Chapter 3. Integration. 3.1 Indefinite Integration

Transcription

1 Chapter 3 Integration 3. Indefinite Integration Integration is the reverse of differentiation. Consider a function f(x) and suppose that there exists another function F (x) such that df f(x). (3.) For example, if f(x) x, thenf (x) x satisfies Eq. (3..). But note that, for example, F (x) x +also satisfies Eq. (3..). More generally, F (x) x + c where c is any constant satisfies Eq. (3..). So clearly F (x) is not unique. More generally therefore, we could replace Eq. (3..) by d (F (x)+c) f(x). (3.) In words, f(x) is the derivative of F (x)+c with respect to x. Reversing this process, we say that F (x)+c is the indefinite integral of f(x) with respect to x. The notation for this is f(x) F (x)+c. (3.3) Integration is the reverse of differentiation. If we integrate f(x), this means we are finding another function F (x)+c that has a gradient of f(x). Using the example above, we can write x x + c. Some more examples are x n xn+ since d xn+ (n +)x n. n + + c d ln(x)+c since x ln(x) x 4

2 and Examples 3, Q e x e x + c since d ex e x sin(x) cos(x)+c since d cos(x) sin(x) cos(x) sin(x)+c since d sin(x) cos(x). Examples 3, Q Integration is HARDER than differentiation. With a few tricks (chain rule, product rule, etc.) you can differentiate pretty much anything. You can t integrate everything. To make life easier, we often use standard integrals. 3. Some Standard Integrals It is commonplace to use a list of well-known or standard integrals. Useful examples include +x tan x + c; (3.4) x sin x + c; (3.5) sec (x) tan(x)+c; (3.6) cosec (x) cot(x)+c; (3.7) a + x x a tan + c; (3.8) a a x x sin a + c (3.9) Examples 3, Q3 3.3 Evaluation of Integrals by Substitution Some integrals can be evaluated by substituting x with another variable. This is best explained by example. Consider x + a. Let t x + a. Thenx t a and. For the purposes of making the substitution dt (and at no other time) we can write this last expression as dt by multiplying through by dt. Then the integral becomes x + a t dt ln(t)+c ln(x + a)+c. 5

3 On closer inspection, you might notice that integration by substitution is the reverse of the chain rule for differentiation. 3.4 Integration by Parts A useful formula is Examples 3, Q4, Q5 u dv uv This comes from a rearrangment of the product rule, eqn (.6). u dv d du (uv) v Integrating this term-by-term gives eqn (3.). To see how this is useful, consider an example: x sin(x). du v. (3.) In this integral, we try to split it up into a product of something we can differentiate easily, and something that we can integrate easily. The trick is that the result should be simpler than what we started with. In this case, we let u x and we know we can differentiate this easily to give du. We also let dv sin(x), knowingthatwecanintegratethiseasilytoget v sin(x) cos(x). (3.) (Notethatwedon tneedthe+c here.). Note that u dv x sin(x), whichiswhatweare trying to integrate. So, substituting these into equation (3.) x sin(x) x cos(x)+ cos(x) but, now, the final integral is easy to do (because we chose a good u and a good v for the trick to work!). So x sin(x) x cos(x) + cos(x) x cos(x)+sin(x)+c. Examples 3, Q6 6

4 3.5 Partial Fractions This method is for dealing with integrals of the form (x + a)(x + b). Suppose that it is possible to write (x + a)(x + b) A x + a + B x + b A(x + b)+b(x + a) (x + a)(x + b) (A + B)x +(Ab + Ba) (x + a)(x + b) In order to make the two sides of this equation equal we must have A + B and so Also so that Hence so that B A. Ab + Ba Ab Aa A(b a). A b a and B b a (x + a)(x + b) (b a)(x + a) (b a)(x + b) Therefore (x + a)(x + b) (b a)(x + a) (b a)(x + b) ln(x + a) ln(x + b)+c b a b a (ln(x + a) ln(x + b)) + c b a µ x + a b a ln + c x + b In practice, rather then remembering this formula, we work it out by the method above for each specific example. NB bx + c (x + a) b (x + a) + c ab (x + a) Examples 3, Q7 7

5 3.6 Definite Integrals Integration can tell you about the area underneathacurveonagraph. Todothis,weuse something called definite integration. Let s start with the definition. Recall the previous definition of integration as the reverse of differentiation: if df f(x), (3.) then the indefinite integral of f(x) is f(x) F (x)+c. (3.3) This is indefinite because of the arbitrary constant, c. The definite integral of f(x) between x a and x b is written as b a f(x) [F (x)] b a F (b) F (a) where a and b are just numbers - they are called the limits of the integral. The funny notation [F (x)] b a is just a shorthand for F (b) F (a), but it is useful because it allows us to do the integral first, then substitute the values of a and b in later. To see how this works in practice, it is best to use examples. Here are two: x - do the integral and write it in the square brackets - substitute in the limits of the integral 3. - evaluate the result π/ sin(x) [ cos(x)] π/ - do the integral and write it in the square brackets h ³ π i cos cos() - substitute in the limits of the integral [ ]. - evaluate the result Notice that the result of a definite integration is just a number. Thisisalwaystruewhen the limits of the integral are numbers. Now we shall show how definite integrals relate to the area under curves. Recall that f (x) df lim δf δx δx. Now, looking at Fig. 3., the area of the small column of width δx between y and y f(x) is approximately f(x)δx. Thus the shaded area is Area Xxb f(x)δx. xa 8

6 Figure 3.: The figure shows (shaded) the area between the function y f(x) and the x-axis. We can make this more accurate by taking ever larger numbers of columns of ever smaller width δx. In the limit as δx, Xxb Area lim f(x)δx δx xa Xxb lim δf since fδx δf δx xa lim [(F (a + δx) F (a)) + (F (a +δx) F (a + δx)) δx +(F (a +3δx) F (a +δx)) +...+(F (b δx) F (b δx)) (F (b) F (b δx))] F (b) F (a) But this is just the definite integral as defined above! So, Area b a f(x) [F (x)] b a F (b) F (a). Note that where f(x) <, thisgivesanegative contribution to the definite integral (the shaded area is below the x-axis). Hence: x. x " # ( ). 9

7 x. " # ( ) Examples 3, Q8 3