Midpoint Riemann sum for f(x) = |x - 2| with n = 4

With $n = 4$ subintervals on $[0,4]$, the width is $\Delta x = \frac{4-0}{4} = 1$. The subintervals are $[0,1], [1,2], [2,3], [3,4]$, with midpoints $0.5, 1.5, 2.5, 3.5$ respectively.

Evaluating $f(x) = |x-2|$ at each midpoint gives $f(0.5) = 1.5$, $f(1.5) = 0.5$, $f(2.5) = 0.5$, $f(3.5) = 1.5$. The midpoint sum is $M_4 = \Delta x \left[f(0.5) + f(1.5) + f(2.5) + f(3.5)\right] = 1 \cdot (1.5 + 0.5 + 0.5 + 1.5) = 4$.

Yes. The midpoints zero point five and three point five both gave height one point five, and the midpoints one point five and two point five both gave height zero point five. This happens because the graph of absolute value of x minus two is symmetric about x equals two, so midpoints equally spaced from the vertex produce matching heights.