Higher. Integration 89

Transcription

1 hsn.uk.net Higher Mathematics UNIT UTCME Integration Contents Integration 89 Indefinite Integrals 89 Preparing to Integrate 9 Differential Equations 9 Definite Integrals 9 Geometric Interpretation of Integration 96 6 Areas between Curves 0 7 Integrating along the -ais 06 HSN00 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For more details about the copright on these notes, please see

2 Unit Integration UTCME Integration Indefinite Integrals In integration, our aim is to undo the process of differentiation. Later we will see that integration is a useful tool for evaluating areas and solving a special tpe of equation. We have alread seen how to differentiate polnomials, so we will now look at how to undo this process. The basic technique is: n+ n d = + c n, c is the constant of integration. n + Stated simpl: raise the power (n) b one (giving n + ), divide b the new power ( n + ), and add the constant of integration ( c ). EXAMPLES. Find. Find. Find d. d = + c = + c. d. d = + c = + c d. 9 9 d = + = 9 c 9 + c.. We use the smbol for integration. The must be used with d in the eamples above, to indicate that we are integrating with respect to. The constant of integration is included to represent an constant term in the original epression, since this would have been zeroed b differentiation. Integrals with a constant of integration are called indefinite integrals. Page 89 HSN000

3 Unit Integration Checking the answer Since integration and differentiation are reverse processes, if we differentiate our answer we should get back to what we started with. For eample, if we differentiate our answer to Eample above, we do get back to the epression we started with. + c differentiate integrate Integrating terms with coefficients The above technique can be etended to: n+ n n a a d = a d = + c n, a is a constant. n + Stated simpl: raise the power (n) b one (giving n + ), divide b the new power ( n + ), and add on c. EXAMPLES 6 d.. Find 6 6 d = + c = + c.. Find d. d = + c = 8 + c 8 = + c. Note It can be eas to confuse integration and differentiation, so remember: d = + c k d = k + c. Page 90 HSN000

4 Unit Integration ther variables Just as with differentiation, we can integrate with respect to an variable. EXAMPLES p dp. 6. Find p = + c. p p dp = + c 7. Find p d. p d = p + c. Integrating several terms The following rule is used to integrate an epression with several terms: ( f ( ) + g ( )) d = f ( ) d + g ( ) d. Stated simpl: integrate each term separatel. EXAMPLES 8. Find ( ) ( ) d. d = + c 9. Find 8 ( ) = + c = + c. d. 8 ( ) d = c 8 8 = c 8 = c. Note dp tells us to integrate with respect to p. Note Since we are integrating with respect to, we treat p as a constant. Page 9 HSN000

5 Unit Integration Preparing to Integrate As with differentiation, it is important that before integrating all brackets are multiplied out and there are no fractions with an term in the denominator (bottom line), for eample: = EXAMPLES = = d. Find for 0. d is just a short wa of writing d, so:. Find d d = d = d = + c = + c. d for > 0. = = d d = + c = + c. = =. 7. Find p dp where p 0. 7 p 7 dp = p dp = 7 p + c 7 = + c. p Page 9 HSN000

6 Unit Integration. Find d. = ( ) d d 6 = + c 6 6 = 8 + c 6 = c. Differential Equations A differential equation is an equation d involving derivatives, e.g. d =. A solution of a differential equation is an epression for the original function; in this case = + c is a solution. + c differentiate integrate In general, we obtain solutions using integration: d = d or f ( ) = f ( ) d. d This will result in a general solution since we can choose the value of c, the constant of integration. The general solution corresponds to a famil of curves, each with a different value for c. The graph to the left illustrates some of the curves = + c for particular values of c. If we have additional information about the function (such as a point its graph passes through), we can find the value of c and obtain a particular solution. Page 9 HSN000

7 Unit Integration EXAMPLES = passes through the point (, ). The graph of f ( ) d If =, epress in terms of. d d = d d ( ) = d = + c. We know that when =, = so we can find c: So = + c = ( ) ( ) + c = 9 + c c = = +... The function f, defined on a suitable domain, is such that f ( ) = + +. Given that f ( ) =, find a formula for f ( ) in terms of. f ( ) = f ( ) d = + + ( ) d = + + d = + + c = + + c. We know that f ( ) =, so we can find c: ( ) = f + + c = ( ) + ( ) + c = + + c c =. So ( ) f = + +. Page 9 HSN000

8 Unit Integration Definite Integrals If F( ) is an integral of f ( ), then we define: b b [ ] ( ) ( ) f ( ) d = F ( ) = F b F a a a where a and b are called the limits of the integral. Stated simpl: Work out the integral as normal, leaving out the constant of integration. Evaluate the integral for = b (the upper limit value). Evaluate the integral for = a (the lower limit value). Subtract the lower limit value from the upper limit value. Since there is no constant of integration and we are calculating a numerical value, this is called a definite integral. EXAMPLES. Find d. d = ( ) ( ) = = = =. 0. Find ( + ) ( ) 0 d. + d = + = = = = + 8 =. Page 9 HSN000

9 Unit Integration. Find d. d = d = = = ( ) = + = Geometric Interpretation of Integration We will now consider the meaning of integration in the contet of areas. Consider ( ) d = 8 ( ) = 8 0 =. 0 0 n the graph of = : = 0 The shaded area is given b ( ) d. Therefore the shaded area is square units. In general, the area enclosed b the graph = f ( ) and the -ais, between = a and = b, is given b b f ( ) d. a Page 96 HSN000

