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Dr J Frost (jfrost@tiffin.kingston.sch.uk) C2 Chapter 11 Integration Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 17th October 2013.

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Presentation on theme: "Dr J Frost (jfrost@tiffin.kingston.sch.uk) C2 Chapter 11 Integration Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 17th October 2013."β€” Presentation transcript:

1 Dr J Frost (jfrost@tiffin.kingston.sch.uk)
C2 Chapter 11 Integration Dr J Frost Last modified: 17th October 2013

2 Recap 2 π‘₯ 2 +3π‘₯ 𝑑π‘₯ = 2 3 π‘₯ π‘₯ 2 +𝑐 1+π‘₯+ π‘₯ 2 + π‘₯ 3 𝑑π‘₯ =π‘₯ π‘₯ π‘₯ π‘₯ 4 +𝑐 π‘₯ βˆ’ 𝑑π‘₯ =βˆ’ 3 4 π‘₯ βˆ’ 𝑐 2+ π‘₯ π‘₯ 2 𝑑π‘₯ =βˆ’2 π‘₯ βˆ’1 βˆ’2 π‘₯ βˆ’ 𝑐 π‘₯ π‘₯ π‘₯ 4 𝑑π‘₯ =βˆ’ 2 3 π‘₯ βˆ’ 𝑐 ? ? ? ? ?

3 Definite Integration 𝑦 Suppose you wanted to find the area under the curve between π‘₯=π‘Ž and π‘₯=𝑏. 𝛿π‘₯ π‘₯ π‘Ž 𝑏 We could add together the area of individual strips, which we want to make as thin as possible…

4 Definite Integration What is the total area between π‘₯ 1 and π‘₯ 7 ?
𝑦 𝑦=𝑓(π‘₯) 𝛿π‘₯ π‘₯ π‘₯ 1 π‘₯ 2 π‘₯ 3 π‘₯ 4 π‘₯ 5 π‘₯ 6 π‘₯ 7 π‘Ž 𝑏 What is the total area between π‘₯ 1 and π‘₯ 7 ? 𝑖=1 7 𝑓 π‘₯ 𝑖 𝛿π‘₯ π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ As 𝛿π‘₯β†’0

5 Definite Integration π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯  οƒΌ    οƒΌ οƒΌ   βˆ’ + βˆ’ + βˆ’ +
π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ You could think of this as β€œSum the values of 𝑓(π‘₯) between π‘₯=π‘Ž and π‘₯=𝑏.” 𝑦 Reflecting on above, do you think the following definite integrals would be positive or negative or 0? 𝑦= sin π‘₯ 0 πœ‹ 2 sin π‘₯ 𝑑π‘₯  βˆ’ + οƒΌ  π‘₯ πœ‹ 2πœ‹ 0 2πœ‹ sin π‘₯ 𝑑π‘₯  βˆ’ +  οƒΌ πœ‹ 2 2πœ‹ sin π‘₯ 𝑑π‘₯ βˆ’ οƒΌ +  

6 Evaluating Definite Integrals
1 2 3 π‘₯ 2 𝑑π‘₯ ? = π‘₯ We use square brackets to say that we’ve integrated the function, but we’re yet to involve the limits 1 and 2. = 2 3 βˆ’ 1 3 =7 ? Then we find the difference when we sub in our limits. ? π‘Ž 𝑏 𝑓 β€² π‘₯ 𝑑π‘₯= 𝑓 π‘₯ π‘Ž 𝑏 =𝑓 𝑏 βˆ’π‘“(π‘Ž)

7 Evaluating Definite Integrals
1 2 2 π‘₯ 3 +2π‘₯ 𝑑π‘₯ = π‘₯ 4 + π‘₯ = 8+4 βˆ’ = 21 2 βˆ’2 βˆ’1 4π‘₯ 3 +3 π‘₯ 2 𝑑π‘₯ = π‘₯ 4 + π‘₯ 3 βˆ’2 βˆ’1 = 1βˆ’1 βˆ’ 16βˆ’8 =βˆ’8 ? ? Bro Tip: Be careful with your negatives, and use bracketing to avoid errors.

