx

5

dy
12x
dx

6

dy
3x2
dx

7

dy
4x3
dx

8

dy
5x4
dx

9

dy
9x2
dx

10

dy
4x 2
dx

Looking at the answers to Exercise 1, we can form a rule for anti-differentiation.
If we describe the process of “undoing” differentiation for this type of function, we
could say “add 1 to the power and divide by that new power”.
In mathematical notation this is
This symbol means “the
integral of”.

冮x

n

dx

xn 1
n 1
This means “with
respect to x” – the
variable we are
concerned with.

14.2 Constant of integration
dy
4. Geometrically, this means that the gradient of
dx
the original function, y, is constant and equal to 4.
Again consider the situation of

So y 4x c (c is the y-intercept of the line from the general equation of a line
y mx c ).
Consider the lines y 4x 3, y 4x and y 4x 5.
dy
4 (as the gradient is 4 each time). Remember that, when
dx
differentiating, a constant “disappears” because the gradient of a horizontal line is zero
For each line

or alternatively the derivative of a constant is zero.
So

冮4 dx 4x c.

Unless more information is given (a point on the line), then the value of c remains
dy
4. In fact
unknown. So any of the lines below could be the original function with
dx
there are an infinite number of lines that could have been the original function.
y 4x 2
y 4x 1
y 4x
y 4x 2

y

2
1

0.5
2

374

x

14 Integration 1

This “c” is called the constant of integration. It must be included in the answer of any
integral, as we do not know what constant may have “disappeared” when the function
was differentiated.

Example
Find y if

dy
8x.
dx



y 8x dx 4x2 c

Example
Find

冮10x 7dx.

冮10x 7dx 冮10x dx 冮 7 dx
So 冮10x 7dx 5x 7x c Example
So

2

We established in
Chapter 8 that a function
could be differentiated
“term by term”. In a
similar way, this can be
integrated term by term.

Example
Integrate 8x 2 x4.
3

冮8x

2

x4 dx 8x 1
3

4 74
x c
7

The coefficient of a term has no effect on the process of integration, and so a constant
can be “taken out” of the integral. This is demonstrated in the next example.

Example
Find the solution of

dy
2 1
x 3.
dx
9

So

冮9 x
2
y 冮x
9

y

2

13

dx

13

dx

This is asking to find y
by integrating.

2 # 3 23
x c
9 2
1 2
y x3 c
3
y

As with the expressions differentiated in Chapters 8–10, it is sometimes necessary to
simplify the function prior to integrating.

Example

375

14 Integration 1

Example
x2 x5

冮 32x dx.

Find

x2 x5

冮 32x dx

The integral sign and the dx
remain until the integration is
performed.



1 12 2

x 1x x5 2 dx
3




1 32
9
x x2 dx
3
The c can remain outside all
brackets as it is an arbitrary
constant and so, when
multiplied by another constant,
it is still a constant.

1 2 52
2 112
B x
x R c
3 5
11
2 112
2 52

x
x c
15
33


Example

冮p dp.
4

Find

Here p is the variable and so
the integral is with respect to p.

2

冮p dp
4

2



4p 2 dp
4p 1 c
4
c
p

Exercise 2
Integrate these expressions.
1 2x 1

2 x2

3 x3

4 x4

5 6x2 5

6 8x3 4x 3

7 5x2 4

8 x 2

10 x 3

1

11 7 4x 3

2

9 x2
Find these integrals.
12

冮x

15

冮4x

12

3

dx

13

冮2x

4x 9 dx

16

冮1 2x 6x

19

dy
1
2x
dx
2x

6

5x4 dx
2

x3 dx

14

冮5x 4x

17

冮x dx

20

dy
3
8 1
dx
x3

3
2

3

6

3

Find the solution of these.
18

376

dy
2
5
dx
x

dx

14 Integration 1

dy
8x12x2 32
dx
dy
x2 5
24

dx
x5
21

dy
1x 92 12x 32
dx
dy
4x3 7x
25

dx
2x
22

dy
13x 42 2
dx
3
dy
7x4 6x4
26

1
dx
2x4
23

Find y by integrating with respect to the relevant variable.
27

dy
12
3
dp
p

28

dy
5
8k4
dk

30

1t 32 2
dy

dt
t4

31

dy
2t 4t3

dt
3t

29

dy
1
z3 ¢z2 2 ≤
dz
z

14.3 Initial conditions
In all of the integrals met so far, it was necessary to include the constant of integration,
c. However, if more information is given (often known as initial conditions), then the
value of c can be found.
dy
Consider again the example of
4. If the line passes through the point (1, 3) then
dx
c can be evaluated and the specific line found.
So

