Methods in calculus.

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Presentation transcript:

Methods in calculus

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 5𝑥 3+ 𝑥 2 𝑑𝑥 =5 3+ 𝑥 2 +𝐶 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 −2𝑥+2 𝑒 𝑥 +𝐶 sin 𝑥 cos 𝑥 1+3 𝑠𝑖𝑛 2 𝑥 𝑑𝑥 =𝑙𝑛 1+3 𝑠𝑖𝑛 2 𝑥 +𝐶

The integral 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥 is improper if Notes 𝐴𝑟𝑒𝑎= 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥 A definite integral represents the area enclosed by a continuous function 𝑦=𝑓(𝑥), the x-axis and line 𝑥=𝑎 and 𝑥=𝑏 𝑜 𝑦 𝑥 𝑦= 1 𝑥 2 𝑎 (1) 𝐴𝑟𝑒𝑎= 𝑎 ∞ 1 𝑥 2 𝑑𝑥 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 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥 is improper if one or both of the limits is infinite f(x) is undefined at 𝑥=𝑎 or 𝑥=𝑏 or at a point in the interval [a, b] 𝑜 𝑦 𝑥 𝑦= 1 𝑥 3 (2) 𝐴𝑟𝑒𝑎= 0 3 1 𝑥 𝑑𝑥 If the improper integral exists it is said to be convergent. If it does not exist it is said to be divergent

WB A1- limits of inifinity Evaluate each improper integral 𝑎) 1 ∞ 1 𝑥 2 𝑑𝑥 𝑏) 1 ∞ 1 𝑥 𝑑𝑥 Use limit notation to rewrite the integral as 𝑙𝑖𝑚 𝑡→∞ 1 𝑡 1 𝑥 2 𝑑𝑥 = 𝑙𝑖𝑚 𝑡→∞ − 1 𝑥 𝑡 1 = 𝑙𝑖𝑚 𝑡→∞ − 1 𝑡 +1 𝑎𝑠 𝑡→∞ , 1 𝑡 →0 =𝟏 b) 𝑙𝑖𝑚 𝑡→∞ 1 𝑡 1 𝑥 𝑑𝑥 = 𝑙𝑖𝑚 𝑡→∞ ln 𝑥 𝑡 1 = 𝑙𝑖𝑚 𝑡→∞ ln 𝑡 − ln 1 𝑎𝑠 𝑡→∞ , ln 𝑡 →∞ 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆

WB A2 - undefined limits Evaluate each improper integral 𝑎) 1 0 1 𝑥 2 𝑑𝑥 𝑏) 0 2 𝑥 4− 𝑥 2 𝑑𝑥 Use limit notation to rewrite the integral as 𝑙𝑖𝑚 𝑡→0 𝑡 1 1 𝑥 2 𝑑𝑥 = 𝑙𝑖𝑚 𝑡→0 − 1 𝑥 1 𝑡 = 𝑙𝑖𝑚 𝑡→0 −1+ 1 𝑡 𝑎𝑠 𝑡→0 , 1 𝑡 →∞ −1+ 1 𝑡 →∞ as 𝑡→0 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆 b) 𝑙𝑖𝑚 𝑡→0 0 𝑡 𝑥 4− 𝑥 2 𝑑𝑥 = 𝑙𝑖𝑚 𝑡→∞ − 4− 𝑥 2 𝑡 0 The function is undefined for 𝑥=2 = 𝑙𝑖𝑚 𝑡→∞ − 4− 𝑡 2 +− 4− 0 2 𝑎𝑠 𝑡→2 , 4− 𝑡 2 →0 = (0)+ 4 = 2

NOW DO EX 3A WB A3 – both limits are infinity a) Find 𝑥 𝑒 − 𝑥 2 𝑑𝑥 b) Hence show that converges and find its value −∞ ∞ 𝑥 𝑒 − 𝑥 2 𝑑𝑥 −∞ ∞ 𝑥 𝑒 − 𝑥 2 𝑑𝑥 =− 1 2 𝑒 − 𝑥 2 +C b) −∞ ∞ 𝑥 𝑒 − 𝑥 2 𝑑𝑥 = −∞ 0 𝑥 𝑒 − 𝑥 2 𝑑𝑥 + 0 ∞ 𝑥 𝑒 − 𝑥 2 𝑑𝑥 Split into two improper integrals = 𝑙𝑖𝑚 𝑡→∞ − 1 2 𝑒 − 𝑥 2 0 −𝑡 + 𝑙𝑖𝑚 𝑡→∞ − 1 2 𝑒 − 𝑥 2 𝑡 0 = 𝑙𝑖𝑚 𝑡→∞ − 1 2 −− 1 2 𝑒 − 𝑡 2 + 𝑙𝑖𝑚 𝑡→∞ − 1 2 𝑒 − 𝑡 2 −− 1 2 𝑎𝑠 𝑡→∞ , 𝑒 − 𝑡 2 →0 both integrals converge = − 1 2 + 1 2 = 0 NOW DO EX 3A

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 –

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