The Riemann-sum limit for a changing height, integrals from easy to advanced lesson

Under the line $y = x$ from $0$ to $1$ the height changes at every point, so no single rectangle fits the region. A rectangle of height $1$ covers the region and the empty space above the line, and a rectangle of height $0$ misses the region entirely, which only shows that

0 \le \int_0^1 x\,dx \le 1.

This page cuts the interval into $n$ equal strips, adds the areas of rectangles whose heights come from the ends of the strips, and takes the limit as the strips get thinner. The limit is exactly $\tfrac12$ , and the same construction defines the integral when the graph is curved.

Key terms

Riemann sum

Definition Let $f$ be a function on $[a, b]$ with $a < b$ . A partition of $[a, b]$ is a finite list of points

a = x_0 < x_1 < \dots < x_n = b,

which cuts $[a, b]$ into the $n$ pieces $[x_{k-1}, x_k]$ with widths $\Delta x_k = x_k - x_{k-1}$ . Choose a sample point $x_k^{*}$ in each piece $[x_{k-1}, x_k]$ . The number

\sum_{k=1}^{n} f(x_k^{*})\,\Delta x_k

is a Riemann sum of $f$ for this partition and these sample points.

In plain words Replace the region by a row of rectangles, each as wide as its piece and as tall as the graph at the chosen point, and add their areas. Where $f$ is negative the rectangle hangs below the axis and its term is negative, which matches the signed area that an integral measures.

Example For $f(x) = x$ on $[0, 1]$ , take the partition $0 < \tfrac14 < \tfrac12 < 1$ and the sample points $\tfrac14$ , $\tfrac12$ and $\tfrac34$ , one in each piece. The widths are $\tfrac14$ , $\tfrac14$ and $\tfrac12$ , so the Riemann sum is

\begin{aligned} &\tfrac14 \cdot \tfrac14 + \tfrac12 \cdot \tfrac14 + \tfrac34 \cdot \tfrac12 \\[0.4em] &\quad = \tfrac{1}{16} + \tfrac{2}{16} + \tfrac{6}{16} = \tfrac{9}{16}. \end{aligned}

This estimate, $0.5625$ , is close to the area $\tfrac12$ of the triangle under the line but not equal to it.

Regular partition and mesh width

Definition The mesh width of a partition is the width of its widest piece,

\max_{1 \le k \le n} \Delta x_k.

A regular partition of $[a, b]$ into $n$ pieces uses the equal width $\Delta x = (b - a)/n$ and the points $x_k = a + k\,\Delta x$ for $k = 0, 1, \dots, n$ . Its mesh width is $\Delta x$ itself, so for regular partitions the mesh width approaches $0$ exactly when $n \to \infty$ .

In plain words The mesh width measures how coarse a partition is at its coarsest place. Equal pieces are the simplest way to make every piece thin at once. With unequal pieces, adding more pieces does not by itself shrink the mesh width. A partition that keeps $[0, \tfrac12]$ as one piece and cuts $[\tfrac12, 1]$ into a thousand pieces still has mesh width $\tfrac12$ .

Example The regular partition of $[0, 1]$ into $n$ pieces has width $\Delta x = \tfrac1n$ and points $x_k = \tfrac{k}{n}$ . For $n = 4$ the points are $0$ , $\tfrac14$ , $\tfrac12$ , $\tfrac34$ and $1$ , and the mesh width is $\tfrac14$ . The partition $0 < \tfrac14 < \tfrac12 < 1$ of the Riemann sum example has two pieces of width $\tfrac14$ , but its mesh width is $\tfrac12$ . In this notation $n$ counts the pieces, while $k$ labels one point or one piece.

Right-endpoint sum

Definition For the regular partition of $[a, b]$ into $n$ pieces, the right-endpoint sum uses the right end $x_k$ of each piece $[x_{k-1}, x_k]$ as its sample point,

R_n = \sum_{k=1}^{n} f(x_k)\,\Delta x.

If $f$ is nondecreasing on $[a, b]$ , each right end gives the largest height on its piece, so $R_n \ge \int_a^b f(x)\,dx$ . If $f$ is nonincreasing, each right end gives the smallest height, so $R_n \le \int_a^b f(x)\,dx$ .

In plain words Each rectangle touches the graph at its upper right corner. On a rising graph the rest of its top sits on or above the graph, so the right-endpoint rectangles cover the region and overestimate its area.