dy
4
dx



1 y 4 dx
1 y 4x c
Since we know that when x 1, y 3 these values can be substituted into the
equation of the line.
So

3 4 1 c
1 c 1

The equation of the line is y 4x 1.
When the value of c is unknown, this is called the general solution.
If the value of c can be found, this is known as the particular solution.

Example
Given that the curve passes through (1, 3) and
of the curve.

dy
2x 1, find the equation
dx

dy
2x 1
dx



1 y 2x 1 dx
1 y x2 x c
Using (1, 3)

1 3 12 1 c
1c 3

So the equation of the curve is y x2 x 3.

377

14 Integration 1

Example
Find P given that

dP
1

and P 7 when t 100
dt
2t

1
dP

dt
2t
1P

冮 2t dt
1



1 P t 2 dt
1

1

1 P 2t 2 c
Since P 7 when t 100, 7 2# 2100 c
1 c 13
1
2

Hence P 2t 13

Exercise 3
Given the gradient of each curve, and a point on that curve, find the equation
of the curve.
1

dy
6, 12, 82
dx

2

dy
4x, 11, 5 2
dx

3

dy
8x 3, 1 2, 42
dx

4

dy
2x 5, 14, 42
dx

5

dy
4x3 6x2 7, 11, 92
dx

6

dy
6
4x2 2 , 14, 12
dx
x

7

dy
8

, 19, 22
dx
2x

Find the particular solution, using the information given.

378

8

dy
t2 1t 4 3t 2 42, y 6 when t 1
dt

9

p3 4p5
dQ

, Q 2 when p 0
dp
32p

14 Integration 1

14.4 Basic results
Considering the basic results from differentiation, standard results for integration can
now be produced. For polynomials, the general rule is:



xn dx

However, consider

冮 x dx 冮x
1

1

dx.

Using the above rule, we would obtain
that

xn 1
c
n 1

x0
, but this is not defined. However, it is known
0

d
1
1ln x2 .
x
dx
This provides the result that

冮 x dx ln冟x冟 c
1

Remembering that ln x is defined only for positive values of x, we recognize that
1
1
is defined for all x H ⺢, x 0
dx ln冟x冟 c, taking the absolute value of x. As
x
x
and ln x is defined only for x 7 0, the absolute value sign is needed so that we can
1
integrate
for all values of x for which it is defined.
x



Similarly

d x
1e 2 ex so
dx

冮e dx e c
x

x

When differentiating sine and cosine functions the following diagram was used and is
now extended:

Differentiate

S
C
S
C

Integrate

The integrals of other trigonometric functions can be found by reversing the basic rules
for differentiation, and will be discussed further in Chapter 15.

Standard results
Function

Integral ⴙ c

f(x)

冮f1x2 dx

xn 1n 12

xn 1
n 1

1
x
ex
sin x
cos x

ln x
ex
cos x
sin x

379

14 Integration 1

Example
Integrate sin x ex.

冮sin x e dx
x

cos x ex c

Example
Integrate

3
4 cos x.
x

冮 x 4 cos x dx
3

3 ln冟x冟 4 sin x c

Exercise 4
Integrate these functions.
2
1 x3
x
4 6 sin x 6x4
7

ex
5

7 sin x
3
2x

5
cos x
x

2 4ex sin x

3

5 8 sin x 7ex

6 5ex 2 sin x

8

3
x

ex
152x cos x
15

14.5 Anti-chain rule
When functions of the type 12x 12 5, e8x and sin¢2x

p
≤ are differentiated, the
3

chain rule is applied. The chain rule states that we multiply the derivative of the outside
function by the derivative of the inside function. So to integrate functions of these types
we consider what we started with to obtain that derivative.

