Math Core

Derivatives answer the question "how fast is something changing?" This unit asks the reverse question: if you know how fast something changes, how much does it change in total? The first tool for answering that is the Riemann sum, which adds up many small rectangles to estimate the area under a rate curve.

Area under a rate curve is accumulated change

Suppose a car drives at a constant 60 miles per hour for 3 hours. It travels $60 \times 3 = 180$ miles. On a graph of velocity against time, that product is the area of a rectangle with height 60 and width 3.

The same idea works when the rate is not constant. If $r(t)$ is a rate of change, then the area between the graph of $r$ and the $t$ -axis over an interval equals the total change in the quantity over that interval. The units confirm it: the height has units like gallons per minute, the width has units of minutes, so the area has units of gallons.

The trouble is that curved regions don't have easy area formulas. So you slice the region into thin vertical strips, approximate each strip with a rectangle, and add up the rectangles.

Left and right Riemann sums

Split the interval $[a, b]$ into subintervals. On each one, build a rectangle whose width is the length of the subinterval and whose height is a value of the function somewhere on it.

A left Riemann sum uses the function value at the left endpoint of each subinterval.
A right Riemann sum uses the function value at the right endpoint.

Definition

Riemann sum

A Riemann sum for $f$ on $[a, b]$ is a sum of the form

f(x_1^*)\,\Delta x_1 + f(x_2^*)\,\Delta x_2 + \cdots + f(x_n^*)\,\Delta x_n,

where the interval is split into $n$ subintervals with widths $\Delta x_1, \ldots, \Delta x_n$ and each $x_k^*$ is a point chosen in the $k$ th subinterval. The sum estimates the signed area between the graph of $f$ and the $x$ -axis.

Here is a left Riemann sum for $f(x) = \dfrac{x^2}{2}$ on $[0, 4]$ with four subintervals of width 1. Each rectangle's height is the curve's height at the left edge.

Left Riemann sum for f(x) = x²/2 on [0, 4] with four rectangles. Each rectangle's height is f at its left edge, so every rectangle sits under the rising curve.Open in grapher →

Worked example: Left and right sums for a curve

Estimate the area under $f(x) = \dfrac{x^2}{2}$ on $[0, 4]$ using four equal subintervals, first with a left sum and then with a right sum.

Each subinterval has width $\Delta x = \dfrac{4 - 0}{4} = 1$ . The endpoints are $0, 1, 2, 3, 4$ .

Left sum (use $x = 0, 1, 2, 3$ ):

L_4 = 1\,[f(0) + f(1) + f(2) + f(3)] = 0 + 0.5 + 2 + 4.5 = 7.

Right sum (use $x = 1, 2, 3, 4$ ):

R_4 = 1\,[f(1) + f(2) + f(3) + f(4)] = 0.5 + 2 + 4.5 + 8 = 15.

The true area is between 7 and 15. Because $f$ is increasing on $[0, 4]$ , the left rectangles fall short and the right rectangles overshoot.

Midpoint and trapezoidal sums

Two other estimates are usually much more accurate.

A midpoint sum uses the function value at the midpoint of each subinterval.
A trapezoidal sum replaces each rectangle with a trapezoid whose top connects the two endpoint heights. One trapezoid over a subinterval of width $\Delta x$ has area $\dfrac{f(\text{left}) + f(\text{right})}{2}\,\Delta x$ .

For the same function, the midpoint sum is $1\,[f(0.5) + f(1.5) + f(2.5) + f(3.5)] = 0.125 + 1.125 + 3.125 + 6.125 = 10.5$ , and the trapezoidal sum is $\dfrac{0 + 0.5}{2} + \dfrac{0.5 + 2}{2} + \dfrac{2 + 4.5}{2} + \dfrac{4.5 + 8}{2} = 11$ . The exact area turns out to be $\dfrac{32}{3} \approx 10.67$ , so both are close.

