The Area Problem, calculus lesson

Rectangles, triangles and any shape with straight sides have areas you can compute from a formula. A region under the curve $y = x^2$ has a curved edge, and no formula from geometry covers it. Yet you still feel it has an area. The way out is the same move that gave us tangent lines. Approximate with simple pieces, then take a limit.

Key terms

Nonnegative regions

Definition Let $f$ be continuous on a closed interval $[a, b]$ with $a < b$ and $f(x) \ge 0$ for all $x$ in $[a, b]$ . The region under the graph of $f$ from $a$ to $b$ is the set

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

It is bounded by the graph above, the $x$ -axis below and the vertical lines $x = a$ and $x = b$ on the sides.

In plain words Shade everything between the curve and the $x$ -axis over the interval. For now the curve stays on or above the axis, so the area we are after is an ordinary, nonnegative area.

Example For $f(x) = 2x + 1$ on $[0, 3]$ the region $S$ is a trapezoid with parallel sides $f(0) = 1$ and $f(3) = 7$ and width $3$ . Its area is

\frac{1 + 7}{2} \cdot 3 = 12.

A straight top edge makes this easy. The rest of the lesson handles curved ones.

Area by polygons and rectangles

Definition A rectangle of width $w$ and height $h$ has area $wh$ . A polygon has the area obtained by cutting it into triangles and adding their areas, where a triangle has area

\frac{1}{2} \cdot \text{base} \cdot \text{height}.

Area has two basic properties that every later definition must keep. It is additive, so a region split into pieces that overlap only along their edges has the sum of the pieces’ areas. It is monotone, so a region contained in another has area no larger than the other.

In plain words Straight-edged shapes are solved. Monotonicity is the key to curved shapes, since a stack of rectangles inside a region gives a number that is too small and a stack covering it gives a number that is too big.

Example Under $f(x) = x^2$ on $[0, 1]$ , the rectangle $[\frac{1}{2}, 1] \times [0, \frac{1}{4}]$ lies inside the region, because $x^2 \ge \frac{1}{4}$ there. So the area is at least $\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$ . The square $[0, 1] \times [0, 1]$ contains the region, so the area is at most $1$ .

Area approximation with rectangles

Definition Divide $[a, b]$ into $n$ subintervals of equal width

\Delta x = \frac{b - a}{n}, \qquad x_i = a + i \, \Delta x \quad (i = 0, 1, \ldots, n).

The left endpoint sum and the right endpoint sum are

L_n = \sum_{i=1}^{n} f(x_{i-1}) \, \Delta x, \qquad R_n = \sum_{i=1}^{n} f(x_i) \, \Delta x.

Each term is the area of a rectangle of width $\Delta x$ whose height is the value of $f$ at one end of the strip.

In plain words Slice the region into thin vertical strips, replace each strip by a rectangle whose height you read off the curve, and add the rectangle areas.

Example For $f(x) = x^2$ on $[0, 1]$ with $n = 4$ , $\Delta x = \frac{1}{4}$ and the points are $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1$ .

R_4 = \frac{1}{4} \left( \frac{1}{16} + \frac{4}{16} + \frac{9}{16} + \frac{16}{16} \right) = \frac{15}{32} = 0.46875,

L_4 = \frac{1}{4} \left( 0 + \frac{1}{16} + \frac{4}{16} + \frac{9}{16} \right) = \frac{7}{32} = 0.21875.

To see these rectangles, choose $x^2$ on $[0, 1]$ with $n = 4$ in the Riemann sum explorer and switch between the left and right rules.

Lower and upper sums

Definition Let $f$ be continuous and nonnegative on $[a, b]$ , divided into $n$ equal subintervals as above. On the $i$ th subinterval let $m_i$ be the minimum and $M_i$ the maximum value of $f$ , which exist by the extreme value theorem. The lower sum and upper sum are

s_n = \sum_{i=1}^{n} m_i \, \Delta x, \qquad S_n = \sum_{i=1}^{n} M_i \, \Delta x.

The lower rectangles lie inside the region and the upper rectangles cover it, so whatever the area $A$ is, it must satisfy $s_n \le A \le S_n$ .

In plain words The lower sum uses the shortest height on each strip and always comes in under the true area. The upper sum uses the tallest height and always comes in over it.

Example The function $f(x) = x^2$ is increasing on $[0, 1]$ , so its minimum on each strip is at the left end and its maximum at the right end. That makes $s_4 = L_4 = 0.21875$ and $S_4 = R_4 = 0.46875$ . So the area under the parabola lies between $0.21875$ and $0.46875$ .

