Lecture_4 - CALCULUS II Spring 2009 LECTURE 4 This lecture...

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CALCULUS II Spring 2009 LECTURE 4 This lecture provides eight examples of improper integrals. 1
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Recall We recall from Lecture 3 every polynomial can (theoretically) be written as a product of linear and quadratic factors integration of rational functions can be effected by partial fraction decomposition of the integrand End of Recall 2
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Mathematics Interlude I Improper Integrals We recall briefly the definition of the Riemann Integral. Definition . Let f denote a real-valued function defined on the closed and bounded interval [ a,b ]. Let L denote a real number such that for each ± > 0 there exists δ > 0 such that if { x i } n i =0 is a partition of [ a,b ] with max 1 i n Δ x i < δ , then ± ± ± ± ± n X i =1 f ( ξ i x i - L ± ± ± ± ± < ± for every choice of the ξ i ’s constrained by x i - 1 ξ i x i , 1 i n . The number L is called the integral of f , or the Riemann integral of f . One writes the integral of f by Z b a f ( x ) dx. If a real-valued function defined on [ a,b ] has a Riemann integral then it is said to be integrable or Riemann integrable . Most functions that we deal with are Riemann integrable, but not all. The function f ( x ) = ( 1 x 0 < x 1 0 x = 0 is not Riemann integrable. But it has an integral in another sense, called an improper integral. (This is our first example below.) Definition . For a real-valued function, f , defined on [ a,b ], a Riemann sum is an expression of the form n X i =1 f ( ξ i x i 3
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n is a positive integer, { x i } n i =0 is a partition of [ a,b ], x i - 1 ξ i x i , 1 i n and Δ x i = x i - x i - 1 , 1 i n . Roughly speaking, one says that the integral of f is the limit, as n → ∞ , of its Riemann sums. Abusing our definition of a limit we write Z b a f ( x ) dx = lim maxΔ x i 0 n X i =1 f ( ξ i x i . The next example is not that of a Riemann integral but of an improper Rie- mann integral . The integrand, 1 x , is not Riemann integrable on [0 , 1], since it is not bounded there. So an additional limiting process is imposed. We work through the example and then look back to see what happened. EXAMPLE. Evaluate the improper integral Z 1 0 1 x dx. It is understood that the function 1 x is not defined at x = 0. But as is seen below, the expression R 1 0 1 x dx is still meaningful. Solution. One writes Z 1 0 1 x dx = lim t 0 + Z 1 t t - 1 2 dx = lim t 0 + 2 x 1 2 ± ± ± ± 1 t = lim t 0 + ² 2 - 2 t 1 2 ³ = 2 - 2 lim t 0 + t 1 2 = 2 . Thus, the value of the improper integral is given by
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Lecture_4 - CALCULUS II Spring 2009 LECTURE 4 This lecture...

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