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Integration By parts
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ππ¦ ππ₯ =3 (π₯β3) 2 π β2π₯ β2 (π₯β3) 3 π β2π₯ ππ¦ ππ₯ = 9β2π₯ (π₯β3) 2 π β2π₯
A2 Integration II KUS objectives BAT use Integration by parts formula to find integrals Starter: Differentiate ππ¦ ππ₯ = π 2π₯ (2π₯+2 π₯ 2 ) π¦ = π₯ 2 π 2π₯ ππ¦ ππ₯ =3 (π₯β3) 2 π β2π₯ β2 (π₯β3) 3 π β2π₯ ππ¦ ππ₯ = 9β2π₯ (π₯β3) 2 π β2π₯ π¦ = (π₯β3) 3 π β2π₯ π¦ = (4π₯β2) β2 π π₯/2 ππ¦ ππ₯ =β8 4π₯β2 β1 π π₯ (4π₯β2) β2 π π₯/2 ππ¦ ππ₯ = 32β64π₯ 2 4π₯β π π₯/2
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Notes 1 In Differentiation you met the product rule This is the differential of two functions multiplied together ο You could think of it as: Rearrange by subtracting vdu/dx Integrate each term with respect to x This is the formula used for Integration by parts! ο You get given this in the booklet The middle term is just a differential ο Integrating a differential cancels them both out! The other terms do not cancel as only part of them are differentiatedβ¦ As a general rule, it is easiest to let u = anything of the form xn. The exception is when there is a lnx term, in which case this should be used as u
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π£= π π₯ π’=4π₯β9 (4π₯β9) π π₯ ππ₯ ππ£ ππ₯ = π π₯ ππ’ ππ₯ =4 = π π₯ 4π₯β9 β π π₯ 4 ππ₯
WB Find the following Integral a) π₯β9 π π₯ ππ₯ b) βπ₯ π 2π₯ ππ₯ c) π₯β3 π π₯ ππ₯ π’=4π₯β9 ππ£ ππ₯ = π π₯ π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ π£= π π₯ (4π₯β9) π π₯ ππ₯ ππ’ ππ₯ =4 = π π₯ 4π₯β9 β π π₯ 4 ππ₯ = 4π₯β9 π π₯ β4 π π₯ +πΆ = 4π₯β13 π π₯ +πΆ
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π’=2π₯β3 π£=β π βπ₯ ππ£ ππ₯ = π βπ₯ ππ’ ππ₯ =2 WB 1c c) Find 2π₯β3 π π₯ ππ₯
π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ π’=2π₯β3 ππ£ ππ₯ = π βπ₯ π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ π£=β π βπ₯ =β(2π₯β3) π βπ₯ β 2 π βπ₯ ππ₯ ππ’ ππ₯ =2 =β 2π₯β3 π βπ₯ +2 π βπ₯ +πΆ = 3β2π₯ π βπ₯ +2 π βπ₯ +πΆ = 5β2π₯ π βπ₯ +πΆ
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π£= 1 3 π₯ 3 π’= ln π₯ π₯ 2 ln π₯ ππ₯ ππ£ ππ₯ = π₯ 2 ππ’ ππ₯ = 1 π₯
WB Find π₯ 2 ln π₯ ππ₯ π’= ln π₯ ππ£ ππ₯ = π₯ 2 π£= 1 3 π₯ 3 π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ π₯ 2 ln π₯ ππ₯ ππ’ ππ₯ = 1 π₯ = ln π₯ π₯ 3 β π₯ π₯ ππ₯ = π₯ 3 ln π₯ β 1 9 π₯ 3 +πΆ
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π’= ln π₯ π£=π₯ π₯ ln π₯ ππ₯ ππ£ ππ₯ =1 ππ’ ππ₯ = 1 π₯ = ln π₯ π₯ β π₯ 1 π₯ ππ₯
WB Find ln π₯ ππ₯ π’= ln π₯ ππ£ ππ₯ =1 π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ π£=π₯ π₯ ln π₯ ππ₯ ππ’ ππ₯ = 1 π₯ = ln π₯ π₯ β π₯ 1 π₯ ππ₯ =π₯ ln π₯ βπ₯+πΆ
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WB 4 use Integration by parts Twice Find a) π₯ 2 π 4π₯ ππ₯ b)
