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SCHOOL OF MATHEMATICS
MATHEMATICS FOR PART I ENGINEERING
Self-paced Course
MODULE 4
INTEGRATION I
Module Topics
1. Definition of integral
2. Standard integrals
3. Use of simple substitutions
4. Integration by parts
5. Numerical integration
A:
Work Scheme based on JAMES (FIFTH EDITION)
1. This module covers basic material on integration and will be mainly revision for you. Read the introductory sentence to section 8.7 on p.622. Then study section 8.7.1 on defining an integral as a limit, up to
and including Example 8.37. You should note that areas above the x-axis are positive whereas those below
the axis are negative.
2. Study section 8.7.4 on definite and indefinite integrals, and then study section 8.7.5 on the fundamental theorem of calculus. Some simple integrals can be determined from first principles, using the
definition of the integral as a limit. From these results and corresponding ones for simple derivatives it
follows that differentiation and integration are inverse processes, as stated in the key result at the bottom
of p.633 and proved more mathematically in theorem 8.1 on p.634.
3. Read the introduction to section 8.8, and then study section 8.8.1.
The fundamental theorem of calculus allows the results for derivatives on the formula sheet (and in the Data
d
Book) to be re-interpreted as results for integrals. For example, you know
(sin x) = cos x and this leads
dx
Z
to the equivalent result
cos x dx = sin x + c , the only extra feature being the constant of integration
which appears for all indefinite integrals.
Rules 1 and 2 on p.636 are very important, but you should note that rule 3 is not usually remembered in
that form and rule 4 is not normally used at all, although it is a neat result. Work through parts (a), (b)
and (d) of Example 8.42. If you need further practice study Example 8.43.
4. Example 8.42(c) uses the linear composite rule which, as I have said above, is not usually remembered.
The normal way to evaluate integrals of this type is to use a simple substitution so that the original integral
can be expressed
in a standard form. For this example introduce u = 5x + 2. Then the original integral
Z
√
becomes
u dx. However, it is only possible to integrate a function of u with respect to u and hence a
du
= 5, which
further change to the integral is required. Since u = 5x + 2 it follows immediately that
dx
du
can be arranged to give du = 5 dx or dx =
. This rather casual approach to the quantities du and
5
–1–
Z
dx can be justified, if necessary. The original integral can therefore be written
!
3
3
1 u2
2 3
2
+c=
u2 + c =
(5x + 2) 2 + c , as obtained in J.
5 3/2
15
15
√
u
du
, with solution
5
Similar methods can be used to extend other standard results listed on the formula sheet (and in the Data
Book). To integrate the function cos(4x), for instance, put u = 4x, which implies du = 4 dx. It then follows
that
Z
Z
du
1
1
cos(4x) dx = cos u
= sin u + c = sin(4x) + c.
4
4
4
Inverse functions will be studied in detail in module 7, but for the present let us assume that if y = tan x
then reversing the operation gives x = tan−1 y, i.e. x is the angle whose tangent is y. It will be shown
d
1
, as stated on the formula
in module 7, if you have not met the result before, that
(tan−1 x) =
2
dx
1 + xZ
1
dx = tan−1 x + c.
sheet (and in the Data Book), with the obvious corresponding integral result
1 + x2
Simple substitutions can be used to deal with slight variations of the integrand. For example, to integrate
1
you need to put u = 2x so that the denominator becomes 1 + u2 , in which case the
the function
1 + 4x2
integrand reduces to the standard form. From the stated substitution du = 2 dx, so the calculation proceeds
as follows:
Z
Z
dx
1
(du/2)
1
=
= tan−1 u + c = tan−1 (2x) + c.
1 + 4x2
1 + u2
2
2
1
An example which looks very similar has the integrand
. For this case first note that 4 + x2 =
4 + x2
x 2
x2
x
4 1+
= 4 1+
. This time introduce u = , so that the denominator equals 4(1 + u2 ). The
4
2
2
dx
present substitution leads to du =
and hence
2
Z
dx
=
4 + x2
Z
1
1
2du
−1
−1 x
=
tan
u
+
c
=
tan
+ c.
4(1 + u2 )
2
2
2
[In the above three cases you should check the answers by differentiating the final expressions, using the
chain rule if necessary, to make sure you obtain the original functions.]
5. When you are using a substitution to evaluate a definite integral, i.e. an integral with prescribed limits,
then you must always
Z remember to change the limits to the values appropriate to the new variable. For
1
(2x + 1)3 dx . Many of you will write down the answer immediately for this simple
example, consider
0
du
integral but if you use the substitution u = 2x + 1, then
= 2. The original limits are x = 0 and x = 1
dx
and in the new variable these become u = 1 and u = 3 respectively. Hence, the calculation proceeds as
below
4
Z 1
Z 3
u
1
80
3
3 du
(2x + 1) dx =
u
=
= (34 − 14 ) =
= 10.
2
8
8
8
0
1
6. Study Example 8.55 part (a) on p.655. As shown in this Example the identities on the formula sheet
(and in the Data Book) connecting products and sums of trigonometric functions are very useful in evaluating
some integrals.
