Consider the integral - 2 1 (1/t) dt in Example 1. By dividing the interval 1 t 2 into
Chapter 5, Problem 45(choose chapter or problem)
Consider the integral - 2 1 (1/t) dt in Example 1. By dividing the interval 1 t 2 into 10 equal parts, we can show that 0.1 1 1.1 + 1 1.2 + ... + 1 2 , 2 1 1 t dt and , 2 1 1 t dt 0.1 1 1 + 1 1.1 + ... + 1 1.9 . (a) Now divide the interval 1 t 2 into n equal parts to show that +n r=1 1 n + r < , 2 1 1 t dt < +n1 r=0 1 n + r . (b) Show that the difference between the upper and lower sums in part (a) is 1/(2n). (c) The exact value of - 2 1 (1/t) dt is ln 2. How large should n be to approximate ln 2 with an error of at most 5 106, using one of the sums in part (a)?
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