Example For the integral $\int_0^1 x\,dx$ , use $n$ equal strips and right endpoints, so the mesh width is $\tfrac1n$ . The $k$ th strip has height $f(\tfrac{k}{n}) = \tfrac{k}{n}$ and width $\tfrac1n$ , so

\begin{aligned} R_n &= \sum_{k=1}^{n} \frac{k}{n} \cdot \frac{1}{n} \\[0.4em] &= \frac{1}{n^2}\,(1 + 2 + \dots + n). \end{aligned}

Write $1 + 2 + \dots + n$ once forward and once backward, one row above the other. Each of the $n$ columns adds to $n + 1$ , so twice the sum is $n(n + 1)$ , for odd $n$ as well as even $n$ . Hence

\begin{aligned} R_n &= \frac{n(n + 1)}{2n^2} \\[0.4em] &= \frac12 + \frac{1}{2n}. \end{aligned}

The term $\tfrac{1}{2n}$ is the total overshoot. Each rectangle sticks out above the line by a small triangle with both legs $\tfrac1n$ , of area $\tfrac{1}{2n^2}$ , and there are $n$ of them.

Left-endpoint sum

Definition For the same regular partition, the left-endpoint sum uses the left end $x_{k-1}$ of each piece $[x_{k-1}, x_k]$ ,

L_n = \sum_{k=1}^{n} f(x_{k-1})\,\Delta x.

For every $f$ the two endpoint sums share every height except $f(a)$ , which only $L_n$ uses, and $f(b)$ , which only $R_n$ uses, so $R_n - L_n = \bigl(f(b) - f(a)\bigr)\,\Delta x$ . Every monotone function on $[a, b]$ is integrable, and if $f$ is nondecreasing on $[a, b]$ , then

L_n \le \int_a^b f(x)\,dx \le R_n.

If $f$ is nonincreasing on $[a, b]$ , both inequalities reverse.

In plain words Each rectangle touches the graph at its upper left corner. On a rising graph the left rectangles fit under the graph and the right rectangles cover it, so the true area is trapped between the two sums, in a gap proportional to the strip width.

Example For $f(x) = x$ on $[0, 1]$ , the left heights are $0, \tfrac1n, \dots, \tfrac{n-1}{n}$ , so

\begin{aligned} L_n &= \frac{0 + 1 + \dots + (n - 1)}{n^2} \\[0.4em] &= \frac{(n - 1)\,n}{2n^2} \\[0.4em] &= \frac12 - \frac{1}{2n}. \end{aligned}

Because $f(x) = x$ is increasing, $L_n \le \tfrac12 \le R_n$ , and the gap is $R_n - L_n = \tfrac1n$ , which is $\bigl(f(1) - f(0)\bigr) \cdot \tfrac1n$ . For $n = 4$ the bracket is $\tfrac38 \le \tfrac12 \le \tfrac58$ .

Limit of Riemann sums

Definition Let $f$ be a function on $[a, b]$ with $a < b$ . The Riemann sums of $f$ converge to a number $I$ as the mesh width approaches $0$ if for every $\varepsilon > 0$ there is a $\delta > 0$ such that

\Bigl\lvert \sum_{k=1}^{n} f(x_k^{*})\,\Delta x_k - I \Bigr\rvert < \varepsilon

for every partition with mesh width less than $\delta$ and every choice of sample points. Then $f$ is integrable on $[a, b]$ and $\int_a^b f(x)\,dx = I$ . Every continuous function on $[a, b]$ is integrable. For an integrable $f$ , the sums of regular partitions approach the integral as $n \to \infty$ with any sample points, and in particular

\begin{aligned} \int_a^b f(x)\,dx &= \lim_{n \to \infty} R_n \\[0.4em] &= \lim_{n \to \infty} L_n. \end{aligned}

Convergent endpoint sums alone do not prove that a function is integrable. On $[0, 1]$ , the function equal to $1$ at rational numbers and $0$ at irrational numbers has $L_n = R_n = 1$ for every $n$ , because every point $\tfrac{k}{n}$ is rational, yet sample points chosen irrational give sums equal to $0$ , so the function is not integrable.

In plain words The integral is the number the rectangle sums settle on as every strip becomes thin, whichever heights inside the strips are used. No sum with finitely many strips has to equal it, and no strip of width zero is ever drawn. The $dx$ of the integral takes the place of the positive widths $\Delta x$ of the sums.

