Riemann Sums

When a rate graph is curved, you can’t find the area under it with triangles and rectangles exactly. But you can approximate it: slice the region into thin strips, approximate each strip with a rectangle or trapezoid, and add up the areas. These approximations are called Riemann sums, and they show up on almost every AP exam, often with a table of real data.

Key ideas

Subintervals

Split the interval $[a, b]$ into pieces called subintervals. If there are $n$ pieces of equal width, each has width

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

Subintervals don’t have to be equal. With data from a table, the widths are often different, so find each width separately.

Four kinds of sums

On each subinterval, build a shape and find its area:

Sum	Height of each rectangle (or shape)
Left	the function value at the left end of the subinterval
Right	the function value at the right end
Midpoint	the function value at the middle of the subinterval
Trapezoidal	a trapezoid joining the values at both ends

Each rectangle’s area is height × width. A trapezoid with parallel sides $f(x_{k-1})$ and $f(x_k)$ and width $\Delta x_k$ has area

$$\frac{f(x_{k-1}) + f(x_k)}{2} \cdot \Delta x_k .$$

The trapezoidal sum is always the average of the left and right sums.

Four approximations of the area under $f(x) = \tfrac{1}{4}x^2 + 1$ from $0$ to $4$. The exact area is $\tfrac{28}{3} \approx 9.333$.

Overestimate or underestimate?

For a function that is increasing on the interval:

• the left sum is an underestimate (each rectangle uses the smallest height on its strip);
• the right sum is an overestimate.

For a decreasing function, it’s the other way around: left over, right under.

Concavity tells you about trapezoids: if the graph is concave up, the trapezoids sit above the curve, so the trapezoidal sum is an overestimate (and the midpoint sum is an underestimate). If it’s concave down, the trapezoidal sum is an underestimate.

Units and context

A Riemann sum of a rate is an approximation of an accumulated change, so it has the same units as the area under a rate graph. With a table, you can only use the values you’re given: a midpoint sum needs a data value at the middle of each subinterval.

Worked examples

Example 1: Four sums from a formula

Approximate the area under $f(x) = \tfrac{1}{4}x^2 + 1$ from $x = 0$ to $x = 4$ using $n = 4$ equal subintervals, with left, right, midpoint, and trapezoidal sums.

Solution. $\Delta x = \tfrac{4 - 0}{4} = 1$. The function values you need are:

$x$	$0$	$0.5$	$1$	$1.5$	$2$	$2.5$	$3$	$3.5$	$4$
$f(x)$	$1$	$1.0625$	$1.25$	$1.5625$	$2$	$2.5625$	$3.25$	$4.0625$	$5$

$$\begin{aligned} L_4 &= 1\,(1 + 1.25 + 2 + 3.25) = 7.5 \\ R_4 &= 1\,(1.25 + 2 + 3.25 + 5) = 11.5 \\ M_4 &= 1\,(1.0625 + 1.5625 + 2.5625 + 4.0625) = 9.25 \\ T_4 &= \tfrac{1}{2}(L_4 + R_4) = \tfrac{1}{2}(7.5 + 11.5) = 9.5 \end{aligned}$$

Since $f$ is increasing on $[0, 4]$, $L_4$ is an underestimate and $R_4$ is an overestimate. The exact area turns out to be $\tfrac{28}{3} \approx 9.333$, which is between them.

Example 2: A table with uneven subintervals

Water flows into a tank at the rate $r(t)$ litres per minute. Some values are shown below.

$t$ (min)	$0$	$2$	$5$	$9$	$10$
$r(t)$ (L/min)	$4$	$7$	$9$	$6$	$5$

Use a right Riemann sum and a trapezoidal sum, with the four subintervals in the table, to approximate how much water flows in from $t = 0$ to $t = 10$.

