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I know that in a positive and increasing function, the right riemann sum is an overestimate and the left is an underestimate, but what about if the function is negative and increasing like this? Which one would be an overestimate and underestimate?

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  • $\begingroup$ Did you calculate the corresponding sums? You should be able to see which is bigger. $\endgroup$ – Paul K Jan 20 '18 at 19:01
  • $\begingroup$ Wait so the one that is bigger would be an overestimate for this table? $\endgroup$ – deezy Jan 20 '18 at 19:07
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    $\begingroup$ Also, conceptually why would it be an overestimate or underestimate? $\endgroup$ – deezy Jan 20 '18 at 19:23
  • $\begingroup$ To get an idea what happens you could draw a graph and try to understand what the left/right riemann sum actually are. $\endgroup$ – Paul K Jan 20 '18 at 19:38
  • $\begingroup$ But the points don't really connect all that well though. $\endgroup$ – deezy Jan 20 '18 at 19:43
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It makes no difference whether the values of a function are positive or negative, if you always choose the smallest value of the function on each interval, the Riemann sum will be an underestimate. If you choose the largest value of the function on each interval, you will get an overestimate:

$$\sum_i \left(\min_{t_{i-1} \le t \le t_i} f(x)\right)\Delta t_i \le \int_a^b f(t)\,dt \le \sum_i \left(\max_{t_{i-1} \le t \le t_i} f(x)\right)\Delta t_i $$

If $f$ is increasing, then its minimum will always occur on the left side of each interval, and its maximum will always occur on the right side of each interval. So for increasing functions, the left Riemann sum is always an underestimate and the right Riemann sum is always an overestimate.

If $f$ is decreasing, this is reversed.

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  • $\begingroup$ So the right sum would be an overestimate because the function is increasing? $\endgroup$ – deezy Jan 21 '18 at 5:24
  • $\begingroup$ Yes, because $f$ is increasing, it takes on its maximum on the right side of each interval, so it will be greater than or equal to any other Riemann sum on the same partition (or any refinement of that partition). $\endgroup$ – Paul Sinclair Jan 21 '18 at 6:00

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