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Methods in calculus.

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Presentation on theme: "Methods in calculus."β€” Presentation transcript:

1 Methods in calculus

2 FM Methods in Calculus: inverse trig functions
KUS objectives BAT integrate using partial fractions Starter: express in partial fractions 1 π‘₯ βˆ’ 1 π‘₯+1 1 π‘₯ π‘₯+1 2π‘₯βˆ’1 π‘₯ 2 βˆ’4 5 4(π‘₯+2) + 3 4(π‘₯βˆ’2) 3π‘₯βˆ’2 2 π‘₯ 2 +3π‘₯+1 5 π‘₯+1 βˆ’ 7 2π‘₯+1

3 Don’t mix up 1 π‘Ž 2 βˆ’ π‘₯ 2 𝑑π‘₯ and 1 π‘Ž 2 + π‘₯ 2 𝑑π‘₯
WB E1 a) show that π‘Ž 2 βˆ’ π‘₯ 2 𝑑π‘₯= 1 2π‘Ž ln π‘Ž+π‘₯ π‘Žβˆ’π‘₯ +𝐢 where a is a constant 1 π‘Ž 2 βˆ’ π‘₯ 2 𝑑π‘₯= 1 (π‘Ž+π‘₯)(π‘Žβˆ’π‘₯) 𝑑π‘₯ 1 (π‘Ž+π‘₯)(π‘Žβˆ’π‘₯) = 1 2π‘Ž 1 (π‘Ž+π‘₯) + 1 (π‘Žβˆ’π‘₯) = 1 2a (π‘Ž+π‘₯) 𝑑π‘₯ + 1 2π‘Ž (π‘Žβˆ’π‘₯) 𝑑π‘₯ = 1 2π‘Ž ln π‘Ž+π‘₯ + ln π‘Žβˆ’π‘₯ +𝐢 = 1 2π‘Ž ln π‘Ž+π‘₯ π‘Žβˆ’π‘₯ +𝐢 Don’t mix up π‘Ž 2 βˆ’ π‘₯ 2 𝑑π‘₯ and π‘Ž 2 + π‘₯ 2 𝑑π‘₯

4 5 π‘₯ 2 +π‘₯βˆ’10 (π‘₯+1) π‘₯ 2 βˆ’3 = 𝑨 π‘₯+1 + 𝑩π‘₯+𝐢 π‘₯ 2 βˆ’3
Notes partial fractions and quadratic factors When a partial fraction includes a quadratic factor of the form π‘₯ 2 +𝑐 we need to put a linear numerator of the form 𝐴π‘₯+𝐡 on the partial fraction 5 π‘₯ 2 +π‘₯βˆ’10 (π‘₯+1) π‘₯ 2 βˆ’3 = 𝑨 π‘₯+1 + 𝑩π‘₯+𝐢 π‘₯ 2 βˆ’3 WB E2 5 π‘₯ 2 +π‘₯βˆ’10=𝑨 π‘₯ 2 βˆ’3 + 𝑩π‘₯+π‘ͺ π‘₯+1 Let x=βˆ’1 then 5+(βˆ’1)βˆ’10=𝑨 1βˆ’3 gives 𝑨=πŸ‘ Look at coefficients of π‘₯ 2 5 π‘₯ 2 =𝑨 π‘₯ 2 +𝑩 π‘₯ gives 𝑩=𝟐 π‘₯=𝑩π‘₯+π‘ͺπ‘₯ gives π‘ͺ=βˆ’πŸ Look at coefficients of π‘₯ 5 π‘₯ 2 +π‘₯βˆ’10 (π‘₯+1) π‘₯ 2 βˆ’3 = πŸ‘ π‘₯+1 + 𝟐π‘₯βˆ’πŸ π‘₯ 2 βˆ’3

