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Setting up the limit of Riemann sums for a function with a corner

Consider the function f(x)=∣x−2∣f(x) = |x-2| on the interval [0,4][0,4]. Note that ff has a corner at x=2x=2, where it switches from decreasing to increasing, but ff is still continuous on [0,4][0,4], so the definite integral still exists and can be written as a limit of Riemann sums.

Suppose [0,4][0,4] is divided into nn subintervals of equal width, and xi∗x_i^* denotes a sample point in the ii-th subinterval. Write the expression lim⁡n→∞∑i=1nf(xi∗) Δx\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta x explicitly for f(x)=∣x−2∣f(x) = |x-2|, giving Δx\Delta x in terms of nn and f(xi∗)f(x_i^*) explicitly in terms of xi∗x_i^*. Then explain in a sentence why the presence of the corner at x=2x=2 does not prevent this limit from existing and equaling ∫04∣x−2∣ dx\int_0^4 |x-2|\,dx.

Hint

First find Δx=(b−a)/n\Delta x = (b-a)/n for this interval. Then substitute f(xi∗)=∣xi∗−2∣f(x_i^*) = |x_i^* - 2| directly into the sum — you don't need to split into cases. For the explanation, think about what property of ff (not differentiability) is actually required for the Riemann sum limit to exist.

Step-by-step solution

Since [0,4][0,4] is split into nn equal subintervals, Δx=4−0n=4n\Delta x = \dfrac{4-0}{n} = \dfrac{4}{n}. Substituting f(xi∗)=∣xi∗−2∣f(x_i^*) = |x_i^* - 2| into the general Riemann sum formula gives lim⁡n→∞∑i=1n∣xi∗−2∣⋅4n.\lim_{n\to\infty}\sum_{i=1}^{n} |x_i^*-2|\cdot \frac{4}{n}. This expression represents the exact area under the graph of ff on [0,4][0,4], including both pieces where the graph slopes down to (2,0)(2,0) and back up.

The corner at x=2x=2 is not an issue because the existence of the Riemann sum limit only requires ff to be continuous (or at least bounded with finitely many discontinuities) on [a,b][a,b] — differentiability is not required. Since ∣x−2∣|x-2| is continuous everywhere, including at x=2x=2 where the two linear pieces meet, the limit defining ∫04∣x−2∣ dx\int_0^4 |x-2|\,dx exists regardless of the sharp corner in the graph.

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