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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: Improper integrals
KUS objectives BAT evaluate improper integrals; the mean of a function; Integrate using trig substitution; integrate using partial fractions Starter: find 5π‘₯ 3+ π‘₯ 2 𝑑π‘₯ =5 3+ π‘₯ 2 +𝐢 π‘₯ 2 𝑒 π‘₯ 𝑑π‘₯ = π‘₯ 2 βˆ’2π‘₯+2 𝑒 π‘₯ +𝐢 sin π‘₯ cos π‘₯ 1+3 𝑠𝑖𝑛 2 π‘₯ 𝑑π‘₯ =𝑙𝑛 1+3 𝑠𝑖𝑛 2 π‘₯ +𝐢

3 The integral π‘Ž 𝑏 𝑓(π‘₯) 𝑑π‘₯ is improper if
Notes π΄π‘Ÿπ‘’π‘Ž= π‘Ž 𝑏 𝑓(π‘₯) 𝑑π‘₯ A definite integral represents the area enclosed by a continuous function 𝑦=𝑓(π‘₯), the x-axis and line π‘₯=π‘Ž and π‘₯=𝑏 π‘œ 𝑦 π‘₯ 𝑦= 1 π‘₯ 2 π‘Ž (1) π΄π‘Ÿπ‘’π‘Ž= π‘Ž ∞ 1 π‘₯ 2 𝑑π‘₯ 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 π‘Ž 𝑏 𝑓(π‘₯) 𝑑π‘₯ is improper if one or both of the limits is infinite f(x) is undefined at π‘₯=π‘Ž or π‘₯=𝑏 or at a point in the interval [a, b] π‘œ 𝑦 π‘₯ 𝑦= 1 π‘₯ 3 (2) π΄π‘Ÿπ‘’π‘Ž= π‘₯ 𝑑π‘₯ 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
π‘Ž) 1 ∞ 1 π‘₯ 2 𝑑π‘₯ 𝑏) 1 ∞ 1 π‘₯ 𝑑π‘₯ Use limit notation to rewrite the integral as π‘™π‘–π‘š π‘‘β†’βˆž 1 𝑑 1 π‘₯ 2 𝑑π‘₯ = π‘™π‘–π‘š π‘‘β†’βˆž βˆ’ 1 π‘₯ 𝑑 1 = π‘™π‘–π‘š π‘‘β†’βˆž βˆ’ 1 𝑑 +1 π‘Žπ‘  π‘‘β†’βˆž , 1 𝑑 β†’0 =𝟏 b) π‘™π‘–π‘š π‘‘β†’βˆž 1 𝑑 1 π‘₯ 𝑑π‘₯ = π‘™π‘–π‘š π‘‘β†’βˆž ln π‘₯ 𝑑 1 = π‘™π‘–π‘š π‘‘β†’βˆž ln 𝑑 βˆ’ ln 1 π‘Žπ‘  π‘‘β†’βˆž , ln 𝑑 β†’βˆž 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 π’„π’π’π’—π’†π’“π’ˆπ’†

5 WB A2 - undefined limits Evaluate each improper integral
π‘Ž) π‘₯ 2 𝑑π‘₯ 𝑏) 0 2 π‘₯ 4βˆ’ π‘₯ 2 𝑑π‘₯ Use limit notation to rewrite the integral as π‘™π‘–π‘š 𝑑→0 𝑑 π‘₯ 2 𝑑π‘₯ = π‘™π‘–π‘š 𝑑→0 βˆ’ 1 π‘₯ 1 𝑑 = π‘™π‘–π‘š 𝑑→0 βˆ’1+ 1 𝑑 π‘Žπ‘  𝑑→0 , 1 𝑑 β†’βˆž βˆ’1+ 1 𝑑 β†’βˆž as 𝑑→0 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 π’„π’π’π’—π’†π’“π’ˆπ’† b) π‘™π‘–π‘š 𝑑→0 0 𝑑 π‘₯ 4βˆ’ π‘₯ 2 𝑑π‘₯ = π‘™π‘–π‘š π‘‘β†’βˆž βˆ’ 4βˆ’ π‘₯ 2 𝑑 0 The function is undefined for π‘₯=2 = π‘™π‘–π‘š π‘‘β†’βˆž βˆ’ 4βˆ’ 𝑑 2 +βˆ’ 4βˆ’ 0 2 π‘Žπ‘  𝑑→2 , 4βˆ’ 𝑑 2 β†’0 = (0) = 2

6 NOW DO EX 3A WB A3 – both limits are infinity a) Find π‘₯ 𝑒 βˆ’ π‘₯ 2 𝑑π‘₯
b) Hence show that converges and find its value βˆ’βˆž ∞ π‘₯ 𝑒 βˆ’ π‘₯ 2 𝑑π‘₯ βˆ’βˆž ∞ π‘₯ 𝑒 βˆ’ π‘₯ 2 𝑑π‘₯ =βˆ’ 1 2 𝑒 βˆ’ π‘₯ 2 +C b) βˆ’βˆž ∞ π‘₯ 𝑒 βˆ’ π‘₯ 2 𝑑π‘₯ = βˆ’βˆž 0 π‘₯ 𝑒 βˆ’ π‘₯ 2 𝑑π‘₯ + 0 ∞ π‘₯ 𝑒 βˆ’ π‘₯ 2 𝑑π‘₯ Split into two improper integrals = π‘™π‘–π‘š π‘‘β†’βˆž βˆ’ 1 2 𝑒 βˆ’ π‘₯ βˆ’π‘‘ π‘™π‘–π‘š π‘‘β†’βˆž βˆ’ 1 2 𝑒 βˆ’ π‘₯ 2 𝑑 0 = π‘™π‘–π‘š π‘‘β†’βˆž βˆ’ 1 2 βˆ’βˆ’ 1 2 𝑒 βˆ’ 𝑑 π‘™π‘–π‘š π‘‘β†’βˆž βˆ’ 1 2 𝑒 βˆ’ 𝑑 2 βˆ’βˆ’ 1 2 π‘Žπ‘  π‘‘β†’βˆž , 𝑒 βˆ’ 𝑑 2 β†’ both integrals converge = βˆ’ = 0 NOW DO EX 3A

7 One thing to improve is –
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 – One thing to improve is –

8 END


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