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

Right sum (use the value at the right end of each subinterval):

$$2(7) + 3(9) + 4(6) + 1(5) = 14 + 27 + 24 + 5 = 70$$

Trapezoidal sum:

$$\begin{aligned} &2 \cdot \frac{4 + 7}{2} + 3 \cdot \frac{7 + 9}{2} + 4 \cdot \frac{9 + 6}{2} + 1 \cdot \frac{6 + 5}{2} \\ &= 11 + 24 + 30 + 5.5 = 70.5 \end{aligned}$$

About $70$ L (right sum) or $70.5$ L (trapezoidal sum) flows in. You can’t say whether these are over- or underestimates, because $r$ is neither always increasing nor always decreasing.

Example 3: Is it an overestimate?

Find the left Riemann sum for $f(x) = \dfrac{1}{x}$ on $[1, 3]$ with $n = 4$ equal subintervals. Is it an overestimate or an underestimate?

Solution. $\Delta x = \tfrac{3 - 1}{4} = 0.5$, and the left endpoints are $1$, $1.5$, $2$, $2.5$:

$$L_4 = 0.5\left(1 + \frac{2}{3} + \frac{1}{2} + \frac{2}{5}\right) = 0.5 \cdot \frac{77}{30} = \frac{77}{60} \approx 1.283$$

Since $f(x) = \tfrac{1}{x}$ is decreasing on $[1, 3]$, each left rectangle is as tall as the highest point of its strip. The left sum is an overestimate.

Example 4: A midpoint sum from a table

A runner’s velocity $v(t)$, in m/s, is recorded every $2$ seconds.

$t$ (s)	$0$	$2$	$4$	$6$	$8$
$v(t)$ (m/s)	$0$	$3$	$5$	$6$	$6.5$

Use a midpoint sum with two subintervals of equal width to approximate the distance the runner covers from $t = 0$ to $t = 8$.

Solution. Two equal subintervals are $[0, 4]$ and $[4, 8]$, each $4$ s wide. Their midpoints are $t = 2$ and $t = 6$, which are in the table:

$$M_2 = 4\,v(2) + 4\,v(6) = 4(3) + 4(6) = 36$$

The runner covers about $36$ m.

Common mistakes

Using the same width for uneven subintervals. In a table, check every gap. In Example 2 the widths are $2$, $3$, $4$, and $1$, not all $2.5$.

Using too many or too few function values. With $n$ subintervals, a left sum uses the first $n$ values and a right sum uses the last $n$. If you’ve added $n + 1$ heights, something is wrong.

Making up values for a midpoint sum. With a table, you can only use a midpoint sum when the table gives the value at each midpoint. Don’t average neighbouring values and call it a midpoint sum.

Mixing up over and under. Picture it: for an increasing function, right rectangles poke above the curve, so the right sum is too big. Decide using whether $f$ is increasing or decreasing on the whole interval; if it changes direction, you usually can’t tell.

Forgetting units. A sum of (L/min) × (min) is in litres. AP free-response answers need units.

Practice

1. (Warm-up) The interval $[2, 8]$ is split into $3$ equal subintervals. What is $\Delta x$, and what are the endpoints of the subintervals?

Solution

$\Delta x = \dfrac{8 - 2}{3} = 2$. The subintervals are $[2, 4]$, $[4, 6]$, and $[6, 8]$.

2. (Warm-up) Find the left Riemann sum for $f(x) = x^2$ on $[0, 3]$ with $3$ equal subintervals.

Solution

$\Delta x = 1$, and the left endpoints are $0$, $1$, $2$:

$$L_3 = 1\,(0^2 + 1^2 + 2^2) = 5$$

3. (Warm-up) The function $g$ is increasing on $[0, 5]$. Is a left Riemann sum for $g$ on $[0, 5]$ an overestimate or an underestimate of the area under $g$?

Solution

An underestimate. On each subinterval, the left endpoint gives the smallest value of $g$, so each rectangle fits under the curve.