Example
dy
12x 12 3. Find y.
dx



So y 12x 12 3 dx
When integrating, 1 is added to the power, so y must be connected to
12x 12 4. Since we multiply by the power and by the derivative of the inside
function when differentiating, we need to balance this when finding y.

380

So y

1#1
12x 12 4 c
4 2

1y

1
12x 12 4 c
8

14 Integration 1

Example
Find

冮 cos 3x dx.

S
C
Using Diff
Int this begins with sin 3x.
S
C
Balancing to obtain cos 3x when differentiating
1

冮 cos 3x dx 3 sin 3x c
1

Example
Find

冮6e

4x

dx.

This started with e4x as



d 4x
1e 2 4e4x.
dx



So 6e4x dx 6 e4x dx
6#


1 4x
e c
4

3 4x
e c
2

Example
Find y given that

So y
As

dy
1
p

sin¢4x ≤.
dx
3x 4
2

冮3x 4 sin¢4x 2 ≤ dx
1

p

1
13x 42 1
3x 4

we recognize that this comes from ln冟3x 4冟 as
So y

d
3
1ln冟3x 4冟 2
.
dx
3x 4

1
p
1
ln冟3x 4冟 cos¢4x ≤ c
3
4
2

For these simple cases of the “anti-chain rule”, we divide by the derivative of the
inside function each time. With more complicated integrals, which will be met in
the next chapter, this is not always the case, and at that point the results will be
formalized. This is why it is useful to consider these integrals as the reverse of
differentiation.

381

14 Integration 1

Exercise 5
Find these integrals.
1
5
9
13

冮 sin 5x dx
冮8 cos 4x dx
冮e dx
冮8x e dx
5x

2x

2
6
10
14

冮 cos 6x dx
冮 6 sin 3x dx
冮4e dx
冮4e dx
4x

3
7
11

冮 sin 2x dx
冮 5 cos 2x dx
冮8e dt
6t

4
8
12

冮 sin 2 x dx
冮e dx
冮 5e dp
1

6x

6p

2x

Find y if:
dy
1

dx
2x 3
dy
13x 12 5
18
dx
dy
21
13 2x2 4
dx
15

dy
4

dx
3x 1
dy
3

27
dt
6 t
24

dy
1

dx
8x 7
dy
14x 72 6
19
dx
dy
4
22

dx
13x 22 3

dy
4

dx
2x 5
dy
14x 32 3
20
dx
dy
3
23

dt
12t 12 2

16

25

17

dy
6

dx
3x 5

26

dy
8

dp
4 p

Integrate these functions.
28 6e4x
1
13x 42 5
31
2x 1

30 4e 8x 4 cos 2x

sin 3x 4x
2
32 6x2
3x 2
29

14.6 Definite integration
Definite integration is where the integration is performed between limits, and
this produces a numerical answer.
Upper limit 1x b2
b

A definite integral is of the form

冮 f1x2 dx

a

Lower limit 1x a2

Example
3

冮 2x 1 dx

2

c x2 x d

3

2

382

This notation means that the
integration has taken place.
The two values are now
substituted into the function
and subtracted.

When a definite integral
is created, the lower
limit is always smaller
than the upper limit.

14 Integration 1

132 32 122 22
6 2
4
There is no constant of integration used here. This is because it cancels itself out
and so does not need to be included.
c x2 x c d

3

2

132 3 c2 122 2 c2
6 c 2 c
4

Example
4

冮 12x 32

2

dx

0

4

1
B 12x 32 3R
6
0


1
1
18 32 3 ¢ 10 32 3≤
6
6



125
27

6
6



152
6



76
3

Example
p
4

0

冮 sin 2x 1 dx
p

4
1
B cos 2x xR
2
0

1
p
p
1
¢ cos ≤ ¢ cos 0 0≤
2
2
4
2


p
1

4
2



p 2
4

To differentiate trigonometric
functions, we always use
radians, and the same is true
in integration.