–2–
N.B. In some versions of the Data Book there are errors on p.3-8. Correct results are
2 sin A sin B = cos(A − B) − cos(A + B)
1
1
sin A + sin B = 2 sin (A + B) cos (A − B)
2
2
Z
Two simple integrals which occur frequently in applications and fall into this category are
cos2 x dx
Z
and
sin2 x dx . To evaluate these integrals you use the double angle formulae cos 2x = 2 cos2 x − 1
or cos 2x = 1 − 2 sin2 x to write cos2 x or sin2 x as an expression involving cos 2x, which can be easily
integrated.
The latter procedure can be repeated twice if you want to integrate cos4 x , for instance. You can write
1
cos2 x = (1 + cos 2x) and hence cos4 x = (cos2 x)2 = 41 (1 + cos 2x)2 = 41 (1 + 2 cos 2x + cos2 2x) . The
2
quantity cos2 2x can be replaced, using the double angle formula with x replaced by 2x, by 12 (1 + cos 4x)
and the original function has then been written in terms of simple functions which can be integrated using
minor extensions of the listed standard integrals.
7. Study the general comments at the top of p.644, section 8.8.2.
***Do Exercises 104(a),(b),(c),(d),(f ) on p.647, Exercises 119(b), 120(a) and 120(b) on p.658
and Exercises 105(c), 106(b) on p.647***
***Do Exercise A:
Find the indefinite integral of cos4 x ***
8. Study section 8.8.4 on integration by parts and then work through all three parts of Example 8.50.
Integrals of the type considered in part (c), where you have to integrate by parts twice to obtain an integral
similar to the starting one, are important and most of you will meet them later in your studies in calculating
Fourier coefficients.
***Do Exercises 110(a),(b),(c),(d (harder)), 111(a) on p.651***
Z
***Do Exercise B: (harder)
Use integration by parts (twice) to evaluate
e2x cos x dx
9. The next topic to be briefly considered is numerical integration. Read the introduction to section
8.10 on p.679.
Study section 8.10.1 on the trapezium rule. The area under the graph is divided into a number of strips.
The area of each strip is approximated by the area of a trapezium, as shown in Figure 8.71. The area of
1
this trapezium is (fr + fr+1 )h, and hence the formula (8.47) can be obtained. The latter appears on the
2
Formula sheet and in the Data Book.
Work through Example 8.69. Efficient ways of calculating the answer are explored in this solution, but in
this module you will only be expected to use straightforward methods of evaluation and, in fact, the number
of strips to be used in any Example will normally be prescribed.
10. Study section 8.10.2 on Simpson’s rule. You should look through the derivation of formula (8.49)
on p.685, which uses the extrapolation formula second from the top of p.683. It is more usual to derive
Simpson’s rule using the method mentioned briefly in the middle of p.685, joining three points using a
parabola and not the straight lines used in deriving the trapezium rule. Once again the derivation is not
very important, but you must be able to use formula (8.49), which is stated on the Formula sheet and in
the Data Book.
–3–
Work through Example 8.71, which illustrates a more practical use of Simpson’s rule.
***Do Exercise 142 on p.688***
***Do Exercise C:
in the latter).
B:
Using the trapezium rule evaluate again the integral in Exercise 142 (with h = 0.1
Work Scheme based on STROUD (SIXTH EDITION)
This module covers some of Programme 15 in S. Turn to p.824 and read frame 1. Move on to the next
frame and results 1-4, 6-8 and 13 in the table. Note that inverse functions will be studied in module 7, and
for the present it is only necessary to know that if y = tan x then reversing the operation gives x = tan−1 y,
with tan−1 being an inverse function. Try examples (a) − (d), (f ), (j) in frame 4.
Study frames 6-10, omitting questions 4, 9 and 10 in frame 9.
Move forward to frame 21 on integration by parts, and work through frames 21-30.
Next consider the integration of some trigonometric functions. Work through frames 44, 46(iv), 47, 49-53
remembering that the formulae written in frame 50 can be found on the formula sheet (and in the Data
Book, noting that the fourth identity in frame 50 is wrongly stated in some versions of the Data Book).
Finally turn to p.987 of S. and work through frames 16–34 in programme 21. This covers Simpson’s rule
for numerically evaluating integrals. For the trapezium rule you need to study section 9 in the above scheme
based on J. (Copies of James are available in the library.)
Specimen Test 4
1.
Determine the following indefinite integrals:
Z
Z
1
1
(i)
dx
(ii)
dx
(iii)
x2
x+4
Z
(v)
2.
Z
cos(2x) cos(4x) dx
(vi)
Evaluate
Z 1
(i)
(x2 + 1)2 dx
Z
(iv)
sin(1 + 4x) dx
1
dx
9 + x2
sin2 x dx
(ii)
0
Z
3.
e2x+1 dx
π
Z
0
Z
Evaluate
2
x ln x dx
1
Z
4.
Estimate the value of
1
x
2
1
dx
x
1.00
using Simpson’s rule, after completing the table below,
1.25
1.50
1
x
–4–
1.75
2.00