Example The function $f(x) = x$ is continuous on $[0, 1]$ , so it is integrable, and the regular partitions with right endpoints have mesh width $\tfrac1n$ , which approaches $0$ . The right-endpoint sums $R_4 = 0.625$ , $R_{10} = 0.55$ and $R_{100} = 0.505$ are each above $\tfrac12$ by $\tfrac{1}{2n}$ . Since $\tfrac{1}{2n} \to 0$ ,

\begin{aligned} \int_0^1 x\,dx &= \lim_{n \to \infty} \left(\frac12 + \frac{1}{2n}\right) \\[0.4em] &= \frac12. \end{aligned}

The left-endpoint sums $\tfrac12 - \tfrac{1}{2n}$ have the same limit. The triangle with base $1$ and height $1$ has area $\tfrac12 \cdot 1 \cdot 1 = \tfrac12$ , which confirms the value.

Check yourself

Compute the right-endpoint sum $R_4$ and the left-endpoint sum $L_4$ for $\int_0^2 x\,dx$ . Between which two numbers must the integral lie?

Show the answer

The width is $\Delta x = \tfrac24 = \tfrac12$ and the points are $0$ , $\tfrac12$ , $1$ , $\tfrac32$ and $2$ . The right heights $\tfrac12$ , $1$ , $\tfrac32$ and $2$ add to $5$ , so $R_4 = \tfrac12 \cdot 5 = \tfrac52$ . The left heights $0$ , $\tfrac12$ , $1$ and $\tfrac32$ add to $3$ , so $L_4 = \tfrac12 \cdot 3 = \tfrac32$ . Because $f(x) = x$ is increasing,

\tfrac32 \le \int_0^2 x\,dx \le \tfrac52.

The gap is $\bigl(f(2) - f(0)\bigr) \cdot \tfrac12 = 1$ . The triangle with base $2$ and height $2$ has area $2$ , which lies inside the bracket.

Use right-endpoint sums of regular partitions to find $\int_0^2 3x\,dx$ as a limit.

Show the answer

With $n$ pieces, $\Delta x = \tfrac2n$ and $x_k = \tfrac{2k}{n}$ , so the $k$ th height is $f(x_k) = \tfrac{6k}{n}$ . Then

\begin{aligned} R_n &= \sum_{k=1}^{n} \frac{6k}{n} \cdot \frac{2}{n} \\[0.4em] &= \frac{12}{n^2} \cdot \frac{n(n + 1)}{2} \\[0.4em] &= 6 + \frac{6}{n}. \end{aligned}

The function $3x$ is continuous on $[0, 2]$ , so it is integrable, and the mesh width $\tfrac2n$ approaches $0$ . Hence

\begin{aligned} \int_0^2 3x\,dx &= \lim_{n \to \infty} \left(6 + \frac{6}{n}\right) \\[0.4em] &= 6. \end{aligned}

The triangle with base $2$ and height $6$ has area $6$ as well.

A classmate says that right-endpoint sums always overestimate an integral. Test the claim on $\int_0^1 (1 - x)\,dx$ by finding $R_n$ and $L_n$ .

Show the answer

The right heights are $1 - \tfrac{k}{n}$ for $k = 1, \dots, n$ , so

\begin{aligned} R_n &= \frac1n \sum_{k=1}^{n} \left(1 - \frac{k}{n}\right) \\[0.4em] &= 1 - \frac{n + 1}{2n} \\[0.4em] &= \frac12 - \frac{1}{2n}. \end{aligned}

The left heights are $1 - \tfrac{k-1}{n}$ , and the same steps give $L_n = \tfrac12 + \tfrac{1}{2n}$ . Both sums approach $\tfrac12$ , the area of the triangle under $y = 1 - x$ , but here $R_n < \tfrac12 < L_n$ for every $n$ . The function is decreasing, so each right end gives the smallest height on its piece and the right rectangles fit under the graph. The claim is false. Whether a right-endpoint sum overestimates depends on how the function behaves on each piece, and for a function that rises and then falls it can go either way.

The Riemann-sum limit for a changing height

Key terms

Riemann sum

Regular partition and mesh width

Right-endpoint sum

Left-endpoint sum

Limit of Riemann sums

Check yourself

How this lesson connects

Builds on

Used later in