383

14 Integration 1

Example
2

冮 x 4 dx
2

2

c 2 ln冟x 4冟 d

2
2

2 ln冟 2冟 2 ln冟 6冟
2 ln 2 2 ln 6
ln 4 ln 36

An answer could have
been approximated earlier,
but if an exact answer
were required, this form
would need to be given.

4
ln
36
ln

1
9

2.20

Example
Although there is no value for
the upper limit, an answer
can still be found that is an
expression in a.

a

冮e

4x

dx where a 7 2

2

a

1
B e4xR
4
2


1 4a 1 8
e e
4
4



1 4a
1e e8 2
4

Exercise 6
Find the value of these definite integrals
2

1

3

冮 2x dx

2

1

冮 8x 4x

dx

5

冮e

dx

8

0



冮 5 dx

0

1

4
dx
x2

6

冮 x dx
6

3

3

冮 sin 3x dx

1

冮 4e

3x

dx

9

冮 cos 2u du

0

4x

p
6

p
2

11

冮 2e

dx

2

2

p
3

384

3

4
2x

0

dx

2

3

10

2

3
3

0

7

冮 6x

2

2

4

4

12

冮 cos 3t 6 dt

0

14 Integration 1
3

13

0

冮 12x 12

2

dx

14



2
dx
12x 12 3

1

17

2

冮 13 2x2

15

20

4

dx

5




4

1

冮 2x 1 dx
1

18

2

1

4
dx
3x 4

p

22

dx

3
dx
13x 42 2

2

4



3

1

4

19

冮 13x 12

2

0

16

2

2

5
dt
t 2

冮 6p 4p 1 dp
3

21

1

k

冮 sin 4x 6x dx

p

23

冮 2x 1 dx
4

0

14.7 Geometric significance of integration
When we met differentiation, it was considered as a technique for finding the gradient
of a function at any point. We now consider the geometric significance of integration.
5

Consider

冮 3 dx.

Remember the limits
are values of x.

y

1
5

冮 3 dx

3

y 3

1

c 3x d

5

0

1

1

5

x

15 3
12
This is the same as the area enclosed by the function, the x-axis and the vertical lines
x 1 and x 5.
This suggests that the geometric interpretation of integration is the area between the
curve and the x-axis.
Consider y 3x2.
y
y 3x2

0

1

3 x

In the previous example where y 3, it was easy to find the area enclosed by the
function, the x-axis and the limits. However, to find the area under a curve is less obvious.
This area could be approximated by splitting it into rectangles.

385

14 Integration 1
y
27

y 3x2

A ⯝ 1 3 1 12
12

15 square units

3
0

1 2

3 x

This is clearly not a very
accurate approximation,
so to make it more
accurate thinner
rectangles are used.

Four rectangles would make this more accurate.
y
27
75
4

y 3x2

A⯝

1
1
27
1
1
75
3
12
2
2
4
2
2
4

A⯝

81
4

12

27
4

3

0

1 2 3 x
1.5
2.5

As the rectangles become thinner, the approximation becomes more accurate.

1

x

3

x

This can be considered in a more formal way. Each strip has width dx with height y and
area dA.
So dA ⯝ y # dx
1 A a dA
1 A ⯝ a y dx
As dx gets smaller, the approximation improves
x b

and so A lim a ydx.
dxS0
x a

dA
Now y ⯝
dx
1 y lim

dxS0

1y

dA
dx

Integrating gives

386

dA
dx

冮y dx 冮 dx dx 冮dA A c
dA

Limits are needed here to
specify the boundaries
of the area.

14 Integration 1

With the boundary conditions,

c can now be ignored.
b

冮 y dx

A

This is the basic formula
for finding the area
between the curve and
the x-axis.

a

b
x b

Hence lim a y # dx
dxS0
x a

冮 y dx.

a

This now shows more formally that the geometric significance of integration is that it
finds the area between the curve and the x-axis.
If the two notations for summation are compared, we find that sigma notation is used
for a discrete variable and that integral notation is used for a continuous variable.

So the

冮 sign actually means

“sum of” (it is an elongated S).