8 Exercise 11B 1 Find the area between the curve with equation 𝑦=𝑓 π‘₯ the π‘₯-axis and the lines π‘₯=π‘Ž and π‘₯=𝑏. 𝑓 π‘₯ =3 π‘₯ 2 βˆ’2π‘₯ π‘Ž=0, 𝑏=2 𝑓 π‘₯ = π‘₯ +2π‘₯ π‘Ž=1, 𝑏=2 𝑓 π‘₯ = 8 π‘₯ π‘₯ π‘Ž=1, 𝑏=4 The sketch shows the curve with equation y=π‘₯( π‘₯ 2 βˆ’4). Find the area of the shaded region (hint: first find the roots). Find the area of the finite region between the curve with equation 𝑦=(3βˆ’π‘₯)(1+π‘₯) and the π‘₯-axis. Find the area of the finite region between the curve with equation 𝑦= π‘₯ 2 2βˆ’π‘₯ and the π‘₯-axis. πŸ– πŸ’.𝟐𝟐 𝟏𝟎𝟏 𝟏𝟐 ? a ? c ? e 2 πŸ’ ? 𝟏𝟎 𝟐 πŸ‘ ? 4 6 𝟏 𝟏 πŸ‘ ?

9 Harder Examples Sketch: ? ? ?
Find the area bounded between the curve with equation 𝑦= π‘₯ 3 βˆ’π‘₯ and the π‘₯-axis. Sketch: (Hint: factorise!) ? 𝑦 π‘₯ βˆ’1 1 Looking at the sketch, what is βˆ’1 1 π‘₯ 3 βˆ’π‘₯ 𝑑π‘₯ and why? 0, because the positive and negative region cancel each other out. ? What therefore should we do? Find the negative and positive region separately. βˆ’1 0 π‘₯ 3 βˆ’3 𝑑π‘₯ =βˆ’ π‘₯ 3 βˆ’3 𝑑π‘₯ =+ 1 4 So total area is 𝟏 πŸ’ + 𝟏 πŸ’ = 𝟏 𝟐 ?

10 Harder Examples ? ? The Sketch The number crunching
Sketch the curve with equation 𝑦=π‘₯ π‘₯βˆ’1 π‘₯+3 and find the area between the curve and the π‘₯-axis. The Sketch The number crunching ? ? 𝑦 π‘₯ π‘₯βˆ’1 π‘₯+3 = π‘₯ 3 βˆ’2 π‘₯ 2 βˆ’3π‘₯ βˆ’3 0 π‘₯ 3 βˆ’2 π‘₯ 2 βˆ’3π‘₯ 𝑑π‘₯ =11.25 0 1 π‘₯ 3 βˆ’2 π‘₯ 2 βˆ’3π‘₯ 𝑑π‘₯ =βˆ’ 7 12 Adding: =11 5 6 π‘₯ -3 1

11 Exercise 11C Find the area of the finite region or regions bounded by the curves and the π‘₯-axis. 1 1 3 ? 𝑦=π‘₯ π‘₯+2 𝑦= π‘₯+1 π‘₯βˆ’4 𝑦= π‘₯+3 π‘₯ π‘₯βˆ’3 𝑦= π‘₯ 2 π‘₯βˆ’2 𝑦=π‘₯ π‘₯βˆ’2 π‘₯βˆ’5 1 20 5 6 ? 2 40 1 2 ? 3 1 1 3 ? 4 5 ?

12 Curves bound between two lines
𝑦=𝑓(π‘₯) 𝛿π‘₯ π‘₯ π‘Ž 𝑏 Remember that 𝒂 𝒃 𝒇(𝒙) meant the sum of all the 𝑦 values between π‘₯=π‘Ž and π‘₯=𝑏 (by using infinitely thin strips).

13 Curves bound between two lines
𝑦=𝑔(π‘₯) 𝑦=𝑓(π‘₯) π‘₯ 𝑏 π‘Ž How could we use a similar principle if we were looking for the area bound between two lines? What is the height of each of these strips? 𝑔 π‘₯ βˆ’π‘“(π‘₯) 𝐴= π‘Ž 𝑏 𝑔 π‘₯ βˆ’π‘“ π‘₯ 𝑑π‘₯ ? ? therefore area…

14 Curves bound between two lines
Find the area bound between 𝑦=π‘₯ and 𝑦=π‘₯ 4βˆ’π‘₯ . 𝑦 𝑦=π‘₯ 0 3 π‘₯ 4βˆ’π‘₯ βˆ’π‘₯ 𝑑π‘₯=4.5 ? 𝑦=π‘₯ 4βˆ’π‘₯ π‘₯ Bro Tip: Always do the function of the top line minus the function of the bottom line. That way the difference in the 𝑦 values is always positive, and you don’t have to worry about negative areas. Bro Tip: We’ll need to find the points at which they intersect.

