Area Approximation with Rectangles, precalculus lesson

A car that drives at a steady 60 kilometers per hour for 2 hours covers 120 kilometers. On a graph of speed against time that distance is the area of a rectangle 2 wide and 60 tall. Real speed changes all the time, so the region under the speed graph has a curved top. Cut it into thin strips, treat each strip as a rectangle and add them up. That plain idea is where integral calculus begins.

Key terms

Area intuition

Definition Any sensible measure of area for regions in the plane follows three rules. A rectangle of width $w$ and height $h$ has area $wh$ . Area is additive, so a region cut into pieces that overlap only along their edges has the sum of the areas of the pieces. Area is monotone, so a region that lies inside another has no larger area. Together these rules mean that rectangles placed inside a region without overlapping, except along edges, give a lower bound for its area, and rectangles that together cover the region give an upper bound.

In plain words You may not know the area of a curved region yet, but you can trap it. Rectangles inside it give no more than the area, rectangles covering it give no less, and the true area lies in between.

Example The region under $f(x) = 4 - x^2$ from $x = 0$ to $x = 2$ fits inside the rectangle of width $2$ and height $4$ , so its area $A$ is at most $8$ . It contains the rectangle from $x = 0$ to $x = 1$ with height $3$ , because $4 - x^2 \ge 3$ for $0 \le x \le 1$ . So $3 \le A \le 8$ . The rest of the lesson narrows this down.

Nonnegative function

Definition A function $f$ is nonnegative on an interval $[a, b]$ with $a < b$ if $f(x) \ge 0$ for every $x$ in $[a, b]$ . The region under the graph of $f$ from $a$ to $b$ is then the set

\{ (x, y) \mid a \le x \le b, \ 0 \le y \le f(x) \},

bounded above by the graph, below by the $x$ -axis and on the sides by the lines $x = a$ and $x = b$ . Rectangle sums estimate the area of this region only when $f$ is nonnegative. Where $f$ is negative a product $f(x) \, \Delta x$ is negative, and a sum counts that part of the picture with a minus sign.

In plain words For now the curve stays on or above the $x$ -axis, so every rectangle has a genuine height that is zero or positive.

Example The function $4 - x^2$ is nonnegative on $[0, 2]$ , since $x^2 \le 4$ there. It is not nonnegative on $[0, 3]$ , because its value at $3$ is $4 - 9 = -5$ . A rectangle of width $1$ built on that value would add $1 \cdot (-5) = -5$ to a sum, which no area can be.

Finite partition

Definition A partition of $[a, b]$ , with $a < b$ , is a finite list of points

a = x_0 < x_1 < x_2 < \cdots < x_n = b.

It cuts $[a, b]$ into $n$ subintervals $[x_{i-1}, x_i]$ with widths $\Delta x_i = x_i - x_{i-1} > 0$ , which add up to $b - a$ . In a regular partition all widths are equal,

\begin{gathered} \Delta x = \frac{b - a}{n}, \\[0.8em] x_i = a + i \, \Delta x \quad (i = 0, 1, \ldots, n). \end{gathered}

In plain words Choose finitely many cut points between $a$ and $b$ in increasing order. The pieces between neighboring cut points become the bases of the rectangles.

Example The regular partition of $[1, 3]$ into $n = 4$ pieces has $\Delta x = \frac{1}{2}$ and the points $1, 1.5, 2, 2.5, 3$ . The list $0, 0.5, 2$ is a partition of $[0, 2]$ that is not regular, with widths $0.5$ and $1.5$ .

Left and right rectangles

Definition For a partition of $[a, b]$ , the left rectangle on $[x_{i-1}, x_i]$ has height $f(x_{i-1})$ , the value at the left end, and the right rectangle has height $f(x_i)$ , the value at the right end. The left sum and the right sum are

\begin{aligned} L &= \sum_{i=1}^{n} f(x_{i-1}) \, \Delta x_i, \\[0.8em] R &= \sum_{i=1}^{n} f(x_i) \, \Delta x_i. \end{aligned}

For a regular partition with $n$ pieces they are written $L_n$ and $R_n$ , and every $\Delta x_i$ equals $\Delta x$ .

