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Published byΞΟΟΞΏΟΞΏΟ ΞΞΏΟ ΟαλιανΟΟ Modified over 5 years ago
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Dr J Frost (jfrost@tiffin.kingston.sch.uk)
C2 Chapter 11 Integration Dr J Frost Last modified: 17th October 2013
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Recap 2 π₯ 2 +3π₯ ππ₯ = 2 3 π₯ π₯ 2 +π 1+π₯+ π₯ 2 + π₯ 3 ππ₯ =π₯ π₯ π₯ π₯ 4 +π π₯ β ππ₯ =β 3 4 π₯ β π 2+ π₯ π₯ 2 ππ₯ =β2 π₯ β1 β2 π₯ β π π₯ π₯ π₯ 4 ππ₯ =β 2 3 π₯ β π ? ? ? ? ?
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Definite Integration π¦ Suppose you wanted to find the area under the curve between π₯=π and π₯=π. πΏπ₯ π₯ π π We could add together the area of individual strips, which we want to make as thin as possibleβ¦
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Definite Integration What is the total area between π₯ 1 and π₯ 7 ?
π¦ π¦=π(π₯) πΏπ₯ π₯ π₯ 1 π₯ 2 π₯ 3 π₯ 4 π₯ 5 π₯ 6 π₯ 7 π π What is the total area between π₯ 1 and π₯ 7 ? π=1 7 π π₯ π πΏπ₯ π π π π₯ ππ₯ As πΏπ₯β0
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Definite Integration π π π π₯ ππ₯ ο ο» οΌ ο ο» ο ο» ο ο» οΌ οΌ ο ο» ο ο» β + β + β +
π π π π₯ ππ₯ You could think of this as βSum the values of π(π₯) between π₯=π and π₯=π.β π¦ Reflecting on above, do you think the following definite integrals would be positive or negative or 0? π¦= sin π₯ 0 π 2 sin π₯ ππ₯ ο ο» β + οΌ ο ο» π₯ π 2π 0 2π sin π₯ ππ₯ ο ο» β + ο ο» οΌ π 2 2π sin π₯ ππ₯ β οΌ + ο ο» ο ο»
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Evaluating Definite Integrals
1 2 3 π₯ 2 ππ₯ ? = π₯ We use square brackets to say that weβve integrated the function, but weβre yet to involve the limits 1 and 2. = 2 3 β 1 3 =7 ? Then we find the difference when we sub in our limits. ? π π π β² π₯ ππ₯= π π₯ π π =π π βπ(π)
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Evaluating Definite Integrals
1 2 2 π₯ 3 +2π₯ ππ₯ = π₯ 4 + π₯ = 8+4 β = 21 2 β2 β1 4π₯ 3 +3 π₯ 2 ππ₯ = π₯ 4 + π₯ 3 β2 β1 = 1β1 β 16β8 =β8 ? ? Bro Tip: Be careful with your negatives, and use bracketing to avoid errors.
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Exercise 11B 1 Find the area between the curve with equation π¦=π π₯ the π₯-axis and the lines π₯=π and π₯=π. π π₯ =3 π₯ 2 β2π₯ π=0, π=2 π π₯ = π₯ +2π₯ π=1, π=2 π π₯ = 8 π₯ π₯ π=1, π=4 The sketch shows the curve with equation y=π₯( π₯ 2 β4). Find the area of the shaded region (hint: first find the roots). Find the area of the finite region between the curve with equation π¦=(3βπ₯)(1+π₯) and the π₯-axis. Find the area of the finite region between the curve with equation π¦= π₯ 2 2βπ₯ and the π₯-axis. π π.ππ πππ ππ ? a ? c ? e 2 π ? ππ π π ? 4 6 π π π ?
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Harder Examples Sketch: ? ? ?
Find the area bounded between the curve with equation π¦= π₯ 3 βπ₯ and the π₯-axis. Sketch: (Hint: factorise!) ? π¦ π₯ β1 1 Looking at the sketch, what is β1 1 π₯ 3 βπ₯ ππ₯ 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. β1 0 π₯ 3 β3 ππ₯ =β π₯ 3 β3 ππ₯ =+ 1 4 So total area is π π + π π = π π ?
