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Methods in calculus
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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 ππ₯ =5 3+ π₯ 2 +πΆ π₯ 2 π π₯ ππ₯ = π₯ 2 β2π₯+2 π π₯ +πΆ sin π₯ cos π₯ 1+3 π ππ 2 π₯ ππ₯ =ππ 1+3 π ππ 2 π₯ +πΆ
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The integral π π π(π₯) ππ₯ is improper if
Notes π΄πππ= π π π(π₯) ππ₯ A definite integral represents the area enclosed by a continuous function π¦=π(π₯), the x-axis and line π₯=π and π₯=π π π¦ π₯ π¦= 1 π₯ 2 π (1) π΄πππ= π β 1 π₯ 2 ππ₯ 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 π π(π₯) ππ₯ is improper if one or both of the limits is infinite f(x) is undefined at π₯=π or π₯=π or at a point in the interval [a, b] π π¦ π₯ π¦= 1 π₯ 3 (2) π΄πππ= π₯ ππ₯ If the improper integral exists it is said to be convergent. If it does not exist it is said to be divergent
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WB A1- limits of inifinity Evaluate each improper integral
π) 1 β 1 π₯ 2 ππ₯ π) 1 β 1 π₯ ππ₯ Use limit notation to rewrite the integral as πππ π‘ββ 1 π‘ 1 π₯ 2 ππ₯ = πππ π‘ββ β 1 π₯ π‘ 1 = πππ π‘ββ β 1 π‘ +1 ππ π‘ββ , 1 π‘ β0 =π b) πππ π‘ββ 1 π‘ 1 π₯ ππ₯ = πππ π‘ββ ln π₯ π‘ 1 = πππ π‘ββ ln π‘ β ln 1 ππ π‘ββ , ln π‘ ββ π
πππ πππ ππππππππ
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WB A2 - undefined limits Evaluate each improper integral
π) π₯ 2 ππ₯ π) 0 2 π₯ 4β π₯ 2 ππ₯ Use limit notation to rewrite the integral as πππ π‘β0 π‘ π₯ 2 ππ₯ = πππ π‘β0 β 1 π₯ 1 π‘ = πππ π‘β0 β1+ 1 π‘ ππ π‘β0 , 1 π‘ ββ β1+ 1 π‘ ββ as π‘β0 π
πππ πππ ππππππππ b) πππ π‘β0 0 π‘ π₯ 4β π₯ 2 ππ₯ = πππ π‘ββ β 4β π₯ 2 π‘ 0 The function is undefined for π₯=2 = πππ π‘ββ β 4β π‘ 2 +β 4β 0 2 ππ π‘β2 , 4β π‘ 2 β0 = (0) = 2
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NOW DO EX 3A WB A3 β both limits are infinity a) Find π₯ π β π₯ 2 ππ₯
b) Hence show that converges and find its value ββ β π₯ π β π₯ 2 ππ₯ ββ β π₯ π β π₯ 2 ππ₯ =β 1 2 π β π₯ 2 +C b) ββ β π₯ π β π₯ 2 ππ₯ = ββ 0 π₯ π β π₯ 2 ππ₯ + 0 β π₯ π β π₯ 2 ππ₯ Split into two improper integrals = πππ π‘ββ β 1 2 π β π₯ βπ‘ πππ π‘ββ β 1 2 π β π₯ 2 π‘ 0 = πππ π‘ββ β 1 2 ββ 1 2 π β π‘ πππ π‘ββ β 1 2 π β π‘ 2 ββ 1 2 ππ π‘ββ , π β π‘ 2 β both integrals converge = β = 0 NOW DO EX 3A
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One thing to improve is β
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 β One thing to improve is β
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