15 Curves bound between two lines
Edexcel C2 May 2013 (Retracted) ? π‘₯=βˆ’4, π‘₯=2 ? Area = 36

16 More complex areas C A B ? 𝑨𝒓𝒆𝒂=πŸπŸ” 𝟏 πŸ‘
Bro Tip: Sometimes we can subtract areas from others. e.g. Here we could start with the area of the triangle OBC. y = x(x-3) C y = 2x A B 𝑨𝒓𝒆𝒂=πŸπŸ” 𝟏 πŸ‘ ?

17 Exercise 11D ? 𝐴 βˆ’2,6 𝐡 2,6 π΄π‘Ÿπ‘’π‘Ž=10 2 3 1 A region is bounded by the line 𝑦=6 and the curve 𝑦= π‘₯ 2 +2. Find the coordinates of the points of intersection. Hence find the area of the finite region bounded by 𝐴𝐡 and the curve. The diagram shows a sketch of part of the curve with equation 𝑦=9βˆ’3π‘₯βˆ’5 π‘₯ 2 βˆ’ π‘₯ 3 and the line with equation 𝑦=4βˆ’4π‘₯. The line cuts the curve at the points 𝐴 βˆ’1,8 and 𝐡 1,0 . Find the area of the shaded region between 𝐴𝐡 and the curve. Find the area of the finite region bounded by the curve with equation 𝑦= 1βˆ’π‘₯ π‘₯+3 and the line 𝑦=π‘₯+3. The diagram shows part of the curve with equation 𝑦=3 π‘₯ βˆ’ π‘₯ and the line with equation 𝑦=4βˆ’ 1 2 π‘₯. a) Verify that the line and the curve cross at 𝐴 4,2 . b) Find the area of the finite region bounded by the curve and the line. 3 𝐴 𝐡 6 2 3 ? 4 ? 4.5 9 4 𝐴 ? 7.2

18 Exercise 11D (Probably more difficult than you’d see in an exam paper, but you never know…) The diagram shows a sketch of part of the curve with equation 𝑦= π‘₯ 2 +1 and the line with equation 𝑦=7βˆ’π‘₯. a) Find the area of 𝑅 1 . b) Find the area of 𝑅 2 . Q6 𝑦 7 ? 𝑅 1 =20 5 6 𝑅 2 =17 1 6 𝑅 1 𝑅 2 π‘₯ 7

19 Trapezium Rule Instead of infinitely thin rectangular strips, we might use trapeziums to approximate the area under the curve. What is the area here? y1 y2 y3 y4 h ? π΄π‘Ÿπ‘’π‘Ž= 1 2 β„Ž 𝑦 1 + 𝑦 β„Ž 𝑦 2 + 𝑦 β„Ž 𝑦 3 + 𝑦 4

20 Trapezium Rule β„Ž=0.5 π΄π‘Ÿπ‘’π‘Žβ‰ˆ8.75 ? ? In general:
width of each trapezium π‘Ž 𝑏 𝑦 𝑑π‘₯β‰ˆ β„Ž 2 𝑦 𝑦 2 +…+ 𝑦 π‘›βˆ’1 + 𝑦 𝑛 Area under curve is approximately Example We’re approximating the region bounded between π‘₯=1, π‘₯=3, the x-axis the curve 𝑦= π‘₯ 2 x 1 1.5 2 2.5 3 y 2.25 4 6.25 9 β„Ž= π΄π‘Ÿπ‘’π‘Žβ‰ˆ8.75 ? ?

21 Trapezium Rule ? ? May 2013 (Retracted) 0.8571
Bro Tip: You can generate table with Casio calcs . π‘€π‘œπ‘‘π‘’β†’3 (π‘‡π‘Žπ‘π‘™π‘’). Use β€˜Alpha’ button to key in X within the function. Press = ? 0.8571 𝑨𝒓𝒆𝒂= 𝟎.𝟏 𝟐 𝟎.πŸ•πŸŽπŸ•πŸ+𝟐 𝟎.πŸ•πŸ“πŸ—πŸ+𝟎.πŸ–πŸŽπŸ—πŸŽ+𝟎.πŸ–πŸ“πŸ•πŸ+𝟎.πŸ—πŸŽπŸ‘πŸ• +𝟎.πŸ—πŸ’πŸ–πŸ• =𝟎.πŸ’πŸπŸ” ?

22 To add: When do we underestimate and overestimate?


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