Definition Let $f$ be continuous on $[a, b]$ . A division of $[a, b]$ is a finite list of points $a = x_0 < x_1 < \cdots < x_n = b$ . The strips need not have equal widths, so write $\Delta x_i = x_i - x_{i-1}$ . Lower and upper sums make sense for any division,

s = \sum_{i=1}^{n} m_i \, \Delta x_i, \qquad S = \sum_{i=1}^{n} M_i \, \Delta x_i,

with $m_i$ and $M_i$ the minimum and maximum of $f$ on $[x_{i-1}, x_i]$ . A division refines another if it contains all of the other’s points and possibly more. For a refinement the lower sum can only increase or stay the same, and the upper sum can only decrease or stay the same. To see it, add one point $t$ inside a strip $[x_{i-1}, x_i]$ . The minimum of $f$ on each of the two new pieces is at least $m_i$ , so their lower contributions add up to at least $m_i (t - x_{i-1}) + m_i (x_i - t) = m_i \, \Delta x_i$ . The maximum on each piece is at most $M_i$ , which gives the reverse inequality for upper sums. Adding several points is this step repeated.

In plain words Cutting the strips more finely never makes the estimates worse. The lower estimate climbs and the upper estimate drops, so the gap that holds the true area narrows.

Example For $f(x) = x^2$ on $[0, 1]$ , going from $n = 2$ to $n = 4$ adds the points $\frac{1}{4}$ and $\frac{3}{4}$ .

$n$	Lower sum	Upper sum	Gap
$2$	$0.125$	$0.625$	$0.5$
$4$	$0.21875$	$0.46875$	$0.25$

Limiting area

Definition Let $f$ be continuous and nonnegative on $[a, b]$ . The area of the region under the graph of $f$ from $a$ to $b$ is defined as

A = \lim_{n \to \infty} R_n = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \, \Delta x.

For continuous $f$ this limit exists, and $\lim L_n$ , $\lim s_n$ and $\lim S_n$ all exist and equal the same number $A$ . So the definition does not depend on which heights you use.

In plain words The area is the number that every rectangle approximation closes in on as the strips get thinner. This is a definition, and it agrees with the familiar area for rectangles and triangles.

Example For $f(x) = x^2$ on $[0, 1]$ we have $\Delta x = \frac{1}{n}$ and $x_i = \frac{i}{n}$ . The sum of the first $n$ squares is

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

R_n = \sum_{i=1}^{n} \frac{i^2}{n^2} \cdot \frac{1}{n} = \frac{(n+1)(2n+1)}{6n^2}.

$n$	$R_n$
$10$	$0.385$
$100$	$0.33835$
$1000$	$0.3338335$

As $n \to \infty$ ,

\frac{(n+1)(2n+1)}{6n^2} = \frac{1}{6} \left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right) \to \frac{2}{6},

so the area under the parabola from $0$ to $1$ is exactly $\frac{1}{3}$ .

Check yourself

Estimate the area under $f(x) = \frac{1}{x}$ on $[1, 3]$ with $n = 2$ rectangles, once with left endpoints and once with right endpoints. Which one is too big?

Show the answer

Here $\Delta x = 1$ . Left endpoints give $1 \cdot \left(1 + \frac{1}{2}\right) = 1.5$ and right endpoints give $1 \cdot \left(\frac{1}{2} + \frac{1}{3}\right) \approx 0.833$ . The function is decreasing, so left endpoints give the upper sum, and $1.5$ is too big.

Find $R_n$ for $f(x) = x$ on $[0, 2]$ and take the limit. Does it match the triangle?

Show the answer

The width is $\Delta x = \frac{2}{n}$ and the right endpoints are $x_i = \frac{2i}{n}$ , so

R_n = \sum_{i=1}^{n} \frac{2i}{n} \cdot \frac{2}{n} = \frac{4}{n^2} \cdot \frac{n(n+1)}{2} = \frac{2(n+1)}{n},

which tends to $2$ . The triangle has area $\frac{1}{2} \cdot 2 \cdot 2 = 2$ , so yes.

Without computing, explain why every lower sum for a continuous nonnegative $f$ on $[a, b]$ is at most every upper sum, even when the two come from different divisions.

Show the answer

Take a lower sum $s$ from one division and an upper sum $S$ from another. Put all division points of both together. That common division refines each of the two, so its lower sum $s^*$ satisfies $s \le s^*$ and its upper sum $S^*$ satisfies $S^* \le S$ . On the common division $m_i \le M_i$ on every strip, so $s^* \le S^*$ . Together $s \le s^* \le S^* \le S$ .

The Area Problem

Theory, part 95

Practice, part 96

Key terms

Nonnegative regions

Area by polygons and rectangles

Area approximation with rectangles

Lower and upper sums

Refinement

Limiting area

Check yourself

How this lesson connects

Builds on

Used later in

Resources for this lesson

Riemann sum explorer