In plain words On each piece, read a height off the curve at one end, multiply by the width and add everything up. Left sums use the left ends and right sums use the right ends.

Example For $f(x) = 4 - x^2$ on $[0, 2]$ with $n = 4$ , the width is $\Delta x = 0.5$ and the points $0, 0.5, 1, 1.5, 2$ have heights $4, 3.75, 3, 1.75, 0$ .

\begin{aligned} L_4 &= 0.5\,(4 + 3.75 + 3 + 1.75) \\[0.4em] &= 6.25, \end{aligned}

\begin{aligned} R_4 &= 0.5\,(3.75 + 3 + 1.75 + 0) \\[0.4em] &= 4.25. \end{aligned}

Lower and upper estimates for monotone functions

Definition Let $f$ be nonnegative on $[a, b]$ and let $A$ be the area of the region under its graph. Suppose $f$ is nondecreasing on $[a, b]$ , which means $f(s) \le f(t)$ whenever $a \le s < t \le b$ . Then on every piece the left end has the smallest height and the right end the largest. The left rectangles lie inside the region and the right rectangles cover it, so for every partition

L \le A \le R.

This covers every increasing function and also functions with flat stretches. If $f$ is nonincreasing on $[a, b]$ , with $f(s) \ge f(t)$ whenever $a \le s < t \le b$ , the roles switch and $R \le A \le L$ . A function that is nondecreasing or nonincreasing on an interval is called monotone there. Without monotonicity on the whole interval, neither sum has to be a bound.

In plain words Left and right sums are sure to trap the area when the function moves in one direction across the whole interval. Then the low ends of the pieces give the lower estimate and the high ends give the upper estimate.

Example The function $4 - x^2$ is decreasing on $[0, 2]$ , so the right sum is the lower estimate and the left sum is the upper one, $4.25 \le A \le 6.25$ . Now see why monotonicity matters. The upper half of the unit circle, $y = \sqrt{1 - x^2}$ on $[-1, 1]$ , rises and then falls. With $n = 2$ both sums equal

\begin{aligned} L_2 &= 1 \cdot (0 + 1) = 1, \\[0.4em] R_2 &= 1 \cdot (1 + 0) = 1, \end{aligned}

while the half disk has area $\frac{\pi}{2} \approx 1.571$ . Both sums are too small, and neither is an upper estimate. Splitting at the top point $x = 0$ and using the rule on each monotone half repairs this.

Definition A partition refines another partition of $[a, b]$ if it contains every point of the other one, and possibly more. Let $f$ be nonnegative and monotone on $[a, b]$ . If one partition refines another, its lower estimate is at least as large and its upper estimate at most as large as those of the coarser partition. To see it for a nondecreasing $f$ , add one point $t$ inside a piece $[x_{i-1}, x_i]$ . The new left rectangle on $[t, x_i]$ has height $f(t) \ge f(x_{i-1})$ , so the left sum can only grow. The new right rectangle on $[x_{i-1}, t]$ has height $f(t) \le f(x_i)$ , so the right sum can only shrink. Adding several points repeats this step, and a nonincreasing $f$ works the same way with left and right switched. The regular partition with $2n$ pieces refines the one with $n$ pieces, but for $n \ge 2$ the one with $n + 1$ pieces does not.

In plain words Cutting the pieces more finely can only tighten the trap. The lower estimate creeps up, the upper estimate creeps down, and the area stays caught between them.

Example For $4 - x^2$ on $[0, 2]$ each partition below refines the one before, since doubling $n$ keeps all the old points.

$n$	Lower estimate $R_n$	Upper estimate $L_n$	Gap
$2$	$3$	$7$	$4$
$4$	$4.25$	$6.25$	$2$
$8$	$4.8125$	$5.8125$	$1$

The two sums share every height except $f(0) = 4$ and $f(2) = 0$ , so the gap is $(4 - 0) \, \Delta x$ , and it halves with each doubling.

You can check this table in the Riemann sum explorer by choosing $4 - x^2$ on $[0, 2]$ and switching between the left and right rules at $n = 2$ , $4$ and $8$ .

