MATHEMATICS / ACCUMULATION

Riemann integration: from rectangle sums to the area under a curve

Walk at 2 metres per second for 3 seconds and you cover $2 × 3 = 6$ metres. Put time on the horizontal axis and speed on the vertical axis. The multiplication now has a picture: a rectangle of width 3 and height 2. Its numerical area is the distance we just calculated.

But suppose you keep speeding up. Which speed should we multiply by the time? Using the starting speed misses the later increase. Using the final speed treats the whole walk as though it were that fast. Multiplication still works; the difficulty is that one height no longer describes the whole interval. Let us divide the time into shorter intervals. We can use one speed to estimate each part, calculate its rectangle, and add the areas.

This is not exact yet. The gaps between the rectangles and the curve let us see what our estimate misses. Make the intervals narrower and compare both the gaps and the total. Function graphs show the changing speed, and limits let us ask what this process approaches. If you have read the derivative page, recall its direction: position changes gave us velocity. Here, knowing how fast we move helps us build up the distance travelled.

01How much area lies beneath a curve?

Start with $f(x) = x^2$ , the x-axis and the boundaries $a = 0$ and $b = 3$ . The teal region is the area we want. The gold rectangles are an approximation, not the answer. Click a strip to inspect it, or use the strip selector.

f(x)=x^2

Equal-width partition · n = 4

Rectangle width

\Delta x

0.7500

Rectangle sum

S_n

5.9063

Exact integral (reference)9.0000

The curve increases from left to right. A left-endpoint rectangle misses area; a right-endpoint rectangle includes extra area. As n doubles, each width halves, so the rectangles follow the curve more closely.

1. Divide the width

Split the total width $b - a$ into $n$ equal pieces. The Greek letter $\Delta$ means a change or interval width.

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

2. Choose a height

Pick a point $x_i^*$ in strip i. The curve’s height at that point becomes the rectangle’s height.

A_i=f(x_i^*)\Delta x

3. Add the pieces

$\Sigma$ means “add all these terms”. This finite sum is called a Riemann sum.

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

02Same width. Three ways to choose the height.

Compare the same curve using eight rectangles. A sampling point must lie in its own strip. Play the guide to highlight left, midpoint and right sampling in turn; the three results stay visible for comparison.

Left endpoint ●

Riemann sum S₈7.3828

$x_i^*=a+(i-1)\Delta x$

Midpoint

Riemann sum S₈8.9648

$x_i^*=a+(i-\frac12)\Delta x$

Right endpoint

Riemann sum S₈10.7578

$x_i^*=a+i\Delta x$

Left endpoint：Use the left edge of each strip. For an increasing function, this gives the lowest height in that strip and hence an underestimate.

03What if the maximum width tends to zero?

For these continuous functions, all three sampling choices approach the same limit as the equal-width strips become arbitrarily thin. We call that limit the definite integral and write it as $\int_a^b f(x)\,dx$ . The symbol $\int$ is an elongated S for summation, $a$ and $b$ mark the two ends of the interval, and $dx$ says we accumulate along x. The animation stops at 256 rectangles; it illustrates convergence and does not literally reach infinity.

\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\frac{b-a}{n}

What does the stretched S mean?

An elongated S for summation. It denotes the limiting accumulation, not a new area formula unrelated to rectangles.

Where do the bounds and the height come from?

$a$ and $b$ are the lower and upper bounds on $x$ . $f(x)$ supplies the signed height. When $f(x) \ge 0$ throughout, the integral is the geometric area between the curve and the x-axis.

What does dx tell us?

$dx$ identifies $x$ as the integration variable. In this construction it comes from strip widths tending to zero; we never set every finite rectangle’s width equal to zero.

See an exact limit for $x^2$ on [0, 3]

We want the exact area, beyond the last frame of the animation. Divide the interval from 0 to 3 into n equal strips. Their width is $3/n$ . The right edge of strip i is i widths from zero, so $x_i=3i/n$ . Its height is the square of that input. Multiplying height by width and adding gives:

S_n=\sum_{i=1}^n\left(\frac{3i}n\right)^2\frac3n=\frac{27}{n^3}(1^2+2^2+\cdots+n^2)

The $\Sigma$ symbol asks us to add one term for every integer $i$ from 1 to $n$ . We still need the sum of squares; merely naming a formula would leave the calculation unfinished. First add the integers in forward and reverse order. Each of the $n$ pairs is $n+1$ , and every integer has been counted twice.

\begin{aligned}&2(1+2+\cdots+n)\\&=(1+n)+(2+n-1)\\&\quad+\cdots+(n+1)\\&=n(n+1)\end{aligned}

To get squares, subtract neighbouring cubes. Expanding $(i+1)^3$ means multiplying three copies of $i+1$ : there is one $i^3$ term, three $i^2$ terms, three i terms and one 1. When we add the differences, all the middle cubes cancel.

(i+1)^3-i^3=3i^2+3i+1

\begin{aligned}&(2^3-1^3)+(3^3-2^3)\\&\quad+\cdots+((n+1)^3-n^3)\\&=(n+1)^3-1\end{aligned}

\begin{gathered}3\sum_{i=1}^n i^2\\=(n+1)^3-1-\frac{3n(n+1)}2-n\\=\frac{2n^3+3n^2+n}{2}\end{gathered}

We subtracted three times the integer sum and n copies of 1 from both sides. Divide by 3, then factor out n and factor $2n^2+3n+1$ as $(n+1)(2n+1)$ . Now substitute this result into the rectangle sum.