Tip

Each trapezoid is the average of its left and right rectangles, so the trapezoidal sum is always the average of the left and right sums: $T = \dfrac{L + R}{2}$ . Here, $\dfrac{7 + 15}{2} = 11$ . This works even when the subintervals have different widths.

Riemann sums from a table

On the AP exam, the function is often given only as a table of values, and the subintervals are often unequal. Use the widths exactly as the table gives them, one subinterval at a time.

Worked example: Unequal subintervals from a table

Water flows into a tank at a rate of $r(t)$ gallons per minute. Selected values are shown.

$t$ (minutes)	0	2	5	9	10
$r(t)$ (gal/min)	12	15	20	18	14

Use a right Riemann sum and then a trapezoidal sum with the four subintervals in the table to estimate the total water that flows in from $t = 0$ to $t = 10$ .

The widths are $2, 3, 4, 1$ .

\begin{aligned} R &= 15(2) + 20(3) + 18(4) + 14(1) = 30 + 60 + 72 + 14 = 176 \text{ gallons} \\ L &= 12(2) + 15(3) + 20(4) + 18(1) = 24 + 45 + 80 + 18 = 167 \text{ gallons} \\ T &= \frac{176 + 167}{2} = 171.5 \text{ gallons} \end{aligned}

About 171.5 gallons of water flow into the tank during the 10 minutes.

Common mistake

Don't assume the subintervals are equal. With a table like the one above, multiplying every height by the same width gives a wrong answer. Find each width from consecutive $t$ -values.

Overestimate or underestimate?

AP questions often ask whether an approximation is too large or too small. You can tell from the shape of the graph without computing the exact area.

Over or under

If $f$ is increasing, a left sum underestimates and a right sum overestimates. If $f$ is decreasing, it's the reverse.
If the graph of $f$ is concave up, a trapezoidal sum overestimates (the straight tops lie above the curve) and a midpoint sum underestimates. If $f$ is concave down, it's the reverse.

Worked example: Deciding without computing

A function $g$ is positive, decreasing and concave up on $[1, 6]$ . Which of the left, right and trapezoidal sums overestimate the area under $g$ ?

Since $g$ is decreasing, the left endpoint of each subinterval is the highest point on it, so the left sum overestimates and the right sum underestimates. Since $g$ is concave up, every trapezoid's slanted top lies above the curve, so the trapezoidal sum overestimates too. The left and trapezoidal sums are overestimates; the right sum is an underestimate.

Always state your reasoning in terms of the function's behavior ("because $f$ is increasing on the interval..."). A correct conclusion without a reason earns little credit on free-response questions.

Practice

Practice 1

Let $f(x) = x^2 + 1$ . Find the right Riemann sum for $f$ on $[0, 3]$ using three subintervals of equal width.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

For the same function $f(x) = x^2 + 1$ on $[0, 3]$ with three equal subintervals, find the left Riemann sum.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Use a midpoint sum with two subintervals of equal width to estimate the area under $y = x^3$ on $[0, 4]$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

The velocity of a cyclist, in feet per second, is recorded at selected times.

$t$ (seconds)	0	4	6	10
$v(t)$ (ft/s)	0	8	14	20

Use a trapezoidal sum with the three subintervals in the table to estimate the distance, in feet, that the cyclist travels from $t = 0$ to $t = 10$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

A rate $r(t)$ is measured in liters per hour, with $t$ in hours. A right Riemann sum for $r$ on $0 \le t \le 8$ is computed. What does the sum approximate?

Practice 6

A function $f$ is positive, increasing and concave down on $[a, b]$ . Let $L$ , $R$ and $T$ be the left, right and trapezoidal sums with the same subintervals, and let $A$ be the exact area under $f$ . Which ordering is correct?

Practice 7

Estimate the area under $y = \dfrac{1}{x}$ on $[1, 3]$ using a right Riemann sum with four subintervals of equal width. Give an exact fraction.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.