3
3
2

a 3x

and

1

冮 3x

2

This is the sum of a
continuous variable.

dx

1

c x3 d
This is a sum of a
discrete variable.

3

1

27 1

The area required is
26 square units.

26

Example
4

Find the area given by

冮 2x 1 dx.

2
4

冮 2x 1 dx

2

cx xd

4

2

2

116 42 14 22
12 2
10 square units
y 2x 1
y

0
1

2

4

x

387

14 Integration 1

This integration can be performed on a calculator. Although a calculator
cannot perform calculus algebraically, it can calculate definite integrals.
This is shown in the diagram below, and we find the area is still 10 square
units.

Example
p
Find the area enclosed by y cos 2x and the x axis, between x
and
4
p
x .
2

A

p
2
p
4

冮 cos 2x dx
p

2
1
B sin 2xR
p
2
4



1
1
p
sin p sin
2
2
2

0


1
2

1
2

The answer to this definite integral is negative. However, area is a scalar quantity
1
(it has no direction, only magnitude) and so the area required is
square unit.
2
As the negative sign has no effect on the area, the area can be considered to be
b



the absolute value of the integral. So A 2 f1x2 dx2. Whether the calculation
a

is done using the absolute value sign or whether we do the calculation and then
ignore the negative sign at the end does not matter. As can be seen from the
graph, the significance of the negative sign is that the area is contained below
the x-axis.

388

Some definite integration
questions require the use of
a calculator. Where an exact
value is required, one of the
limits is a variable or the
question is in a non-calculator
paper, algebraic methods
must be employed.

14 Integration 1

Example
1

Find the area given by

冮 x dx.
1

2
1

A 2

冮 x dx2
1

2

2 c ln冟x冟 d

1

2

2

冟ln冟 1冟 ln冟 2冟冟
冟ln 1 ln 2冟
冟 ln 2冟
0.693
So the required area is 0.693 square units.

1

It should be noted that it is actually not possible to find the area given by

冮 x dx as
1

2

there is a vertical asymptote at x 0. It is not possible to find the definite integral over
an asymptote of any curve, as technically the area would be infinite.
This example also provides another explanation for the need for the modulus signs in

冮 x dx ln冟x冟. Although logarithms are not defined for negative values of x, in order to
1

1
find the area under a hyperbola like y , which clearly exists, negative values need to
x
be substituted into a logarithm, and hence the absolute value is required. This was
shown in the above example.

Example
Find the area enclosed by y 12x 12 2 4 and the x-axis.
First the limits need to be found.

389

14 Integration 1

These are the roots of the graph (they can be found algebraically or by using a
calculator)
12x 12 2 4 0
1 2x 1 ;2
1x

3
1
or x
2
2

So the area is given by
3
2

冮 12x 12

2

4 dx

12

3

2
1
B 12x 12 3 4xR
6
12



16
3

Exercise 7
Find the area given by these definite integrals.
6

1

3

冮 8 dx

2

2

冮 13x 22

4

dx

5

8

1

2

10



4

冮 sin x dx


2

6

11



冮 e dx
x

2

2
dx
2x 3

9

e2x 1 4x dx

冮 3x 2 dx

4
p
3

12

0

13 y x2 2 and the lines x 1 and x 4
14 y e6x and the lines x 2 and x 1
15 y sin x and the lines x 0 and x p
4
and the lines x 5 and x 2
3x 4

17 y 1 x2 and the lines x 1 and x 1
18 y 14x 12 2 9 and the lines x 1 and x
19 y x3 2x2 and the lines x 0 and x 2
p
20 y e2x sin 2x and the line x and the y-axis
2

390

冮 cos 3x 4x dx

0

Find the area enclosed by the curve and the x-axis.

16 y

1 dx

0

1

4
dx
2x 1

2

2

5

4
dx
x

冮 6x

1

0

4



3

p

0

7

冮 2x 3 dx

0

2

4

3

1
2

14 Integration 1

21 Find an expression in terms of p for this area.

y

2
3x 5

y
0

p x

1

22 Find an expression in terms of p for this area.
y e 2x x2

y

x

p

0

k

23 Find k 1k 7 02 given that
˛

冮 3x

1
2

dx 16.

