SHS3 Additional Mathematics · Semester 2, Week 3
Principle of Calculus
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Curriculum details
- Strand
- Calculus (Strand 3)
- Sub-strand
- Principle of Calculus (3.1)
- Content standard
- 3.3.1.CS.1 - Demonstrate conceptual understanding of the rules and techniques of integration to select and apply them. appropriately to a function. 3.3.1.LO.1 Identify and apply the integration rules to evaluate integrals.
- Indicator
- 3.3.1.LI.2 - Identify and apply appropriate techniques for integration of a function.
- Suggested placement
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Semester 2, Week 3
(Week 23 of the year)
Our suggestion, laid out in curriculum order across three terms of twelve weeks. NaCCA does not fix the week, so follow your school's scheme of learning.
- Source document note
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The official curriculum document has a fault in this entry, so the curriculum text above is transcribed exactly as printed:
- exemplars - p.510: adjacent duplicated CambriaMath characters were collapsed, but this PDF also maps some equation glyphs to the wrong letter or operator; verify every mathematical expression in this row against the rendered page
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- Curriculum reference
- NaCCA curriculum document, p. 510
Exemplars (from the NaCCA curriculum)
Talk for Learning, Think-pair-share, Experiential Learning; and Group Work/Collaborative Learning. Activity 1: Methods of Substitution Learners in collaborative groups brainstorm and discuss special features of techniques or methods of integration. Methods of substitution If an integral could be written in the form ∫ 𝑓[𝑔(𝑥)]𝑔 - (𝑥)𝑑 then we can set 𝑢 = 𝑔(𝑥) and integrate ∫ 𝑓(𝑢)𝑑 . Example 1 (Substitution) For each of the following integrals, learners are to rewrite each integral in the form. ∫ 𝑓[𝑔(𝑥)]𝑔 - (𝑥)𝑑 and clearly identify 𝑔(𝑥) and 𝑔'(𝑥). - ∫ sin(2𝑥) 𝑑 - ∫ 𝑡√𝑡 ) + 4 𝑑 - ∫(𝑥 ) + 1) ∙ 2𝑥 - ∫(𝑥 ) + 1) 7 ∙ 2𝑥 .0 - ∫ 𝑑 √!$0 ! Example 2 (Evaluation of definite integrals) Evaluate the following: - ∫ , 𝑡√𝑡 ) + 4 𝑑 1 . - ∫ ! (𝑥 ) + 1) . 2𝑥 ( - ∫ *! (𝑥 ) + 1) 7 . 2𝑥 , .0 - ∫ ) √!$0 ! 𝑑 Example - Find the function 𝑓(𝑥) for which 4 𝑓 - (𝑥) = 𝑥 ) − 2 𝑎 𝑓(1) = 3 Solution: 𝑓(𝑥) = ∫ 𝑓 - (𝑥)𝑑 = ∫(𝑥 ) − 2)𝑑 = " 𝑥 " − 2𝑥 + 𝐶, 𝐶 𝑖 𝑎 𝑐 ! ! , 𝑓(1) = " (1) " − 2(1) + 𝐶 = " , . 𝐶 = " + " = 3 ∴ 𝑓(𝑥) = " 𝑥 " − 2𝑥 + 3 ! - A rocket is fired vertically into the air. Its velocity at 𝑡 seconds after lift-off is 𝑣(𝑡) = 6𝑡 + 0.5 meter per second. Before launch, the top of the rocket is 8 meters above the launch pad. Find the height of the rocket (measured from the top of the rocket to the launch pad) at time 𝑡. Solution: 𝑠(𝑡) denotes the height of the rocket at time 𝑡, then 𝑠 - (𝑡) = 𝑣(𝑡) 𝑠(𝑡) = ∫ 𝑣(𝑡)𝑑 = ∫(6𝑡 + 0.5)𝑑 = 3𝑡 ) + 0.5𝑡 + 𝐶 When 𝑡 = 0, 𝑠(0) = 8 3(0) ) + 0.5(0) + 𝐶 = 8 𝐶 = 8 ∴ 𝑠(𝑡) = 3𝑡 ) + 0.5𝑡 + 8 - A company's marginal cost function is 0.015𝑥 ) − 2𝑥 + 80 dollars, where 𝑥 denotes the number of units produced in one day. The company has fixed costs of GHS1000 per day. - Find the cost of producing 𝑥 units per day. - Suppose the current production level is 𝑥 = 30. Determine the costs that will rise if the production level is raised to 𝑥 = 60 units. Solution: (a) Let 𝐶(𝑥) be the cost of producing 𝑥 units in one day. The derivative 𝐶 - (𝑥) is the marginal cost. ∴ 𝐶(𝑥) = ∫(0.015𝑥 ) − 2𝑥 + 80)𝑑 = 0.005𝑥 " − 𝑥 ) + 80𝑥 + 𝐶 The GHC1000 fixed costs are the costs incurred when producing 0 units. That is 𝐶(0) = 1000 0.005(0) " − (0) ) + 80(0) + 𝐶 = 1000 𝐶 = 1000 𝐶(𝑥) = 0.005𝑥 − 𝑥 + 80𝑥 + 1000 " ) (b) Increase in cost when production is raised from 𝑥 = 30 𝑡 𝑥 = 60 is 𝐶(60) − 𝐶(30) 𝐶(60) = 0.005(60) " − (60) ) + 80(60) + 1000 = 3280 𝐶(30) = 0.005(30) " − (30) ) + 80(30) + 1000 ∴ 𝐶(60) − 𝐶(30) = 3280 − 2635 = 𝐺645 = 𝐺2635 Teaching and Learning Resources: - Reading resource colour pens - Notebook - Graph sheets - Mathematical sets, - Technological tools. - Curriculum - Card boards Assessment (3.3.1.AS.2). The document marks these depth-of-knowledge levels for this indicator: Level 3 Strategic reasoning; Level 4 Extended critical thinking and reasoning.