1.1 Approximating Areas

Big idea

The derivative answered a question about a single instant: how fast is this changing right here. The question that opens integration is the opposite kind. Over a whole stretch of the input, how much has accumulated. For a region with straight sides you already have formulas, but the region under a curve has a top edge that bends, and no formula from geometry applies to it.

The move is to give up on exactness at first. Cut the interval $[a,b]$ into $n$ pieces of equal width, and over each piece replace the curve by a horizontal line at the height of the function at one sample point inside that piece. The region becomes a row of rectangles, each of which you can measure, and the total of their areas is an estimate of the area you wanted. Where the sample point sits inside each piece is your choice, and the three standard choices are the left endpoint, the right endpoint, and the midpoint.

None of these estimates is the answer, and that is the point of the construction. As $n$ grows, the rectangles get thin, the flat tops track the curve more closely, and the estimates settle toward a single number. That number is what we will call the area. The approximations are not a detour on the way to a formula; they are the definition, and the formula comes later.

The same sum answers questions that have nothing to do with area. Every term is a height times a width, and if the height is a rate and the width is an amount of time, the term is an amount of stuff. Speed times time is distance, flow rate times time is volume, and in each case adding the terms estimates the total. Learning to read a sum of rectangles as a total accumulation is what makes the rest of the chapter feel inevitable.

Decoder

Cut $[a,b]$ into $n$ equal pieces, multiply the width of each piece by the value of the function at one sample point inside it, and add the results. As $n$ grows without bound the sums close in on one number, and that number is the area under the graph.

Three phrases carry the content. “Equal pieces” fixes the common width at $\Delta x = (b-a)/n$, so the width factors out of the sum and only the heights vary. “One sample point inside it” is the free choice, and it is what separates the left sum from the right sum from the midpoint sum. “Close in on one number” is a claim, not a definition: for a continuous function on a closed interval the choice of sample points stops mattering in the limit, which is why the limit deserves a name.

The classic mistake is arithmetic rather than conceptual. A left-endpoint sum on $n$ pieces uses the heights at $x_0$ through $x_{n-1}$, and a right-endpoint sum uses $x_1$ through $x_n$. Both use $n$ heights, and neither uses all $n+1$ grid points. Counting $n+1$ rectangles, or sliding the list by one, is the usual way these estimates go wrong.

One fact about the direction of the error is worth having in hand before you compute anything. If the function only rises across the interval, every left endpoint gives the smallest value on its piece and every right endpoint gives the largest, so the left sum lands under the true area and the right sum lands over it. If the function only falls, the two roles trade places.

Definitions and results

Regular partition. Divide $[a,b]$ into $n$ subintervals of equal width

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

so that $x_0 = a$, $x_n = b$, and the $i$-th subinterval is $[x_{i-1}, x_i]$.

Riemann sum. Choosing a sample point $x_i^{*}$ in each subinterval gives the sum

$$ \sum_{i=1}^{n} f(x_i^{*})\, \Delta x $$

Each term is a signed rectangle area: positive where the function is positive, negative where it dips below the axis.

Left- and right-endpoint sums. Taking $x_i^{*} = x_{i-1}$ gives the left sum, and taking $x_i^{*} = x_i$ gives the right sum:

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

Midpoint sum. Taking the center of each subinterval, $x_i^{*} = \tfrac{1}{2}(x_{i-1} + x_i)$, gives $M_n$. On a smooth graph the flat top crosses the curve near the middle of the piece, so the overshoot on one half roughly cancels the undershoot on the other; midpoint estimates are usually far better than either endpoint estimate at the same $n$.

Direction of the error for a monotone function. If $f$ is increasing on $[a,b]$ then $L_n \le \text{area} \le R_n$. If $f$ is decreasing the inequalities reverse. Knowing which side you are on turns an estimate into a bound.

Sigma notation. The symbol $\sum_{i=1}^{n} a_i$ abbreviates $a_1 + a_2 + \cdots + a_n$. It is linear: constants pull out, and a sum of two lists splits into two sums. A constant summed $n$ times gives $\sum_{i=1}^{n} c = cn$.

