I'm having trouble with theoretical understanding of the Riemann sum with this explanation/definition from Thomas' Calculus. I checked Wikipedia and it seems to state virtually the same.:

On each subinterval we form the product $f(c_k)*∆x_k$. This product is positive, negative, or zero, depending on the sign of $f(c_k)$. When $f(c_k) > 0$, the product $f(c_k)*∆x_k$ is the area of a rectangle with height $f(c_k)$ and width $∆x_k$. When $f(c_k) < 0$, the product $f(c_k)*∆x_k$ is a negative number, the negative of the area of a rectangle of width $∆x_k$ that drops from the x axis to the negative number $ƒ(c_k)$.

Finally we sum all these products to get:

$$ S_p = \sum_{k=1}^{n}{f(c_k)}∆x_k $$

Any Riemann sum associated with a partition of a closed interval [a, b] defines rectangles that approximate the region between the graph of a continuous function ƒ and the x-axis. Partitions with norm approaching zero lead to collections of rectangles that approximate this region with increasing accuracy

To illustrate the problem, suppose we want to approximate the area between $f(x) = -x$ and the x axis on the interval [-1; 1]. The area is 1, but the Riemann sum should give something close to 0:

Riemann sum example

Is the statement that any Riemann sum with the norm approaching 0 approximates the area with increasing accuracy correct? It seems not, since in the example above the area tends to 0 as the norm approaches 0, which is not "increasing accuracy". Does it miss the part that one should take the absolute values of the rectangles' areas?

Thank you.

  • $\begingroup$ the Riemann sum approaches the signed area, which is $-\frac12+\frac12=0$ in your example; cf. the Wikipedia integral page $\endgroup$ May 24, 2020 at 23:17
  • $\begingroup$ @J.W.Tanner what is a signed area? Negative? I looked into an online app to calculate sums and take the screen shot from here. And it says the area is approximated to 0.16 (with my settings). So it's not like it took only the red or green shaded part $\endgroup$ May 24, 2020 at 23:21
  • $\begingroup$ signed area in your picture would be the green area minus the red area $\endgroup$ May 24, 2020 at 23:22
  • $\begingroup$ @J.W.Tanner so the definition is wrong. Since it doesn't always increase accuracy as one take more fine rectangles. In my example it actually decreases accuracy, since the area tends to 0. Am I right? $\endgroup$ May 24, 2020 at 23:24
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    $\begingroup$ The true value is 0 here, which is the signed area mentioned by J. W. Tanner. I wouldn't say the book is wrong. It's just too much trouble to add "signed" every time you write something. And the reason for considering signed areas should be straightforward. $\endgroup$
    – trisct
    May 24, 2020 at 23:31

1 Answer 1


The Riemann sum approaches the signed area. In your picture, the green area is positive, and the red area is negative. The Riemann sum should approach $0$, which is the accurate signed area for $f(x)=-x$ on the interval $[-1,1]$. If you don't like that, try $f(x)=|x|$.

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    $\begingroup$ Thank you, but I'd like to make a stress, in case anyone else searches for this, the Riemann sum does not approximate the area of the region by definition it approximates the signed area $\endgroup$ May 24, 2020 at 23:38

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