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Methods in calculus.

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1 Methods in calculus

2 FM Methods in Calculus: Improper integrals
KUS objectives BAT evaluate improper integrals; the mean of a function; Integrate using trig substitution; integrate using partial fractions Starter: find 5๐‘ฅ 3+ ๐‘ฅ 2 ๐‘‘๐‘ฅ =5 3+ ๐‘ฅ 2 +๐ถ ๐‘ฅ 2 ๐‘’ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘ฅ 2 โˆ’2๐‘ฅ+2 ๐‘’ ๐‘ฅ +๐ถ sin ๐‘ฅ cos ๐‘ฅ 1+3 ๐‘ ๐‘–๐‘› 2 ๐‘ฅ ๐‘‘๐‘ฅ =๐‘™๐‘› 1+3 ๐‘ ๐‘–๐‘› 2 ๐‘ฅ +๐ถ

3 The integral ๐‘Ž ๐‘ ๐‘“(๐‘ฅ) ๐‘‘๐‘ฅ is improper if
Notes ๐ด๐‘Ÿ๐‘’๐‘Ž= ๐‘Ž ๐‘ ๐‘“(๐‘ฅ) ๐‘‘๐‘ฅ A definite integral represents the area enclosed by a continuous function ๐‘ฆ=๐‘“(๐‘ฅ), the x-axis and line ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘ ๐‘œ ๐‘ฆ ๐‘ฅ ๐‘ฆ= 1 ๐‘ฅ 2 ๐‘Ž (1) ๐ด๐‘Ÿ๐‘’๐‘Ž= ๐‘Ž โˆž 1 ๐‘ฅ 2 ๐‘‘๐‘ฅ An Improper integral represents the area where one of the limits is infinite or where the function is not defined at some point The integral ๐‘Ž ๐‘ ๐‘“(๐‘ฅ) ๐‘‘๐‘ฅ is improper if one or both of the limits is infinite f(x) is undefined at ๐‘ฅ=๐‘Ž or ๐‘ฅ=๐‘ or at a point in the interval [a, b] ๐‘œ ๐‘ฆ ๐‘ฅ ๐‘ฆ= 1 ๐‘ฅ 3 (2) ๐ด๐‘Ÿ๐‘’๐‘Ž= ๐‘ฅ ๐‘‘๐‘ฅ If the improper integral exists it is said to be convergent. If it does not exist it is said to be divergent

4 WB A1- limits of inifinity Evaluate each improper integral
๐‘Ž) 1 โˆž 1 ๐‘ฅ 2 ๐‘‘๐‘ฅ ๐‘) 1 โˆž 1 ๐‘ฅ ๐‘‘๐‘ฅ Use limit notation to rewrite the integral as ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž 1 ๐‘ก 1 ๐‘ฅ 2 ๐‘‘๐‘ฅ = ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž โˆ’ 1 ๐‘ฅ ๐‘ก 1 = ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž โˆ’ 1 ๐‘ก +1 ๐‘Ž๐‘  ๐‘กโ†’โˆž , 1 ๐‘ก โ†’0 =๐Ÿ b) ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž 1 ๐‘ก 1 ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž ln ๐‘ฅ ๐‘ก 1 = ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž ln ๐‘ก โˆ’ ln 1 ๐‘Ž๐‘  ๐‘กโ†’โˆž , ln ๐‘ก โ†’โˆž ๐’…๐’๐’†๐’” ๐’๐’๐’• ๐’„๐’๐’๐’—๐’†๐’“๐’ˆ๐’†

5 WB A2 - undefined limits Evaluate each improper integral
๐‘Ž) ๐‘ฅ 2 ๐‘‘๐‘ฅ ๐‘) 0 2 ๐‘ฅ 4โˆ’ ๐‘ฅ 2 ๐‘‘๐‘ฅ Use limit notation to rewrite the integral as ๐‘™๐‘–๐‘š ๐‘กโ†’0 ๐‘ก ๐‘ฅ 2 ๐‘‘๐‘ฅ = ๐‘™๐‘–๐‘š ๐‘กโ†’0 โˆ’ 1 ๐‘ฅ 1 ๐‘ก = ๐‘™๐‘–๐‘š ๐‘กโ†’0 โˆ’1+ 1 ๐‘ก ๐‘Ž๐‘  ๐‘กโ†’0 , 1 ๐‘ก โ†’โˆž โˆ’1+ 1 ๐‘ก โ†’โˆž as ๐‘กโ†’0 ๐’…๐’๐’†๐’” ๐’๐’๐’• ๐’„๐’๐’๐’—๐’†๐’“๐’ˆ๐’† b) ๐‘™๐‘–๐‘š ๐‘กโ†’2 0 ๐‘ก ๐‘ฅ 4โˆ’ ๐‘ฅ 2 ๐‘‘๐‘ฅ = ๐‘™๐‘–๐‘š ๐‘กโ†’2 โˆ’ 4โˆ’ ๐‘ฅ 2 ๐‘ก 0 The function is undefined for ๐‘ฅ=2 = ๐‘™๐‘–๐‘š ๐‘กโ†’2 โˆ’ 4โˆ’ ๐‘ก 2 +โˆ’ 4โˆ’ 0 2 ๐‘Ž๐‘  ๐‘กโ†’2 , 4โˆ’ ๐‘ก 2 โ†’0 = (0) = 2

6 NOW DO EX 3A WB A3 โ€“ both limits are infinity a) Find ๐‘ฅ ๐‘’ โˆ’ ๐‘ฅ 2 ๐‘‘๐‘ฅ
b) Hence show that converges and find its value โˆ’โˆž โˆž ๐‘ฅ ๐‘’ โˆ’ ๐‘ฅ 2 ๐‘‘๐‘ฅ โˆ’โˆž โˆž ๐‘ฅ ๐‘’ โˆ’ ๐‘ฅ 2 ๐‘‘๐‘ฅ =โˆ’ 1 2 ๐‘’ โˆ’ ๐‘ฅ 2 +C b) โˆ’โˆž โˆž ๐‘ฅ ๐‘’ โˆ’ ๐‘ฅ 2 ๐‘‘๐‘ฅ = โˆ’โˆž 0 ๐‘ฅ ๐‘’ โˆ’ ๐‘ฅ 2 ๐‘‘๐‘ฅ + 0 โˆž ๐‘ฅ ๐‘’ โˆ’ ๐‘ฅ 2 ๐‘‘๐‘ฅ Split into two improper integrals = ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž โˆ’ 1 2 ๐‘’ โˆ’ ๐‘ฅ โˆ’๐‘ก ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž โˆ’ 1 2 ๐‘’ โˆ’ ๐‘ฅ 2 ๐‘ก 0 = ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž โˆ’ 1 2 โˆ’โˆ’ 1 2 ๐‘’ โˆ’ ๐‘ก ๐‘™๐‘–๐‘š ๐‘กโ†’โˆž โˆ’ 1 2 ๐‘’ โˆ’ ๐‘ก 2 โˆ’โˆ’ 1 2 ๐‘Ž๐‘  ๐‘กโ†’โˆž , ๐‘’ โˆ’ ๐‘ก 2 โ†’ both integrals converge = โˆ’ = 0 The areas cancel out NOW DO EX 3A

7 One thing to improve is โ€“
KUS objectives BAT evaluate improper integrals; the mean of a function; Integrate using trig substitution; integrate using partial fractions self-assess One thing learned is โ€“ One thing to improve is โ€“

8 END


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