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20-Improper Integrals

# 20-Improper Integrals - Improper Integrals John E Gilbert...

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Improper Integrals John E. Gilbert, Heather Van Ligten, and Benni Goetz If something is improper about integration, we’d better find out what’s proper! You’re already familiar with using definite integrals and the Fundamental Theorem of Calculus to find the area under the graphs as in 1 2 3 4 5 1 x y f ( x ) = 1 x 2 A 1 area ( A 1 ) = 4 1 1 x 2 dx = 3 4 . 2 1 x y 1 4 1 2 3 4 1 f ( x ) = 1 x A 2 area ( A 2 ) = 1 1 4 1 x dx = 1 . In both cases the interval of integration is finite and does not contain a vertical asymptote of f . The area is always finite - things are behaving well: call these Proper Integrals . If they are not, they are Improper !

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Definition: an Improper Integral is one EITHER of the form I 1 = I f ( x ) dx where I = [ a, ) , ( -∞ , b ] , or ( -∞ , ) , OR I 2 = b a g ( x ) dx where g ( x ) has a vertical asymptote in [ a, b ] . As we shall soon see, ones of type I 1 like Int 1 = 1 1 x 2 dx occur when dealing with infinite series, and learning to deal with integrals over an infinite interval is a good ‘warm-up’ for studying infinite series. They are important too in probability and in many applications as you’ll see in later courses. Integrals of type I 2 like Int 2 = 1 0 1 x dx often occur also - some functions blow up! It’s easy to make Int 1 ‘proper’: cut o ff integration at a large t . Since t 1 1 x 2 dx = - 1 x t 1 = 1 - 1 t is always finite, we can think of the area under the graph of 1 /x 2 as being finite on all of [1 , ) in the sense that Int 1 = 1 1 x 2 dx = lim t → ∞ t 1 1 x 2 dx = lim t → ∞ 1 - 1 t = 1 . But the area is not finite for all f ( x ) on [1 , ) ; this is the crucial point of improper integrals! Indeed, lim t → ∞ t 1 1 x dx = ln t t 1 = lim t → ∞ ln t = , so the area under the graph is infinite on [1 , ) when f ( x ) = 1 /x .
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