1 x Solution 1 a) The integrand x  ] has a singularity at x = 0 in the interval 1 Hence 1 x is an improper integral of the second kind. 2 2 1 1 1 1 Clearly dx   dx dx b) 1 1 x 0 1 1 x x x dx and 0 0 0 2 x 2 both diverge; 90 0 1 lnx  dx 1  ln 0 ln  ln 0  does not 1 1 x 2 1 x ln 2 ln( 0 )  ln(0) does not exist. 0 lnx 02 dx  1 Hence 1 x does not exist. 2 dx  1 However 1 1 x dx exist when 0 ε 2 x -  ln ln   ln   1 lnx    1 Since dx 1 1 x  and 1   ln 2 ln lnx 2 dx  x 2 1 dx  Hence 1 x 1 ln  ln 2 ln    dx 2 x - 
1 x Solution a) The integrand 1 x has a singularity at x = 0 in the interval  ] Hence 2 1 1 x is an improper integral of the second kind. b) 2 1 1 x dx  0 1 1 x dx  2 0 1 x dx Clearly 0 1 1 x and 2 0 1 x dx both diverge; 90 0 1 1 x dx  lnx 1  ln 0 ln 1   ln 0 does not 2 0 1 x dx  lnx 02  ln 2 ln( 0 ) ln(0) does not exist. Hence 2 1 1 x does not exist. However -  1 1 x dx  2 1 x dx exist when ε 0 Since  1 1 x dx  lnx 1  ln   ln 1   ln  and 2 1 x dx  lnx 2  ln 2 ln  Hence -  1 1 x dx  2 1 x dx  ln   ln 2 ln 
CALCULUS%20II.DOCX