0
a

24 Find a given that

冮 19 x2
25

a

2

5
dx .
8

14.8 Areas above and below the x-axis

In this case the formula needs to be applied carefully. Consider y x 1x 12 1x 22 and the
area enclosed by this curve and the x-axis.
˛

y x3 3x2 2x

2

If the definite integral

冮x

3

3x2 2x dx is calculated, we obtain an answer of 0 (remember

0

the calculator uses a numerical process to calculate an integral) and so this result is interpreted
as zero.

391

14 Integration 1

However, it is clear that the area is not zero.
We know that a definite integral for an area below the x-axis provides a negative answer.
This explains the zero answer – the two (identical) areas have cancelled each other out.
So although the answer to the definite integral is zero, the area is not zero.
To find an area that has parts above and below the x-axis, consider the parts separately.
1

So in the above example, area 2

冮x

3

3x2 2x dx

0

1
2
4
1
unit2
2
This demonstrates an important point. The answer to finding an area and to finding the
value of the definite integral may actually be different.
The method used in the example below shows how to avoid such problems.

Method
1 Sketch the curve to find the relevant roots of the graph.
2 Calculate the areas above and below the x-axis separately.
3 Add together the areas (ignoring the negative sign).

Example
Find the area enclosed by y x2 4x 3, the x-axis and the y-axis.
y

A
x

0

B

1. Sketch the curve and shade the areas required.
The roots of the graph are given by x2 4x 3 0
1 1x 12 1x 32 0

1 x 1 or x 3

392

These results can be
found using a
calculator.

14 Integration 1

2. Work out the areas separately.
1



3



A 2 x 4x 3 dx2

B 2 x2 4x 3 dx2

2

0

1

1
1
2 B x3 2x2 3xR 2
3
0

3
1
2 B x3 2x2 3xR 2
3
1

1
2 ¢ 2 3≤ 102 2
3

4
2 19 18 92 ¢ ≤ 2
3



4
4
2 2
3
3

4
3

3. These areas can be calculated on a calculator (separately) and then added.
8
So the total area square units.
3

Example
Find the area bounded by y sin x, the y-axis and the line x

5p
.
3

Graphing this on a calculator,

5
3

This area can be split into:

5p
3

p

冮 sin x dx

and

p

0

c cos x d

冮 sin x dx
c cos x d

p

5p
3

p
0

1 cos p2 1 cos 02

1 1 12
2

¢ cos

1
¢ ≤ 112
2


So the total area is

5p
≤ 1 cos p2
3

3
2

7
square units.
2

393

14 Integration 1

Exercise 8
Find the shaded area on the following diagrams.
1 y x2 8x 12
y

0

x

2 y x2 9x 8
y

x

0

3 y 4 sin x
y
4



0

x
2

4

4 y 5x3
y

1

x
0

2

Find the area bounded by the curve and the x-axis in the following cases.
5 y x 1x 32 1x 22
6 y 1x 42 12x 12 1x 32
˛

7 y x3 x2 16x 16
8 y 6x3 5x2 12x 4

394

14 Integration 1

Find the area bounded by the curve, the x-axis and the lines given.
9 y x2 x 6, x 2 and x 4
10 y 6x3, x 4 and x 2
11 y 3 sin 2x, x 0 and x

5p
6

p
≤, x 0 and x p
6
p
5p
13 y 4 cos 2x, x
and x
2
4
12 y cos¢2x

14 y x3 ex, x 0 and x 4

14.9 Area between two curves
The area contained between two curves can be found as follows.
b

The area under f(x) is given by

b

冮 f1x2 dx and under g(x) is given by 冮 g1x2 dx.

a

a

b

So the shaded area is

b

冮 f1x2 dx 冮 g1x2 dx.

a

y
g(x)

a

f (x)
b

Combining these gives



f1x2 g1x2 dx.

0

a

b

This can be expressed as

冮 upper curve – lower curve dx.

a

b

x

As long as we always
take upper – lower, the
answer is positive and
hence it is not necessary
to worry about above
and below the x-axis.

a

Example
Find the shaded area.
y 2x2

y

y 4x

0

The functions intersect where 2x2 4x.
1 2x2 4x 0
1 2x 1x 22 0
1 x 0 or x 2
˛

x

These intersection
points can also be found
using a calculator.

