Integrals: add up a changing quantity

At a constant speed of 6 metres per second, travelling for 2 seconds covers $6\cdot2=12$ metres. Multiplication works because the speed is constant. An integral extends that calculation to a speed that changes.

Distance is an area

On a speed-versus-time graph, height is speed and width is elapsed time. A rectangle’s area has units $(\text{metres}/\text{second})\cdot\text{seconds}=\text{metres}$.

When speed changes, split the interval into short pieces. On each piece, use one speed reading times the piece’s duration as an approximate distance. Add those distances. For continuous speed, the approximation approaches the exact distance as the pieces become finer.

The staircase closes in

To see the arithmetic, find the area under $f(x)=x^2$ from 0 to 1. With four equal pieces, each width is $\Delta x=1/4$. Using each piece’s right endpoint gives

$$\frac14\left[\left(\frac14\right)^2+ \left(\frac24\right)^2+ \left(\frac34\right)^2+1^2\right] =\frac{30}{64}=0.46875.$$

Because $x^2$ increases on this interval, right-endpoint rectangles overestimate the area. Left-endpoint rectangles underestimate it.

Number of pieces	Left-endpoint sum	Right-endpoint sum
4	0.21875	0.46875
10	0.285	0.385
100	0.32835	0.33835

Both approach $1/3$. Their gap is $1/n$, so the area is squeezed into a smaller interval as n increases.

Increase the number of rectangles. For an increasing curve, left sums lie below the integral and right sums lie above it. This ordering does not hold for every curve.

The symbol

We write the limiting total as

$$\int_0^1 x^2\,dx=\frac13.$$

Part	Meaning
$\int$	Accumulate by taking a limit of sums
0 and 1	Start and end of the interval
$x^2$	Function whose values supply rectangle heights
$dx$	Integrate with respect to x

The finite rectangles have width $\Delta x$. The notation dx identifies the integration variable in the limit; it does not mean we set every rectangle’s width to zero and add zeros.

Below the axis counts negative

A definite integral accumulates signed area. For $f(x)=x-1$ on $[0,2]$, the lower triangle contributes $-1/2$ and the upper triangle contributes $1/2$. The integral is zero, although the total geometric area is 1.

This distinction matters for motion. Integrating velocity gives displacement, including direction. Integrating speed, the absolute value of velocity in one dimension, gives distance travelled. A round trip can have zero displacement and positive distance.

Rules for adding areas

Splitting an interval splits the total:

$$\int_a^c f(x)\,dx=\int_a^b f(x)\,dx+\int_b^c f(x)\,dx.$$

Constants can be pulled outside and sums integrated term by term. Reversing the bounds changes the sign. With identical bounds, the integral is zero.

Average height and weighted averages

The average height of f over $[a,b]$, where $a<b$, is

$$\frac1{b-a}\int_a^b f(x)\,dx.$$

This weights equal-width intervals equally. A probability expectation can use unequal weights: $\mathbb E[X]=\int x f_X(x)\,dx$ weights each value by the probability density there. Only for a uniform distribution does it reduce to an unweighted interval average.

Why the area under x² is exactly 1/3

The right-endpoint sum with n equal pieces is

$$\frac{1^2+2^2+\cdots+n^2}{n^3} =\frac{n(n+1)(2n+1)}{6n^3} =\frac{(1+1/n)(2+1/n)}6.$$

Its limit is $2/6=1/3$. This calculation uses the identity for the sum of the first n squares.

Next

The fundamental theorem explains how an antiderivative can calculate this total without adding hundreds of rectangles.