that the improper integral 1 1 x dx is convergent or divergent. 1 dx b). Show that is divergent. 1 x Solution 1 b 1 converges dx dx a). If converges (or exists) we say that lim 1 x x b 1 otherwise we say that it diverges. = 1 lim b). dx 1 x b b ln x1b l 84 lim = ln b ln 1 b – lim = b = ln ln b since ln 1 = 0 1 Since b dx dose not exist, it diverges. 1 x Example 2 1 Show that the improper integral is convergent dx 1 x 2 Solution 1 exists or does not. b 1 dx x x 2 b 2 1 l b
that the improper integral 1 1 x dx is convergent or divergent. b). Show that 1 1 x dx is divergent. Solution a). If lim b b 1 1 x dx converges (or exists) we say that 1 1 x dx converges otherwise we say that it diverges. b). lim b l 1 1 x dx = b ln x1b 84 = lim b ln b – ln 1 lim b = ln = ln b since ln 1 = 0 Since b 1 1 x dx dose not exist, it diverges. Example 2 Show that the improper integral 1 1 x 2 dx is convergent Solution 1 1 x 2 b b 1 x 2 dx exists or does not. b l
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