395

14 Integration 1
2

So the area

冮 4x 2x

2

dx

0

B2x2
¢8


2 3 2
xR
3
0

16
≤ 102
3

8
3

This function can be drawn using Y1-Y2 and then the area calculated.
The two limits are roots of
the resulting function.

Example
Find the area contained between y ex and y sin x from x p to
x p.





Note that these graphs cross twice within the given interval. So, we need to find
the intersections and then treat each part separately (as the curve that is the
upper one changes within the interval).

396

14 Integration 1

So we
we need
need to
to find
find
So
3.096
3.096



1 exx 2

sin xx 1 e 2 dx
dx.
sin

p
p

p

0.589



sin and
xdx and

x

e

p

x
sin xsin x1 e1x 2 edx.
2 dx.

0.589
0.589

3.096

25.7 13 sf2
The total area 0.00976 ... +1.32 p 24.4 p
The total area 0.00976 ... 1.32 p 24.4 p 25.7 13 sf2

Area between a curve and the y-axis
Mostly we are concerned with the area bounded by a curve and the x-axis. However, for
some functions it is more relevant to consider the area between the curve and the y-axis.
This is particularly pertinent when volumes of revolution are considered in Chapter 16.
b

The area between a curve and the x-axis is

冮 y dx

a

To find the area between a curve and the y-axis we calculate

b

冮 x dy

This is where x is a function
of y.

a

This formula is proved in an identical way to the area between the curve and the x-axis,
except that thin horizontal rectangles of length x and thickness dy are used.

Example
Consider y2 9 x.
y
3

x

0
3

397

14 Integration 1

There is no choice here but to use horizontal strips as opposed to vertical strips,
as vertical strips would have the curve at both ends, and hence the length of the
rectangle would no longer be y and the formula would no longer work.
Area
3



冮 x dy

3
3



冮9 y

2

dy

3

1 3 3
yR
3 3
127 92 1 27 92
18 18
36 square units
B9y

Although the integration is performed with respect to y, a calculator can still be
used to find the area (although of course it is not the correct graph).

y 9 x2

Exercise 9
Calculate the area enclosed by the two functions.
1 y x2, y x
2 y 6x2, y 3x
3 y x3, y x
4 y x2, y 2x
5 y 8 x2, y 2 x
6 y x3 24, y 3x2 10x
7 y 10 x2, y 19 2x2
8 y ex, y 4 x2
1
1
9 y x2 6x 10, y 4x x2 and the x-axis. In this case draw the
2
3
graphs and shade the area.

398

14 Integration 1

10 y 3 cos 2x and y 1 produce an infinite pattern as shown.
y
3
y 1

1
0

x

Find the area of each shaded part.
11 Find the area between y e2x, y 2 x, the x-axis and the y-axis as
shown.
y 2 x

y

y e2x

x

0

12 Find the area between y

2
, y 4 x, the x-axis and the y-axis as
4 x

shown.
y

y

2
4 x
0

x

4

y 4 x

13 Find the shaded area.
y
y sin x

1
0
1


2

x
y sin 2x

14 Find the shaded area.
y
3

y 3 cos 2x

1
0
1

x



y cos x

3

399

14 Integration 1

15 Find the area of the “curved triangle” shown below, the sides of which lie
1
4
on the curves with equations y x 1x 32, y x x2 and y 2 .
4
x
˛

y x(x 3)

y

y 42
x
0

x
y x 14 x 2

Find the area enclosed by the y-axis and the following curves.
16 x 4 y2
y
2

x 4 y2
x

2

17 y2 16 x
18 8y2 18 2x
19 Evaluate the shaded area i with respect to x and ii with respect to y.
y 25 x
y
5