5 = 1 18 ln π‘₯ 2 π‘₯ 2 +9 + 1 3 arctan π‘₯ 3 +𝐢 So 𝐴= 1 18 𝐡= 1 3
WB E3 Show that π‘₯ π‘₯ 3 +9π‘₯ 𝑑π‘₯=𝐴 ln π‘₯ 2 π‘₯ π΅π‘Žπ‘Ÿπ‘π‘‘π‘Žπ‘› π‘₯ 3 +𝐢 where A and B are constants to be found 1+π‘₯ π‘₯ 3 +9π‘₯ 𝑑π‘₯= 1+π‘₯ π‘₯(π‘₯ 2 +9) 𝑑π‘₯ 1+π‘₯ π‘₯(π‘₯ 2 +9) = π‘₯ βˆ’ π‘₯βˆ’9 π‘₯ 2 +9 = π‘₯ 𝑑π‘₯ βˆ’ π‘₯ π‘₯ 2 +9 𝑑π‘₯ βˆ’ π‘₯ 2 +9 𝑑π‘₯ 1 π‘Ž 2 + π‘₯ 2 𝑑π‘₯= 1 π‘Ž arctan π‘₯ π‘Ž +𝐢 = 1 9 ln π‘₯ βˆ’ ln π‘₯ arctan π‘₯ 3 +C = ln π‘₯ βˆ’ ln π‘₯ arctan π‘₯ 3 +𝐢 = ln π‘₯ 2 π‘₯ arctan π‘₯ 3 +𝐢 So 𝐴= 𝐡= 1 3

6 Look at coefficients of π‘₯ 4
WB E4 a) express π‘₯ 4 +π‘₯ π‘₯ 4 +5 π‘₯ as partial fractions b) Hence find π‘₯ 4 +π‘₯ π‘₯ 4 +5 π‘₯ 2 +6 𝑑π‘₯ π‘₯ 4 +π‘₯ π‘₯ 4 +5 π‘₯ 2 +6 = π‘₯ 4 +π‘₯ π‘₯ π‘₯ =A+ 𝐡π‘₯+𝐢 π‘₯ 𝐷π‘₯+𝐸 π‘₯ 2 +3 π‘₯ 4 +π‘₯=𝐴 π‘₯ π‘₯ 𝐡π‘₯+𝐢 π‘₯ (𝐷π‘₯+𝐸) π‘₯ 2 +2 Look at coefficients of π‘₯ 4 π‘₯ 4 =𝑨 π‘₯ gives 𝑨=𝟏 Look at coefficients of π‘₯ 3 0=𝐡 π‘₯ 3 +𝐷 π‘₯ gives 𝑩+𝑫=𝟎 Look at coefficients of π‘₯ π‘₯=3𝐡π‘₯+2𝐷π‘₯ gives 3𝑩+πŸπ‘«=𝟏 Solve these simultaneous equations gives 𝑩=𝟏 𝑫=βˆ’πŸ Look at coefficients of π‘₯ 2 0=5𝐴 π‘₯ 2 +𝐢 π‘₯ 2 +𝐸 π‘₯ 2 gives C+𝐸=βˆ’πŸ“ Look at constant terms 0=6𝐴+3𝐢+2𝐸 gives 3C+2𝐸=βˆ’πŸ” Solve these simultaneous equations gives π‘ͺ=πŸ’ 𝑬=βˆ’πŸ—

7 WB E4b a) express π‘₯ 4 +π‘₯ π‘₯ 4 +5 π‘₯ 2 +6 as partial fractions
b) Hence find π‘₯ 4 +π‘₯ π‘₯ 4 +5 π‘₯ 2 +6 𝑑π‘₯ π‘₯ 4 +π‘₯ π‘₯ 4 +5 π‘₯ 2 +6 = π‘₯ 4 +π‘₯ π‘₯ π‘₯ =1+ π‘₯+4 π‘₯ βˆ’π‘₯βˆ’9 π‘₯ 2 +3 𝑏) π‘₯ 4 +π‘₯ π‘₯ 4 +5 π‘₯ 2 +6 𝑑π‘₯ = 𝑑π‘₯+ π‘₯ π‘₯ 𝑑π‘₯ π‘₯ 𝑑π‘₯βˆ’ π‘₯ π‘₯ 𝑑π‘₯ βˆ’ π‘₯ 𝑑π‘₯ = π‘₯ ln π‘₯ arctan π‘₯ βˆ’ 1 2 ln π‘₯ βˆ’ arctan π‘₯ 𝐢 = π‘₯ ln π‘₯ π‘₯ arctan π‘₯ βˆ’ arctan π‘₯ 𝐢 NOW DO EX 3E

8 One thing to improve is –
KUS objectives BAT integrate using partial fractions self-assess One thing learned is – One thing to improve is –

9 END


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