4. (Core) Find the right Riemann sum for $f(x) = x^3$ on $[1, 3]$ with $4$ equal subintervals. Is it an over- or underestimate?

Solution

$\Delta x = 0.5$, and the right endpoints are $1.5$, $2$, $2.5$, $3$:

$$R_4 = 0.5\,(3.375 + 8 + 15.625 + 27) = 0.5\,(54) = 27$$

$f(x) = x^3$ is increasing on $[1, 3]$, so the right sum is an overestimate. (The exact area is $20$.)

5. (Core) Use a trapezoidal sum with the three subintervals in the table to approximate the area under $f$ from $x = 0$ to $x = 4$.

$x$	$0$	$1$	$3$	$4$
$f(x)$	$2$	$5$	$4$	$8$

Solution

The widths are $1$, $2$, and $1$:

$$1 \cdot \frac{2 + 5}{2} + 2 \cdot \frac{5 + 4}{2} + 1 \cdot \frac{4 + 8}{2} = 3.5 + 9 + 6 = 18.5$$

6. (Core) Approximate the area under $y = \sin x$ from $x = 0$ to $x = \pi$ (radians) using a midpoint sum with $2$ equal subintervals. Give the exact value and a decimal to three places.

Solution

$\Delta x = \dfrac{\pi}{2}$, and the midpoints are $\dfrac{\pi}{4}$ and $\dfrac{3\pi}{4}$:

$$M_2 = \frac{\pi}{2}\left(\sin\frac{\pi}{4} + \sin\frac{3\pi}{4}\right) = \frac{\pi}{2}\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) = \frac{\pi\sqrt{2}}{2} \approx 2.221$$

(Remember: in calculus, trig functions use radians. The exact area is $2$.)

7. (Core) During a rainstorm, rain falls at a rate $R(t)$ millimetres per hour. $R$ is decreasing for $0 \le t \le 6$.

$t$ (h)	$0$	$1$	$3$	$4$	$6$
$R(t)$ (mm/h)	$8$	$7$	$5$	$4$	$1$

• (a) Use a right Riemann sum with the four subintervals in the table to approximate the total rainfall from $t = 0$ to $t = 6$. Include units.
• (b) Is your answer an overestimate or an underestimate? Explain.

Solution

(a) Widths $1$, $2$, $1$, $2$:

$$1(7) + 2(5) + 1(4) + 2(1) = 7 + 10 + 4 + 2 = 23 \text{ mm}$$

(b) An underestimate. $R$ is decreasing, so the right endpoint of each subinterval gives the smallest value of $R$ on that subinterval, and each rectangle is below the graph.

8. (Challenge) For $f(x) = x^2$ on $[0, 2]$ with $n$ equal subintervals, show that $R_n - L_n = \dfrac{8}{n}$. How many subintervals do you need so that the right and left sums differ by less than $0.1$?

Solution

The right and left sums share all the middle heights. The right sum has $f(2)$ that the left doesn’t, and the left has $f(0)$ that the right doesn’t:

$$R_n - L_n = \Delta x\,\big(f(2) - f(0)\big) = \frac{2}{n}(4 - 0) = \frac{8}{n}$$

$\dfrac{8}{n} \lt 0.1$ means $n \gt 80$, so you need at least $n = 81$ subintervals.

9. (Challenge) A function $f$ is increasing and concave up on $[a, b]$. Put these in order from smallest to largest: the left sum $L$, the right sum $R$, the midpoint sum $M$, the trapezoidal sum $T$, and the exact area $A$ (all with the same $n$).

Solution

Increasing means $L$ is too small and $R$ is too big. Concave up means $T$ is too big and $M$ is too small. Also, $T$ is the average of $L$ and $R$, so it’s between them, and $M$ uses a height between the left and right heights, so $M \gt L$.

$$L \lt M \lt A \lt T \lt R$$

You can check this against Example 1: $7.5 \lt 9.25 \lt 9.333 \lt 9.5 \lt 11.5$.