MATH128 Lecture 3: Tutorial-May30
16 Jun 2016
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Solution the integral has a discontinuity at x = 0. So we must split it up as: t 0 z t lim. 1 x4 dx + lim t 0+z 3 t. By replace t by t we may change the direction of approach in the rst limit: 2 t3 = this integral does not converge. Problem find the values of p for which the integral converges: Solution again we are faced with a discontinuity at 0. So our integral becomes t 0+z 1 lim t. Hence we observe two ways in which this could be unde ned. This happens when p 1 > 0 aka when p > 1. Thus the integral is divergent when p 1 and hence the integral converges for values of p < 1. lim t 0+ Solution we will split the integral up at 0 for convenience: Which by de nition is equal to t z 0 lim t xdx + lim t z t.
