269 - 2 a) In order to bring (1/n) outside the sum, we put n in the numerator; Then dividing numerator and denominator by n yields (1/n) sum 1/(1+(k/n)) which approaches int 1/(1+x) dx b) Put (1/n) outside and n (which makes kn) in the numerator; divide numerator and denominator by n^2 which leaves (1/n) sum (k/n)/(1+(k/n)^2) which approaches int x/(1+x^2) dx 6 The key observation is that the height of a rectangle is .5(x_k + x_(k-1)) and the coresponding base is (x_k - x_(k-1)) multiplying produces the area of a rectangle is .5((x_k)^2 - (x_(k-1))^2). summing ffom k=1 to n telescopes to .5((x_n)^2 - (x_0)^2) letting x_0 = 0 and x_n = b provides the result. (we may choose a sequence of partitions with the mesh going to 0) 12 a) improper since log(x) goes to -infinity as x goes to 0 convergent since x ln(x) goes to 0. value = -1. b) improper since log(1) =0, hence 1/(x log(x) ) blows up diverges since log(log(x)) goes to -infinity as x goes to 1 c) improper since i/sqrt(x-1) blows up as x goes to 1; convergent since behaves as 1/sqrt(x) at 0. integral = (2/3)(2+x)sqrt(x-1); value = 8/3 d) proper since xlog(x) goes to 0 as x goes to 0. integral = (x^2)/2 log(x) - (x^2)/4; value = -.25

Sun Dec 14 1997