Module 9 - Integration 2 (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

INTEGRATION II

MODULE 9 Module Topics

1. Use of substitution - further examples 2. Applications of integration (volumes of revolution, centroids, centres of gravity, mean values, arclength and surface area)

A:

Work Scheme based on JAMES (THIRD EDITION)

1. This module covers a relatively small portion of J. but requires you to evaluate a large number of integrals. Study first the paragraph at the bottom of p.553 and the top of p.554 and work carefully through Example 8.37. In part (a) the integrand is a product in which one term of the product (2x) is the differential of an expression (x2 + 3) which appears in the other term, whereas in part (b) the integrand is a quotient in which the numerator (x + 1) is a simple multiple of the differential of the denominator (x2 + 2x + 1). Both parts of this example involve types of integral which occur frequently in practice, and you must be able to recognise them. Both types are solved with appropriate substitutions but, with practice, some of you may be able to write down the answers immediately. ***Do Exercise A: Find the following indefinite integrals R 2 R R (a) x (1 + x3 )4 dx, (b) cos x sin3 x dx, (c) x cos(x2 ) dx, Z (e)

Z

2x dx, 1 + x2

(f)

Z

cos x − sin x dx, sin x + cos x

(g)

Z (d)

cosh x dx, sinh x

ln x dx. x

2. The use of more complicated substitutions starts at the top of p.555. Read the√first short paragraph and then work through Example 8.38. Note that after using the substitution t = 1 − x the integrand −2t . This is a quotient of polynomials of equal powers and will be considered in more detail in becomes 2+t module 10. However, for the moment it is sufficient to observe that J. rearranges the expression into one which can be easily integrated. One way of carrying out the simplification is as follows: 2t = −2 − 2+t



t 2+t



 = −2

(2 + t) − 2 2+t



 = −2

2 2+t − 2+t 2+t



 = −2 1 −

2 2+t



 = +2

 2 −1 . 2+t

3. Next work through Example 8.39. Here the substitution x = sin θ reduces the integral to one which has been considered in module 4. A list of further substitutions is given on p.557, with some of the possible functions for which these substitutions are often appropriate. The choice of substitution is usually made so that either the formula cos2 θ + sin2 θ = 1 or cosh2 θ − sinh2 θ = 1 , or a closely related one, can be used to simplify the integrand, –1–

making sure that if a square root appears in the integrand that you end up with taking the root of a positive number. Note carefully the appearance of the constant a, both in the general forms of the functions and in the substitutions. The constant is of course 1 in the simplest situation. The integrals which produce the inverse functions in the list on the Formula Sheet and in the Data Book can be derived in this way. Two are derived below and then you are asked to obtain the remaining ones. Z dx 1 √ = cos θ and Consider the integral dx. Here the substitution x = sin θ can be used, giving dθ 1 − x2 hence dx = cos θ dθ. The integral can now be exaluated: Z Z Z Z Z 1 1 1 cos θ √ p √ dθ = dθ = θ + c = sin−1 x + c. cos θ dθ = dx = cos θ dθ = cos θ 1 − x2 cos2 θ 1 − sin2 θ Z

1 √ dx can be determined from first principles with the substitution 1 + x2 dx = cosh u and hence dx = cosh u du. The integral is now found as x = sinh u. Differentiation leads to du follows: Z Z Z Z 1 1 1 cosh u √ p √ du cosh u du = dx = cosh u du = 2 2 2 cosh u 1+x cosh u 1 + sinh u

In an analogous way the integral

Z

=

du = u + c = sinh−1 x + c.

***Do Exercise B: Find the following indefinite integrals from first principles with the given substitution Z Z 1 1 √ dx with x = cosh u, (b) dx with x = tan θ. (a) 2 1 + x2 x −1 4. Study Example 8.40. Here again the integral derived for u cannot be obtained immediately. The integrand is the quotient of quadratic polynomials and has to be manipulated into an acceptable form. One way to proceed, very similar to the manipulation in section 2 above, is  2   2    4 4 8 (u + 4) − 4 u +4 2u2 = 2 = 2 − = 2 1 − =2− 2 , u2 + 4 u2 + 4 u2 + 4 u2 + 4 u2 + 4 u +4 which can then be easily integrated. ***Do Exercises 89(e), 90(b), 94(a),(c) on p.558*** 5. The module is completed with some applications of integration in calculating quantities of engineering importance. First study section 8.7.1 on finding volumes of revolution. The formula 8.33 is derived from rotating about the x-axis a thin strip initially parallel to the y-axis. 6. Again from using thin strips the general expressions for the centroid of a plane area and the centre of gravity of a solid of revolution are derived in sections 8.7.2 and 8.7.3. Study both these sections. 7. Next study sections 8.7.4 and 8.7.5 on finding the mean values of continuously varying quantities. Note that integration is necessary in these situations - it is not correct simply to average the first and last values. 8. Finally the integral formulae to calculate arclength and surface area are obtained in section 8.7.6. Study the derivation of equations 8.42 and 8.43. –2–

9. Work through Example 8.41. To evaluate the integral for the area in part (a) you could use the substitution u = x − 2, if you wish. The same substitution could also be used to evaluate the integral for x. Work through Example 8.42. Next work through Example Z 1 p 8.43. The substitution u = 4x + 1 could be introduced in this Example to (4x + 1) dx. evaluate the integral 0

Finally work through Example 8.44. The integral here is much harder to calculate and uses the substitution t = sinh u. After this change it is still necessary to use the hyperbolic equivalents of the trigonometric double angle formulae in order to get the answer. ***Do Exercises 100, 101, 104 on p.569***

B:

Work Scheme based on STROUD (FIFTH EDITION)

The programmes in S. on integration contain most of this module. Study first frames 11-20 in Programme 15, which begin on p.809. This covers section 1. of the scheme based on J. S. only covers a little on evaluating integrals using substitution and, therefore, you should work through sections 2-4 of a scheme based on J. On the other hand, the applications of integrals is well-covered in S. Turn to Programme 18 on p.881 and work through frames 1-18 and then frames 23-30. Next move on to Programme 19 on p.901 and work through frames 1-25 and 31-34. (It is sometimes necessary to be able to deal with equations in parametric form and so it is suggested that you also look through Programme 18 frames 19-22 and finally Programme 19 frames 26-30 and 35-38.)

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Specimen Test 9 1.

Find the following indefinte integrals Z Z x dx, (b) cos4 x sin x dx. (a) 1 + x2 Z

2.

Evaluate 0

3.

1

x2 dx. (1 + x3 )2

2

2

Use the substitution u = 1 + x

Z

1

to evaluate 0

√

x3 dx. x2 + 1

4.

1 Find the volume generated when the plane figure bounded by the curve y = x2 + , the x-axis and the x lines x = 1 and x = 2 is rotated about the x-axis through one complete revolution.

5.

An area A is enclosed between the curve y = sin x, the x-axis and the lines x = 0 and x = π. (i) Find the magnitude of A. (ii) Calculate the coordinates of the centroid of A.

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