y 5 x

0

5

x

20 Find the shaded area.
y

1 x
y e
2

y 2 3e x

0

400

x

14 Integration 1

21 Find the shaded area.

y

e 2x

y ex 2

y

3

x

0

22 Find the shaded area.
y
6
y 4 sin x



0

x

y 6 cos x

Review exercise


M
C

7

4

1

M–

M+

CE

%

8

9



5

6

÷

2

3

+

0

ON
X


M
C

7

4

1

M–

M+

CE

%

8

9



5

6

÷

2

3

+

0

1 Integrate these functions.

=

ON
X

=

a 4x2 7

b 9x2 4x 5

2 Solve these equations.
dy
dy
x3 6

p2 13 p5 2
a
b
5
dx
dp
x



3 Given



4 Find these integrals.

M

M–

M+

C

CE

%

7

8

9



4

5

6

÷

1

2

3

+

0

M
C

7

4

1

ON
X

=

M–

M+

CE

%

8

9



5

6

÷

2

3

+

0

ON
X

7

4

1

M–

M+

CE

%

8

9



5

6

÷

2

3

+

0

ON
X

C

7

4

1

M–

M+

CE

%

8

9



5

6

÷

2

3

+

0

M
C

7

4

1

0

ON
X

=

M–

M+

CE

%

8

9



5

6

÷

2

3

+

dy
3t2 2t

dt
4t

冮4e sin x dx
x

b

冮7 cos x x dx
4

c

冮2e

6x



5
4 sin x dx
x

5 Find these integrals.

=



M

c

=


C

d 13 2x2 2

8
x3

dy
3x 8 and the curve passes through (2, 8), find the
dx
equation of the curve.

a
M

c

ON
X

a

冮6 cos 2x dx

b

冮4e

d

冮13x 22

e

冮7e

6

dx

6 Let f 1x2 2x ¢2x

2x

dx

c

2

4
dx
13x 42 5



3x

冮4x 3 dx

冮f 1x2 dx.

3

Find
5≤.
x2
7 Find these definite integrals.
˛

˛

=

p
4

3

a



1

3
4p
dp
12p 12 3

b

冮 sin 4u 1 du

0

k

c

冮 3 cos 2u 4u du

k

401

14 Integration 1


M
C

7

4

1

M–

M+

CE

%

8

9



5

6

÷

2

3

+

0

ON
X

8 Find the area given by these definite integrals.

=

5

a


M

M–

M+

C

CE

%

7

8

9



4

5

6

÷

1

2

3

+

0

ON
X

=

p
6

1

冮 e dx
x

2

b

冮 4x 5 dx

冮 2 cos 3u du

c

3

0

9 Find an expression in terms of p for this shaded area.
y

y

0

M
C
7
4
1

M–

M+

CE

%

8

9



5

6

÷

2

3
+

0

ON
X

5
2

3
2x 5

px

3

10 Find the total area of the two regions enclosed by the curve

=

y x3 3x2 9x 27 and the line y x 3.
M
C
7
4
1

M–

M+

CE

%

8

9



5

6

÷

2

3
+

0

ON
X

=

[IB Nov 04 P1 Q14]

11 The figure below shows part of the curve y x 7x 14x 7.
The curve crosses the x-axis at the points A, B and C.
3

2

y

0


M
C

7

4

1

M–

M+

CE

%

8

9



5

6

÷

2

3

+

0

ON
X

=

A

B

C

x

a Find the x-coordinate of A.
b Find the x-coordinate of B.
c Find the area of the shaded region.
[IB May 02 P1 Q13]
12 Find the area bounded by the curve and the x-axis for:
3p
a y 3 cos 2u, u 0 and u
4
b y x4 2ex, x 1 and x 3

M
C
7
4
1

M–

M+

CE

%

8

9



5

6

÷

2

3
+

0

ON
X

13 Find the area enclosed by:

=

a y ex and y 6 x2
b y
M
C
7
4
1
0

402

M–

M+

CE

%

8

9



5

6

÷

2

3
+

ON
X

=

3
, y 2 x and the x- and y-axes
2 x

14 Find the area between x 8 y2 and the y-axis.



Télécharger le fichier (PDF)










Documents similaires


ibhm 722 728
ibhm 058 085
ibhm 183 216
ch1
tutorial5
ibhm 246 267

Sur le même sujet..