Rectangles first, and they lean a predictable way AB & BC

To approximate the area under $f$ on $[a, b]$ with $n$ rectangles of equal width, cut the interval into $n$ equal pieces. The width of each is

$$\Delta x = \frac{b - a}{n},$$

and the cut points are $x_0 = a$, $x_1 = a + \Delta x$, up to $x_n = b$. Count them: there are $n$ subintervals and $n + 1$ grid points. Almost every setup error in this topic is an off-by-one between those two numbers.

1. $\Delta x$ divides by $n$, not by $n - 1$. Six subintervals on $[2, 14]$ have width $\frac{12}{6} = 2$. Dividing by $5$ gives $2.4$ and pushes every sample point off the grid.
2. A sum has $n$ terms, not $n + 1$. One rectangle per subinterval, so one height per subinterval.
3. Each rectangle takes its height from one sample point in its own subinterval.

Which sample point is what names the sum. On $[2, 14]$ with $n = 6$ the grid is $2, 4, 6, 8, 10, 12, 14$, and:

1. The left sum uses $2, 4, 6, 8, 10, 12$: every grid point except the last.
2. The right sum uses $4, 6, 8, 10, 12, 14$: every grid point except the first.
3. The midpoint sum uses $3, 5, 7, 9, 11, 13$, none of which is a grid point at all.

Then the sum is $\Delta x$ times the total of the chosen heights, since every rectangle has the same width. Forgetting to multiply by $\Delta x$ leaves a total of heights, which is not an area and generally not even close to one.

Take $f$ increasing on a subinterval. The left endpoint is where $f$ is smallest there, so a rectangle at that height fits entirely under the curve and the left sum comes out low. The right endpoint is where $f$ is largest, so that rectangle pokes out above and the right sum comes out high.

Now take $f$ decreasing. Every one of those sentences flips. The left endpoint is now the high side, so the left sum overestimates, and the right sum underestimates.

$$f \text{ increasing} \implies L_n \le \text{area} \le R_n, \qquad f \text{ decreasing} \implies R_n \le \text{area} \le L_n.$$

So "left sums underestimate" is not a rule. It is one half of a rule with its condition removed, and it is wrong exactly half the time. The working question is is $f$ increasing on this interval, and the verdict follows from the answer.

Two boundaries on the claim are worth stating.

1. If $f$ changes direction inside the interval, neither verdict holds. Part of the sum leans one way and part the other, and the totals can land on either side. Nothing about the graph settles it without more work.
2. Concavity does not decide this. Concavity is what governs trapezoids. Using it to rule on a left or right sum is answering with the wrong instrument, and it will be right by luck about half the time.

A trapezoid joins the two endpoint heights with a straight segment instead of a flat one, so it hugs the curve much more closely. Its area on one subinterval is the average of the two heights times the width, and adding those up gives

$$T_n = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right].$$

Every interior height is counted twice because it belongs to two trapezoids, and the two ends are counted once. Equivalently, $T_n$ is the average of $L_n$ and $R_n$, which is a faster way to get it when both are already computed.

The direction of the trapezoid's error has nothing to do with whether $f$ is increasing. The trapezoid's top is the chord joining two points on the curve, and where a chord sits relative to a curve is exactly what concavity means:

1. $f$ concave up: the chord lies above the curve, so $T_n$ overestimates.
2. $f$ concave down: the chord lies below the curve, so $T_n$ underestimates.

"Trapezoids always overestimate" fails for the same reason its rectangle counterpart does: it is one branch of a two-branch rule. And if the concavity changes inside the interval, there is no verdict at all.

The midpoint rule leans the opposite way to the trapezoid, and by about half as much. On a concave-up function the midpoint rectangle's missing corner on one side is more than repaid by its extra on the other, and $M_n$ comes out low while $T_n$ comes out high, with the true value between them. That sandwich is useful: it bounds the answer without ever computing it.

AP free-response questions hand over a table of measured values at times chosen by whoever took the measurements, and those times are almost never evenly spaced. There is no $\Delta x$. There is a $\Delta x$ per piece.

Everything else survives intact. For a left sum, each rectangle uses the value at the left end of its own subinterval and its own width. For trapezoids, each piece is the average of its two endpoint values times its own width. With readings at $t = 0, 2, 5, 6, 10$ the four widths are $2$, $3$, $1$ and $4$, and each term carries the one it belongs to.

Three habits that keep this clean:

1. Write the widths down first, as their own row, before any value is used. They should sum to the length of the whole interval, which is a free check.
2. Never average the readings and multiply by the total time. That silently assumes every reading covered an equal stretch, and here they do not.
3. Do not compute a single $\frac{b-a}{n}$. With five readings and four unequal gaps, $\frac{10}{4} = 2.5$ describes none of them.

The over-or-under question also gets harder with a table, because a handful of readings rarely establishes that $f$ is monotonic or that its concavity keeps one sign. If the question wants a direction, it will usually tell you the function is increasing or concave up. If it does not say, and the data does not force it, the correct response is that the direction cannot be determined.