This document provides an introduction to integration (calculus) as taught in an undergraduate engineering course. It defines integration as the reverse process of differentiation and describes how it can be used to find the area under a curve. The document outlines key integration terminology like indefinite integrals, definite integrals, and the constant of integration. It also provides examples of integrating common functions and using integration to calculate volumes of solids of revolution.

1. Calculus (Integration) © University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License . Mathematics 1 Level 4

2. The following presentation is an introduction to the Algebraic Methods – part one for level 4 Mathematics. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1 st year undergraduate programme. The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments. Contents Integration Terminology Integration of xn Indefinite Integrals Definite Integrals Integrals of Sin x, Cos x, e x and 1/x Integrals of e x and 1/x Area under curves Volumes of Solids of Revolution Worked Example 1 Credits In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see: KA Stroud & DJ Booth, Engineering Mathematics, 8 th Editon, Palgrave 2008. http://www.mathcentre.ac.uk/ Derive 6 Calculus Integration

3. Integration Where differentiation measured the rate of change of a function, integration measures the area under the function between limits. It is the reverse function of differentiation. and therefore if we differentiate a function then integrate it we will get back to where we started. Calculus Integration

4. Terminology A definite integral is typically explained by the area problem as follows. Suppose we attempt to find the area of the region, S, under the curve y =  (x) bounded from lower limit a to upper limit b , where  is a continuous function. S x y  (x) To estimate S, we divide the range [ a, b ] into n subintervals, that is, [ x 0 , x 1 ], [ x 1 , x 2 ], [ x 2 , x 3 ], . . . , [ x n− 1 , x n ], Each of width Δ x = ( b − a ) /n (so x i = a + i Δ x ). a b Calculus Integration

5. Terminology x y  (x) b a x i-1 x i  (x i *) The area S i of the strip between x i−1 and x i can be approximated as the rectangular area with width Δ x and height f(x i *), where x i * is a sample point in [x i-1 , x i ]. So the total area under the curve is approximated: which is called a Riemann Sum. Calculus Integration

6. Terminology The estimation is better as the strips gets thinner, and the exact area under the graph of  with the limit can be identified as: n  ∞ As long as  is continuous, the value of the limit is independent of the sample points x i * used. That limit is represented by and is called definite integral of  from a to b Note that in intervals where  (x) is negative the graph of y =  (x) lies below the x -axis and the definite integral takes a negative value. In general a definite integral gives the net area between the graph of y =  (x) and the x -axis.

7. Integration of x n It has been stated that integration is the reverse process of differentiation. So if y = x 5 dy/dx = 5x 4 Therefore if we start with 5x 4 and we integrate it we will get to x 5 But note y = x 5 + 6 dy/dx = 5x 4 y = x 5 - 12 dy/dx = 5x 4 y = x 5 + constant dy/dx = 5x 4 It is therefore important to note that the integration of 5x 4 gives us x 5 + C where C is the constant of integration In general

8. Examples 1. 2. 3. 4. Calculus Integration

9. Examples cont. 5. 6. 7. 8. Calculus Integration

10. Indefinite Integrals You will note that the integrals we have been working out have no limits – numbers at the top and bottom of the integral sign. These do not generate numerical results and are referred to as indefinite integrals. All indefinite integrals must have a constant of integration. Calculus Integration

11. Definite Integrals Definite integrals will have limits – numbers at the top and bottom of the integral sign. The one on the top is the upper limit and the one on the bottom is the lower limit. Definite integrals will have a value associated with them – to determine this value we integrate then substitute in the upper limit into the new equation and then subtract from this the value when the lower limit is substituted. Due to the subtraction the constant of integration disappears and therefore, for definite integrals it is not normally included. Calculus Integration

12. Example Determine Note where the limits are, the square brackets and no constant Calculus Integration

13. Integrals of Sin x, Cos x, e x and 1/x Sin x If we have y = cos ax then dy/dx = -a sin ax. We can therefore say that ∫-asin ax dx = cos ax Note as well that if y = cos ax +5 then dy/dx = -a sin ax. So ∫sin ax dx = -1/a cos ax + C C = constant of integration Note – the number we divide by is the d/dx of the sin part Likewise Cos x If we have y = sin ax then dy/dx = a cos ax. We can therefore say that ∫acos ax dx = sin ax Note as well that if y = sin ax +5 then dy/dx = a cos ax. So ∫cos ax dx = 1/a sin ax + C C = constant of integration Calculus Integration