Closed forms for the power sums. These three turn a sum with $n$ terms into an expression you can take a limit of:

$$ \sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \qquad \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}, \qquad \sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2 $$

Area as a limit. For $f$ continuous and nonnegative on $[a,b]$, the area under the graph is

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

and the limit is the same for every choice of sample points.

Worked examples

Four rectangles under a rising curve

Estimate the area under $f(x) = x^2$ on $[0,2]$ with $n = 4$. Here $\Delta x = 0.5$ and the grid points are $0, 0.5, 1, 1.5, 2$.

The left sum uses the heights at $0, 0.5, 1, 1.5$, which are $0, 0.25, 1, 2.25$. Their total is $3.5$, so $L_4 = 3.5(0.5) = 1.75$. The right sum uses the heights at $0.5, 1, 1.5, 2$, which are $0.25, 1, 2.25, 4$, totalling $7.5$, so $R_4 = 7.5(0.5) = 3.75$.

The midpoints are $0.25, 0.75, 1.25, 1.75$, with heights $0.0625, 0.5625, 1.5625, 3.0625$. Their total is $5.25$, so $M_4 = 5.25(0.5) = 2.625$.

Since $f$ rises across the whole interval, the true area sits between $1.75$ and $3.75$. It is $8/3 \approx 2.667$, so the left sum is low by about $0.92$, the right sum is high by about $1.08$, and the midpoint sum misses by about $0.04$.

The same setup on a falling curve

Estimate the area under $g(x) = 4 - x^2$ on $[0,2]$, again with $n = 4$ and $\Delta x = 0.5$.

Left heights are $g(0), g(0.5), g(1), g(1.5) = 4, 3.75, 3, 1.75$, totalling $12.5$, so $L_4 = 6.25$. Right heights are $3.75, 3, 1.75, 0$, totalling $8.5$, so $R_4 = 4.25$.

The exact area is $16/3 \approx 5.333$. This time the left sum is the overestimate and the right sum is the underestimate, because $g$ decreases. The labels “left” and “right” carry no information about direction on their own; the behavior of the function does.

Midpoints on a straight line

Estimate the area under $h(x) = 2x + 1$ on $[1,3]$ with $n = 4$, so $\Delta x = 0.5$ and the midpoints are $1.25, 1.75, 2.25, 2.75$.

The heights are $3.5, 4.5, 5.5, 6.5$, totalling $20$, so $M_4 = 10$. The region is a trapezoid with parallel sides $h(1) = 3$ and $h(3) = 7$ and width $2$, so its area is $\tfrac{1}{2}(3+7)(2) = 10$. The midpoint sum is not close; it is exact.

That is not luck. On a straight line the part of the rectangle that pokes above the graph on one side of the midpoint has exactly the area of the gap left below it on the other side. For a line the cancellation is perfect, and for a gently curving graph it is nearly so.

From a sum to an exact area

Find the area under $f(x) = x^2$ on $[0,1]$ as a limit of right sums. Here $\Delta x = 1/n$ and $x_i = i/n$, so

$$ R_n = \sum_{i=1}^{n} \left(\frac{i}{n}\right)^{2}\frac{1}{n} = \frac{1}{n^{3}}\sum_{i=1}^{n} i^{2} = \frac{1}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^{2}} $$

Expanding the numerator gives $2n^2 + 3n + 1$, so

$$ R_n = \frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^{2}} \longrightarrow \frac{1}{3} $$

The area is $1/3$. Two checks: every extra term is positive, so each right sum exceeds $1/3$, which is what an increasing function demands. And at $n = 4$ the formula gives $1/3 + 1/8 + 1/96 \approx 0.4688$, matching the direct computation $(0.0625 + 0.25 + 0.5625 + 1)(0.25)$.

Practice

Start with the sums themselves. These ask you to build a partition, pick left, right or midpoint heights, and evaluate, usually with a small $n$ and clean arithmetic.

Then move to the limit and to what the number means. These ask for exact areas from sigma-notation sums, and for reasoning about whether an estimate is high or low.

Quiz

Five items on partitions, left, right and midpoint sums, sigma-notation formulas, and area as a limit.