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

\begin{aligned}S_n&=\frac{27}{n^3}\frac{2n^3+3n^2+n}{6}\\&=9+\frac{13.5}{n}+\frac{4.5}{n^2}\\&\longrightarrow9\end{aligned}

This was the right-endpoint sum. For $x^2$ , which increases on this interval, every other rectangle height lies between the left and right choices. The right sum minus the left sum is $\frac{3}{n}\left(3^2-0^2\right)=\frac{27}{n}$ : all the intermediate heights cancel. This gap tends to zero, so every choice is squeezed to the same area 9. For the decreasing example $4-x$ , the area is already a trapezium: a 3-by-1 rectangle plus a triangle of base 3 and height 3 gives $3+\frac{3\times3}{2}=7.5$ .

Can the integral be zero while the total area is not?

An integral adds signed heights. A strip below the axis has a negative height, even though its geometric area is positive. For $\sin x$ , shifting x by π changes the sign of its height. The strips on the second half of $[0,2\pi]$ therefore cancel the matching strips on the first half. This explains why the signed integral is zero. To find the total area, we still need the size of one half.

Divide $[0,\pi]$ into n strips of width $h=\pi/n$ and use each midpoint $t_i=\left(i-\tfrac12\right)h$ . Our aim is to express each sine height as a difference of two cosines, so that neighbouring terms cancel. We will first build the relation that allows this.

Adding two arrows means putting the second at the tip of the first and finding the final endpoint. Turn this whole two-segment path as one object. Each segment keeps its length and turns through the same angle, and its endpoint remains the sum of those two segments. This is why we can turn the horizontal and vertical parts separately, then add them.

Before using an angle-addition formula, let us see where it comes from. A unit arrow at angle t has coordinates $(\cos t,\sin t)$ : $\cos t$ steps in the horizontal direction $e_1=(1,0)$ , then $\sin t$ steps in the vertical direction $e_2=(0,1)$ . Turn both directions through b. The first ends at $(\cos b,\sin b)$ . The second stays a quarter-turn ahead, so right becomes up and up becomes left: its destination is $(-\sin b,\cos b)$ . Keep the same two amounts and add their new horizontal and vertical parts.

\begin{gathered}\begin{pmatrix}\cos(t+b)\\\sin(t+b)\end{pmatrix}\\=\cos t\begin{pmatrix}\cos b\\\sin b\end{pmatrix}+\sin t\begin{pmatrix}-\sin b\\\cos b\end{pmatrix}\\=\begin{pmatrix}\cos t\cos b-\sin t\sin b\\\sin t\cos b+\cos t\sin b\end{pmatrix}\end{gathered}

Turning through −b reflects across the horizontal axis: $\cos(-b)=\cos b$ and $\sin(-b)=-\sin b$ . This gives the subtraction formulas by changing only the signs of the sine-b terms.

Subtract the two horizontal coordinates. The $\cos t\cos b$ parts cancel; the two $\sin t\sin b$ parts add.

\cos(t-b)-\cos(t+b)=2\sin t\sin b

h\sin t_i=\frac{h}{2\sin(h/2)}\left[\cos((i-1)h)-\cos(ih)\right]

Here we chose $b=h/2$ because subtracting or adding half a width takes the midpoint to the two edges. On adding the strips, $-\cos h$ cancels $+\cos h$ , and the same happens at every inside edge. Only the first and last edges remain.

M_n=\frac{h}{2\sin(h/2)}[\cos0-\cos\pi]=\frac{h}{\sin(h/2)}

Why does $u/\sin u$ approach 1?

Build three nested regions. Their areas will give us an upper and a lower bound.

OA is one radius, so its length is $1$ . B is on the same circle. Its height above OA is $\sin u$ . Triangle OAB therefore has area $\tfrac12\times1\times\sin u$ .

u=40^\circ\approx0.698

radians

Inside triangle

\frac{\sin u}{2}\approx0.3214

Why does this approach 2? On a unit circle, for a small positive angle u in radians, the triangle inside its sector has area $\tfrac12\sin u$ , the sector has area $\tfrac u2$ , and the triangle bounded by the tangent has area $\tfrac12\tan u$ . Their inclusion gives $\sin u\le u\le\frac{\sin u}{\cos u}$ . Dividing by $\sin u>0$ gives $1\le\frac{u}{\sin u}\le\frac{1}{\cos u}$ . The circle point approaches (1,0) as u approaches zero, so $\cos u$ approaches 1. The two bounds force $u/\sin u$ to approach 1.

\begin{gathered}M_n=2\frac{h/2}{\sin(h/2)}\longrightarrow2\\\int_0^{2\pi}\sin x\,dx=2-2=0\end{gathered}

Thus each half has geometric area 2, giving total area $2+2=4$ . Taking $|\sin x|$ makes both halves positive before adding. Exact cancellation in a finite midpoint picture does not mean its rectangles fit the curve exactly.

We used equal-width partitions for clarity. General Riemann integration allows unequal widths, provided the largest strip width tends to zero and the sums approach a common value independent of sample choices.

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