π’= π₯ 2 π£= 1 4 π 4π₯ ππ’ ππ₯ =2π₯ ππ£ ππ₯ = π 4π₯ Round 1 π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ π₯ 2 ln π₯ ππ₯ = π₯ 2 π 4π₯ β π₯ π 4π₯ ππ₯ π’= 1 2 π₯ π£= 1 4 π 4π₯ ππ’ ππ₯ = 1 2 ππ£ ππ₯ = π 4π₯ Round 2 = π₯ 2 π 4π₯ β π 4π₯ π₯ β π 4π₯ = π₯ 2 π 4π₯ β 1 8 π₯ π 4π₯ π 4π₯ +πΆ
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WB 4b use Integration by parts Twice b) Find 6 (π₯β3) 2 π β2π₯ ππ₯
π’=6 (π₯β3) 2 π£=β 1 2 π β2π₯ ππ’ ππ₯ =12 π₯β3 ππ£ ππ₯ = π β2π₯ Round 1 π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ π’=β6π₯+18 π£=β 1 2 π β2π₯ ππ’ ππ₯ =β6 ππ£ ππ₯ = π β2π₯ Round 2 =β3 (π₯β3) 2 π β2π₯ β β6(π₯β3) π β2π₯ ππ₯ =β3 (π₯β3) 2 π β2π₯ β β 1 2 π β2π₯ β6π₯+18 β 3 π β2π₯ =β3 π₯β3 2 π β2π₯ β 3π₯β9 π β2π₯ π β2π₯ +πΆ =+ π β2π₯ β3 π₯ 2 +18π₯β27β3π₯+9β πΆ =+ π β2π₯ β3 π₯ 2 +15π₯β πΆ
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WB 5a Definite integral a) show that 1 2 π₯ π 1βπ₯ ππ₯ =2β 3 π
π’=π₯ π£= βπ 1βπ₯ ππ’ ππ₯ =1 ππ£ ππ₯ = π 1βπ₯ π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ =π₯ βπ 1βπ₯ β βπ 1βπ₯ ππ₯ =π₯ βπ 1βπ₯ β π 1βπ₯ + C Definite integral = β π 1βπ₯ π₯ = β3 π β1 β β2 π 0 =2β 3 π QED
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WB 5b Definite integral b) show that 1 2 4π₯ 3 ln π₯ ππ₯ =16 ln 2 β 15 4
π£= π₯ 4 ππ’ ππ₯ = 1 π₯ ππ£ ππ₯ = 4π₯ 3 π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ = π₯ 4 ln xβ π₯ 3 ππ₯ = π₯ 4 ln xβ π₯ 4 + C Definite integral = π₯ 4 ln xβ π₯ = 16 ln 2β4 β ln 1β =16 ln 2 β QED
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π£=π₯ π’= ln (3π₯+2) ππ’ ππ₯ = 3 3π₯+2 ππ£ ππ₯ =1
WB 6a Definite integral a) show that ln (3π₯+2) ππ₯ = ln 9 β 2 3 ln 2 β3 π’= ln (3π₯+2) π£=π₯ ππ’ ππ₯ = 3 3π₯+2 ππ£ ππ₯ =1 π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ =π₯ ln 3π₯+2 β 3π₯ 3π₯+2 ππ₯ Algebra division 3π₯ 3π₯+2 = 3π₯+2 3π₯+2 β 2 3π₯+2 = 1β 2 3π₯+2 =π₯ ln 3π₯+2 β 1β 2 3π₯+2 ππ₯
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WB 6a Definite integral. a) show that
WB 6a Definite integral a) show that ln (3π₯+2) ππ₯ = ln 9 β 2 3 ln 2 β3 =π₯ ln 3π₯+2 β 1β 2 3π₯+2 ππ₯ =π₯ ln 3π₯+2 βπ₯ ln (3π₯+2) = π₯ ln 3π₯+2 βπ₯ 3 0 = ln 9 β3 β ln 2 = ln 9 β 2 3 ln 2 β3
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WB 6b Definite integral b) show that 0 3 3π₯+2 ln (2π₯) ππ₯ =6 ln 8 β 4
π£= π₯ 2 +π₯ ππ’ ππ₯ = 1 π₯ ππ£ ππ₯ =2π₯+1 π’ ππ£ ππ₯ = π’π£ β π£ ππ’ ππ₯ = π₯ 2 +π₯ ln 3π₯+2 β 1 π₯ π₯ 2 +π₯ ππ₯ = π₯ 2 +π₯ ln 3π₯+2 β π₯ 2 +π₯ +C = π₯ 2 +π₯ ln 3π₯+2 β π₯ 2 +π₯ = 6 ln 8 β4 β 0 β0 = 6 ln 8 - 4
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Skills 218 homework 218
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One thing to improve is β
KUS objectives BAT use Integration by parts formula to find integrals self-assess One thing learned is β One thing to improve is β
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