10 Unit Integration EXAMPLE. The graph of = is shown below. Calculate the shaded area. = d = ( ) = ( ) ( ) = = 6 8 =. So the shaded area is square units. Areas below the -ais Care needs to be taken if part or all of the area lies below the -ais. For eample if we look at the graph of = : = The shaded area is given b d = ( ) = = 6 + = 6 0 = 0 = 9. In this case, the negative indicates that the area is below the -ais, as can be seen from the diagram. The area is therefore 9 square units. ( ) ( ) Page 97 HSN000

11 Unit Integration Areas above and below the -ais Consider the graph from the eample above, with a different shaded area: From the working above, the total shaded area is: Area + Area = + = 9 square units. Using the method from above, we might tr to calculate the shaded area as follows: d = ( ) = = 0 + = = 8 6. ( ) ( ) Clearl this shaded area is not 6 square units since we alread found it to be square units. This problem arises because Area is above the -ais, while Area is below. To find the true area, we needed to evaluate two integrals: ( ) d and ( ) = Area Area d. We then found the total shaded area b adding the two areas together. We must take care to do this whenever the area is split up in this wa. Page 98 HSN000

12 Unit Integration EXAMPLES. Calculate the shaded area shown in the diagram below. = To calculate the area from = to = : d = ( ) = = ( ( ) ( ) ( ) ) ( ) ( ) ( ) ( ) ( 9 9 9) ( ) = + = + 9 = 0 So the area is 0 square units. We have alread carried out the integration, so we can just substitute in new limits to work out the area from = to = : ( ) d = = ( ( ) ( ) ( ) ) ( ) ( ) ( ) ( ) ( 6 8 ) ( ) = = 8 + Remember =. So the area is square units. So the total shaded area is 0 + = square units. The negative sign just indicates that the area lies below the ais. Page 99 HSN000

13 Unit Integration. Calculate the shaded area shown in the diagram below. = + First, we need to calculate the root between = and = : + = 0 ( )( + 6) = 0 = or = 6. So the root is = To calculate the area from = to = : + d = + ( ) = + = ( ( ) + ( ) ( ) ) ( ) + ( ) ( ) ( ) 6 ( ) ( 8) = + + To calculate the area from = to = : ( ) = 6 6 = 7 So the area is 7 square units. + d = + = ( ( ) + ( ) ( ) ) ( ) + ( ) ( ) ( ) ( 0 6 ) ( 6 96) = + + = 6 = So the area is square units. So the total shaded area is 7 + = square units. Page 00 HSN000

14 Unit Integration 6 Areas between Curves The area between two curves between b ( lower curve) upper curve a So for the shaded area shown below: = f ( ) = a and = b is calculated as: d square units. a = g ( ) b The area is b a ( f ( ) g ( )) d square units. When dealing with areas between curves, areas above and below the -ais do not need to be calculated separatel. However, care must be taken with more complicated curves, as these ma give rise to more than one closed area. These areas must be evaluated separatel. For eample: = g ( ) a b c = f ( ) b In this case we appl ( lower curve) upper curve a So the shaded area is given b: b c ( g ( ) f ( )) d + ( f ( ) g ( )) d d to each area.. a b Page 0 HSN000

15 Unit Integration EXAMPLES. Calculate the shaded area enclosed b the curves with equations = 6 and =. = 6 = To work out the points of intersection, equate the curves: 6 = = 0 9 = 0 ( + )( ) = 0 = or =. Set up the integral and simplif: ( upper curve lower curve) (( 6 ) ( )) = d ( 6 ) ( 9 ) d. = + + d = Carr out integration: 9 d = 9 ( ) d ( ) ( ) = 9( ) 9( ) ( ) ( 7 ) = + = = 6. Therefore the shaded area is 6 square units. Page 0 HSN000

16 Unit Integration. Two functions are defined for R b f ( ) = and g ( ) = +. The graphs of = f ( ) and = g ( ) are shown below. = f ( ) = g ( ) 6 Calculate the shaded area. b Since the shaded area is in two parts, we appl ( upper lower) Area from = to = : ( upper lower) ( ( ) ) = d ( 7 ) = + + d 7 = + + d twice. 7 ( ) 7( ) = ( ) + ( ) ( ) ( ) = = 99 =. d So the first area is square units. a Note The curve is at the top of this area. Page 0 HSN000

17 Unit Integration Area from = to = 6 : 6 ( upper lower) ( ( 7 8 6) ) = + + d 6 6 ( 7 ) = + d 7 = = = ( ) ( ) = 60 =. d So the second area is square units. So the total shaded area is + = 78 square units. 6 Note The straight line is at the top of this area. Page 0 HSN000

18 Unit Integration. A trough is metres long. A cross-section of the trough is shown below. = + = The cross-section is part of the parabola with equation = +. Find the volume of the trough. To work out the points of intersection, equate the curve and the line: + = + = 0 ( )( ) = 0 so = or =. Set up the integral and integrate: ( upper lower) = ( ( )) + = ( + ) d d d = + ( ) ( ) = + ( ) ( ) + ( ) ( ) ( 9 8 9) ( ) = + + = = =. Therefore the shaded area is square units. Volume = cross-sectional area length = = 8 =. Therefore the volume of the trough is cubic units. Page 0 HSN000

19 Unit Integration 7 Integrating along the -ais For some problems, it ma be easier to find a shaded area b integrating with respect to rather than. EXAMPLE The curve with equation = 9 9 = is shown in the diagram below. =6 = Calculate the shaded area which lies between = and = 6. We have = 9 9 = = ± 9 = ±. = The shaded area in the diagram to the right is given b: 6 6 d = d 6 = = 6 = 6 = 6 8 =. = Since this is half of the required area, the total shaded area is square units. =6 Page 06 HSN000