Finite sum versus limiting area

Definition Every left or right sum is a finite sum, and for a curved region it is usually only an approximation. For a continuous nonnegative $f$ on $[a, b]$ , calculus defines the area as the limit

A = \lim_{n \to \infty} R_n,

which exists and equals the limit of $L_n$ . No finite $n$ has to give $A$ exactly. For a monotone $f$ , though, $A$ lies between $L_n$ and $R_n$ , whose difference is

\lvert R_n - L_n \rvert = \lvert f(b) - f(a) \rvert \cdot \frac{b - a}{n},

so the error of either sum is at most this amount, and it shrinks to $0$ as $n$ grows.

In plain words A sum of rectangles is an estimate you can compute. The area is the number those estimates close in on as the rectangles get thinner.

Example For $4 - x^2$ on $[0, 2]$ the width is $\Delta x = \frac{2}{n}$ and the right ends are $x_i = \frac{2i}{n}$ . The sum of the first $n$ squares is

\sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6},

which can be proved by mathematical induction. So

\begin{aligned} R_n &= \sum_{i=1}^{n} \left(4 - \frac{4i^2}{n^2}\right) \frac{2}{n} \\[0.8em] &= 8 - \frac{8}{n^3} \cdot \frac{n(n + 1)(2n + 1)}{6} \\[0.8em] &= \frac{16}{3} - \frac{4}{n} - \frac{4}{3n^2}. \end{aligned}

$n$	$10$	$100$	$1000$
$R_n$	$4.92$	$5.2932$	$5.329332$

Every $R_n$ falls short of $\frac{16}{3}$ , but the last two terms shrink to $0$ as $n$ grows. So the area under $4 - x^2$ from $0$ to $2$ is exactly $\frac{16}{3} \approx 5.333$ , which no single finite sum here reaches.

Check yourself

Estimate the area under $f(x) = x^2 + 1$ on $[0, 3]$ with $n = 3$ left rectangles and $n = 3$ right rectangles. Which sum is a lower estimate?

Show the answer

Here $\Delta x = 1$ and the heights at $0, 1, 2, 3$ are $1, 2, 5, 10$ . So $L_3 = 1 + 2 + 5 = 8$ and $R_3 = 2 + 5 + 10 = 17$ . The function is increasing on $[0, 3]$ , so $L_3 = 8$ is the lower estimate and $8 \le A \le 17$ .

Find lower and upper estimates for the area under $f(x) = \frac{1}{x}$ on $[1, 2]$ with $n = 4$ .

Show the answer

The width is $\Delta x = \frac{1}{4}$ and the heights at $1, 1.25, 1.5, 1.75, 2$ are $1, \frac{4}{5}, \frac{2}{3}, \frac{4}{7}, \frac{1}{2}$ . The function is decreasing on $[1, 2]$ , so the right sum is the lower estimate.

\begin{aligned} R_4 &= \frac{1}{4}\left(\frac{4}{5} + \frac{2}{3} + \frac{4}{7} + \frac{1}{2}\right) \\[0.8em] &= \frac{533}{840} \approx 0.6345, \end{aligned}

\begin{aligned} L_4 &= \frac{1}{4}\left(1 + \frac{4}{5} + \frac{2}{3} + \frac{4}{7}\right) \\[0.8em] &= \frac{319}{420} \approx 0.7595. \end{aligned}

So the area lies between about $0.6345$ and $0.7595$ .

For $f(x) = x^2 + 1$ on $[0, 3]$ , how many equal pieces make sure that $R_n - L_n \le 0.01$ ?

Show the answer

The function is increasing, so the gap is

\begin{aligned} R_n - L_n &= \bigl(f(3) - f(0)\bigr) \cdot \frac{3}{n} \\[0.8em] &= \frac{27}{n}. \end{aligned}

This is at most $0.01$ exactly when $n \ge 2700$ .

Area Approximation with Rectangles

Key terms

Area intuition

Nonnegative function

Finite partition

Left and right rectangles

Lower and upper estimates for monotone functions

Refinement

Finite sum versus limiting area

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus

Resources for this lesson

Riemann sum explorer