Estimate \int_1^4 {1\over x} dx using n = 6 rectangles and
a. the right hand rule.
b. the left hand rule.
c. the average of a and b.

Solution. The width of each interval is \Delta x = {{b-a}\over n}= {1\over 2} … so we have
a. {1\over 2} ({2\over 3}+{2\over 4}+{2\over 5}+{2\over 6}+{2\over 7}+{2\over 8})= {{341}\over{280}}.
b. {1\over 2} ({2\over 2}+{2\over 3}+{2\over 4}+{2\over 5}+{2\over 6}+{2\over 7})= {{223}\over{140}}.
Note that the terms of each sum are the reciprocals of the endpoints 1 (={2\over 2}), {3\over 2}, ...,{8\over2}; this is of course because our integrand (1\over x) is the reciprocal function. Finally, since the integral is “obviously” equal to ln(4) \sim 1.386, and {{787}\over{560}} \sim 1.405, our answer in c is reasonable.


  1. this is still way too much trouble
    since, as far as i’ve been able to
    discover, you can’t just TeX stuff
    a page at a time.

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