AP Calculus AB and BC

Overestimate vs Underestimate

Left and right sums are decided by whether the function increases or decreases; trapezoidal and midpoint sums are decided by concavity instead. On a concave up curve the trapezoidal rule overestimates and the midpoint rule underestimates, whichever direction the function happens to be heading.

Overestimate

Use when: The approximating shapes cover more area than the region under the curve: a right sum on an increasing function, a left sum on a decreasing one, a trapezoidal sum on a concave up curve, or a midpoint sum on a concave down one, where the tangent line at each midpoint lies above the curve.

Underestimate

Use when: The shapes cover less area than the region under the curve: a left sum on an increasing function, a right sum on a decreasing one, a midpoint sum on a concave up curve, where the tangent line at each midpoint lies below it, or a trapezoidal sum on a concave down one.

Side by side

	Overestimate	Underestimate
Left Riemann sum	$f$ decreasing	$f$ increasing
Right Riemann sum	$f$ increasing	$f$ decreasing
Trapezoidal rule	$f$ concave up	$f$ concave down
Midpoint rule	$f$ concave down	$f$ concave up
Example: $f(x) = x^2$ on $[0,2]$	The right sum and the trapezoidal sum	The left sum and the midpoint sum

Left and right sums are settled by direction of travel. On an increasing function the left endpoint holds the smallest value in each subinterval, so every rectangle sits under the graph and the left sum underestimates, while the right endpoint holds the largest and the right sum overestimates. On a decreasing function the two verdicts swap. Concavity never enters this argument.

Trapezoids and midpoints are settled by concavity. A trapezoid joins the two endpoints with a straight chord, and on a concave up arc that chord lies above the curve, so the trapezoidal rule overestimates. The midpoint rectangle crosses the curve rather than sitting above or below it, so the argument runs through area instead: that rectangle has the same area as the trapezoid built on the tangent line at the midpoint, and a concave up curve lies above its tangent, so the midpoint rule underestimates. Concave down reverses both.

L_2 = 1 \;<\; M_2 = 2.5 \;<\; \int_0^2 x^2\,dx = \frac{8}{3} \;<\; T_2 = 3 \;<\; R_2 = 5

Where the guessing starts

The left and right rules get memorised, and then trapezoids and midpoints get judged by the same test. That fails. On $f(x) = x^2$ over $[0,2]$ with two subintervals the function is increasing and concave up, and the two sums land on opposite sides of $\frac{8}{3}$ : the midpoint sum below at $2.5$ , the trapezoidal sum above at $3$ . For those two rules check concavity, never monotonicity.

Frequently asked questions

Does a left Riemann sum always underestimate?

It is guaranteed to underestimate only when the function is increasing across the whole interval. If it decreases the left sum overestimates, and if the direction changes partway monotonicity gives no verdict either way: the sum may still land on either side, but you cannot justify a choice from monotonicity alone.

Why does the trapezoidal rule overestimate a concave up function?

Because each trapezoid is capped by the chord between the two endpoints, and a concave up graph bends below its chords. The trapezoid therefore covers the region under the curve plus a sliver between chord and curve on every subinterval.

How do I justify an overestimate for full credit?

Name the property and tie it to the shapes. For a right sum, state that $f$ is increasing on the interval so each rectangle rises above the curve. For a trapezoidal sum, state that $f$ is concave up so each chord lies above the curve. A bare claim of overestimate with no reason earns no justification point.

In the CED: Unit 6: Integration and Accumulation