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Harder Examples ? ? The Sketch The number crunching
Sketch the curve with equation π¦=π₯ π₯β1 π₯+3 and find the area between the curve and the π₯-axis. The Sketch The number crunching ? ? π¦ π₯ π₯β1 π₯+3 = π₯ 3 β2 π₯ 2 β3π₯ β3 0 π₯ 3 β2 π₯ 2 β3π₯ ππ₯ =11.25 0 1 π₯ 3 β2 π₯ 2 β3π₯ ππ₯ =β 7 12 Adding: =11 5 6 π₯ -3 1
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Exercise 11C Find the area of the finite region or regions bounded by the curves and the π₯-axis. 1 1 3 ? π¦=π₯ π₯+2 π¦= π₯+1 π₯β4 π¦= π₯+3 π₯ π₯β3 π¦= π₯ 2 π₯β2 π¦=π₯ π₯β2 π₯β5 1 20 5 6 ? 2 40 1 2 ? 3 1 1 3 ? 4 5 ?
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Curves bound between two lines
π¦=π(π₯) πΏπ₯ π₯ π π Remember that π π π(π) meant the sum of all the π¦ values between π₯=π and π₯=π (by using infinitely thin strips).
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Curves bound between two lines
π¦=π(π₯) π¦=π(π₯) π₯ π π 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? π π₯ βπ(π₯) π΄= π π π π₯ βπ π₯ ππ₯ ? ? therefore areaβ¦
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Curves bound between two lines
Find the area bound between π¦=π₯ and π¦=π₯ 4βπ₯ . π¦ π¦=π₯ 0 3 π₯ 4βπ₯ βπ₯ ππ₯=4.5 ? π¦=π₯ 4βπ₯ π₯ Bro Tip: Always do the function of the top line minus the function of the bottom line. That way the difference in the π¦ values is always positive, and you donβt have to worry about negative areas. Bro Tip: Weβll need to find the points at which they intersect.
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Curves bound between two lines
Edexcel C2 May 2013 (Retracted) ? π₯=β4, π₯=2 ? Area = 36
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More complex areas C A B ? π¨πππ=ππ π π
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 π¨πππ=ππ π π ?
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Exercise 11D ? π΄ β2,6 π΅ 2,6 π΄πππ=10 2 3 1 A region is bounded by the line π¦=6 and the curve π¦= π₯ 2 +2. Find the coordinates of the points of intersection. Hence find the area of the finite region bounded by π΄π΅ and the curve. The diagram shows a sketch of part of the curve with equation π¦=9β3π₯β5 π₯ 2 β π₯ 3 and the line with equation π¦=4β4π₯. The line cuts the curve at the points π΄ β1,8 and π΅ 1,0 . Find the area of the shaded region between π΄π΅ and the curve. Find the area of the finite region bounded by the curve with equation π¦= 1βπ₯ π₯+3 and the line π¦=π₯+3. The diagram shows part of the curve with equation π¦=3 π₯ β π₯ and the line with equation π¦=4β 1 2 π₯. a) Verify that the line and the curve cross at π΄ 4,2 . b) Find the area of the finite region bounded by the curve and the line. 3 π΄ π΅ 6 2 3 ? 4 ? 4.5 9 4 π΄ ? 7.2
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Exercise 11D (Probably more difficult than youβd see in an exam paper, but you never knowβ¦) The diagram shows a sketch of part of the curve with equation π¦= π₯ 2 +1 and the line with equation π¦=7βπ₯. a) Find the area of π
1 . b) Find the area of π
2 . Q6 π¦ 7 ? π
1 =20 5 6 π
2 =17 1 6 π
1 π
2 π₯ 7
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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΄πππ= 1 2 β π¦ 1 + π¦ β π¦ 2 + π¦ β π¦ 3 + π¦ 4
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Trapezium Rule β=0.5 π΄πππβ8.75 ? ? In general:
width of each trapezium π π π¦ ππ₯β β 2 π¦ π¦ 2 +β¦+ π¦ πβ1 + π¦ π Area under curve is approximately Example Weβre approximating the region bounded between π₯=1, π₯=3, the x-axis the curve π¦= π₯ 2 x 1 1.5 2 2.5 3 y 2.25 4 6.25 9 β= π΄πππβ8.75 ? ?
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Trapezium Rule ? ? May 2013 (Retracted) 0.8571
Bro Tip: You can generate table with Casio calcs . ππππβ3 (πππππ). Use βAlphaβ button to key in X within the function. Press = ? 0.8571 π¨πππ= π.π π π.ππππ+π π.ππππ+π.ππππ+π.ππππ+π.ππππ +π.ππππ =π.πππ ?
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To add: When do we underestimate and overestimate?
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