14. Worked examples Calculus Integration

15. Worked examples Note all angles in radians, cos 0 = 1, cos π /2 = cos 90 = 0 Calculus Integration

16. Integrals of e x and 1/x Recall the results from differentiation. If y = e ax then dy/dx = ae ax And y = ln x then dy/dx = 1/x Therefore ∫ ae ax dx = e ax + C In General ∫e ax dx = 1/a e ax + C Note – the number we divide by is the d/dx of the power part Also ∫ 1/x dx = ln x + C ∫ 1/(ax + b) dx = 1/a ln (ax + b) + C Note – the number we divide by is the d/dx of the reciprocal part Calculus Integration

17. Worked Examples Integrate the following with respect to x Evaluate the following: Calculus Integration

18. Worked Examples Evaluate the following: Calculus Integration

19. Area under curves We are aware that the process of integration is the determining of the area under a curve. e.g. What is the area under the curve y = x 2 + 2 between the vertical lines x = 1 and x = 3. This is simply: Calculus Integration

20. Volumes of Solids of Revolution To find the volume of revolution: Consider the curve y =  (x) shown on the left. Suppose we want the volume of the solid formed by rotating the area ABCD under the curve, about the x-axis. x y y =  (x) a b y dx A B C D Consider the area ABCD as being made up of an infinite number of small strips (there are so many that each approximates to a small rectangle), of which a typical one is y in height and dx in thickness, as shown. Calculus Integration

21. Volumes of Solids of Revolution When rotated the rectangle of area = y dx becomes a cylinder of volume = π y 2 dx. [note volume of a cylinder = π r 2 h where r = y and h = dx] Thus the volume generated in rotating the area ABCD about the x-axis will be the sum of all these small cylindrical volumes, formed by rotating the area of strips about the x-axis from x = a to x = b. i.e. in calculus form N.B. substitute x in terms of y before you integrate

22. Worked Example 1 Find the volume of a solid of revolution formed by revolving the area enclosed by the curve y = 3x 2 + 4, the x-axis and the lines at x = 1 and x = 2, through 360 º about the x-axis. First make a rough sketch of the curve and show the area to be rotated. Y = 3x 2 + 4 a) It is a parabola b) x = 0, y = 4 c) x = 1 y = 7 d) x = 2 y = 16 Area to be rotated y = 3x 2 + 4 y 2 = (3x 2 + 4) 2 y 2 = 9x 4 +24x 2 +16

23. Worked Example 1 Calculus Integration

24. Worked Example 2 Sketch the curve y = x 2 – 2x and find the volume of the solid formed when the area contained between the curve and the x-axis is rotated about the x-axis. Sketch: It is a parabola b) it passes through x =0 y = 0 Where else does it cross the x-axis x 2 - 2x = 0 x(x – 2) = 0 either x = 0 or x = 2 y = x 2 – 2x y 2 = (x 2 – 2x) 2 y 2 = x 4 -4x 3 +4x 2

25. Worked Example 2 Calculus Integration

26. Worked Example 3 What is the equation for the volume of a cone height h and radius r? x y h r We would need to rotate the area shown about the x-axis. What is the equation of the line? y = mx + c (m = slope c = y intersect) Intersect = 0 slope = r/h equation is

27. Examples The curve y = 2x 2 + 3 is rotated about the x-axis through 360 º, between the limits x = 1 and x = 3. Find the volume of the of the solid of rotation formed. [Ans = 991.5 units 3 ] 2. Find the volume of the solid obtained by rotating about the x-axis the part of the curve y = 2x 2 – 7x – 4 lying below the axis. [Ans = 772.9 units 3 ] Calculus Integration

28. This resource was created by the University of Wales Newport and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. © 2009 University of Wales Newport This work is licensed under a Creative Commons Attribution 2.0 License . The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. For Example- The name and logo of University of Wales Newport is a trade mark and all rights in it are reserved. The name and logo should not be reproduced without the express authorisation of the University. Calculus Integration