AREA UNDER A CURVE CALCULUS 8. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Area under a curve 1/15 Adrian Jannetta

Transcription

1 AREA UNDER A CURVE CALCULUS 8 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Area under a curve 1/15 Adrian Jannetta

2 The area beneath a curve =f() The total area beneath the curve is approimatel the sum of all such strips: δ =b Area f()δ (1) =a a b The estimate of area is improved b using narrower strips (i.e. b makingδ smaller). If we allowδ 0 then the area in equation 1 approaches a limit which is equal to the eact area beneath the curve. Consider the curve defined b = f(). The area enclosed b the curve, the -ais and the lines =a and = b can be estimated b dividing it into rectangular strips. The area of each stripδa is given b δa=δ= f()δ =b Area= lim f()δ δ 0 =a This equation can be shown to be the same as the definite integral b Area= f() d (2) a Area under a curve 2/15 Adrian Jannetta

3 Area beneath a curve (above the -ais) Calculate the area bounded b the curve = , the lines = 2, = 2 and the -ais. A picture of the defined area is shown below. In this case we can see that no part of the curve crosses the -ais so the area can be found b evaluating the integral in one step Area = ( ) d 2 = = ( 2) 3 2 = (8+2) ( 8 2) = = 20 The eact area beneath the curve between the limits shown in the picture is 20units 2. It s useful to know what the graph looks like in case an of the area being considered is on the other side of the -ais. Area under a curve 3/15 Adrian Jannetta

4 Area beneath a curve (below the -ais) Calculate the area bounded b the curve = 2 4 and the -ais. Here s a sketch showing the required area. 2 2 A loop of the curve encloses an area below the ais - this is what we must find. Solve the equation 2 4=0 to find the intersection with the -ais. The solutions (= 2, =2) are the limits of the integral. 2 Area = ( 2 4) d 2 = = (2) ( 2) 3 3 4( 2) = = = The negative value occurs because the area is beneath the -ais. However, we do not give a sign when dealing with area so we can ignore it. The area bounded b the curve and the -ais is 32 3 units2. Area under a curve 4/15 Adrian Jannetta

5 Dealing with negative area In the previous eample we had to evaluate 2 Area= 2 ( 2 4)d The graph indicated that the area was below the ais (negative -values) and so we might have anticipated that the integral would have a negative sign. It is better to force the integral to give a positive value because area is defined to be a positive quantit. We could have written Area= 2 2 ( 2 4)d Tr it! We can swap the limits to ensure a positive value. The alternative is to use eisting limit order and take the absolute value of the definite integral. Area under a curve 5/15 Adrian Jannetta

6 Using the area integral correctl The integration formula for calculating areas: b Area= f() d a will give the net area beneath a curve.this means if some of the area is below the -ais (negative values) then this will subtract from the area above the -ais. Consider the area bounded b the curve = cos and the coordinate aes over the interval 0 2π. Here is the picture: Simpl using the limits 0 to 2π will not give the correct area!tr it: 2π 0 cos d=[sin ] 2π 0 = sin 2π=0 But it is clear from the graph that the area is not zero! 0 π 2 π 3π 2 2π The total shaded area above the -ais is identical to the amount of area below it. It cancels out to give zero net area when we do the integral. Area under a curve 6/15 Adrian Jannetta

7 Using the area integral correctl Simpl using the limits 0 to 2π will not give the correct area. Instead we must find the area of each region separatel. A 1 0 π 2 π A 2 3π 2 A 3 2π Area=A 1 + A 2 + A 3 Integrating to find A 2 will give a negative value (because is negative). Just ignore the sign of the value (b taking the modulus, or b reversing the order of the limits). π/2 3π/2 2π Area = cosd + cos d + cos d 0 π/2 3π/2 = [sin ] π/2 0 + [sin ] 3π/2 π/2 +[sin ] 2π 3π/2 = =4 So the total area is actuall 4 units 2. Area under a curve 7/15 Adrian Jannetta

8 Areas above and below the -ais Calculate the area bounded b the curve = 2 5+4, the coordinate aes and the line =5. Sketch the function so that the limits of integration can be found. The intersection with the ais can be found b solving 2 5+4= 0. It factorises to give( 1)( 4)=0 so that =1 and = 4. A 1 A 2 A The areas have to be treated separatel, Area=A 1 + A 2 + A Area = 2 5+4d d d The limits of A 2 were reversed to give a positive value. Area = = ( 8 3 ) ( 8 3 ) = Area = 6 units2 Area under a curve 8/15 Adrian Jannetta

9 Area beneath two different curves Consider the curves = and the line = 3, which are shown on the graph. = Calculate the total area of the shaded region beneath the curves. = 3 This area here is bounded b different curves. To find the total area we should break the region into subregions beneath each curve. O = =3 A 1 A 2 p q The total area is A 1 + A 2 : p q Area= ( 2 + 1) d+ (3 ) d 0 p We must calculate the limits for each integral the values of p and q. Area under a curve 9/15 Adrian Jannetta

10 Therefore: = =3 A 1 A 2 O p q The line and curve intersect at p so we solve: 2 + 1= = 0 (+2)( 1) = 0 So = 1 and = 2. From the picture it s clear that the limit must be p= 1. The line meets the -ais when = 0 so 3 = 0 = 3 So the other limit is q= 3. Now that the limits are known, the total area can be computed. The completed graph is shown below. O = = 3 A 1 A Area = ( 2 + 1) d+ (3 ) d 0 1 = = = 10 3 Area under a curve 10/15 The shaded area is eactl square units. Adrian Jannetta

11 Summar The area enclosed b a curve = f() and the limits =a and =b can be calculated using the definite integral Area= b a f() d Provided the area is completel above (or completel below) the -ais then nothing more needs to be considered. If the area is partiall above and below the -ais then the above integral will onl give the net area. The total area will need to found b considering positive and negative regions separatel. If limits are not given then the will need to be calculated; e.g. b solving f()=0. If the area is defined b more than one function each region must be calculated seperatel. Area under a curve 11/15 Adrian Jannetta

12 Test ourself... Solve the following four area problems using integration. 1. The area enclosed b the curve = 16 2 and the -ais. Answers: 1. Area is Area under a curve 12/15 Adrian Jannetta

13 Test ourself The area enclosed b the curve = 2+4, the -ais and the line =6. 6 Answers: 2. Area is Area under a curve 13/15 Adrian Jannetta

14 Test ourself The area enclosed b the curve = e 1, the -ais and the lines = 2 and =2. Give our answer in terms of e. 2 2 Answers: 3. Area is e e2 Area under a curve 14/15 Adrian Jannetta

15 4. The picture below a shaded area bounded b the curve <k f()= 7 k <7 Find the value of k and hence calculate the eact value of the shaded region. O k 7 Answers: 4. Area is 20. Area under a curve 15/15 Adrian Jannetta