Integrals Turn Rates into Accumulation

Suppose a flow meter reports liters per second. One reading tells us a rate, not the amount of water that has passed. To recover volume, we must combine the rate over time. If the rate changes continuously, ordinary multiplication is only an approximation. The definite integral is the limit of increasingly accurate finite sums.

Learning objectives

By the end of this lesson, you should be able to:

1. construct a Riemann sum from a partition;
2. explain the roles of $f(x_i^*)$ and $\Delta x_i$;
3. distinguish signed accumulation from geometric area;
4. predict the units of an integral;
5. use both parts of the Fundamental Theorem of Calculus;
6. distinguish a definite integral from an indefinite integral.

Activate the prerequisite: finite accumulation

If velocity is constant at $6\,\mathrm{m/s}$ for $4\,\mathrm{s}$, displacement is

$$(6\,\mathrm{m/s})(4\,\mathrm{s})=24\,\mathrm m.$$

If velocity is $2\,\mathrm{m/s}$ for two seconds and then $5\,\mathrm{m/s}$ for three seconds, add the contributions:

$$(2)(2)+(5)(3)=19\,\mathrm m.$$

This is already integration in finite form: rate times interval, summed across intervals.

From rectangles to a Riemann sum

Partition $[a,b]$ at points

$$a=x_0<x_1<\cdots<x_n=b.$$

The $i$th width is $\Delta x_i=x_i-x_{i-1}$. Choose a sample point $x_i^*$ in that subinterval. If $f$ changes little there, the contribution is approximately

$$f(x_i^*)\Delta x_i.$$

Adding every contribution gives the Riemann sum

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

The sample value gives a local rate, height, or density. The width supplies the amount of input over which that value acts. Their product has the units of an accumulated quantity.

Worked construction: integrate $f(x)=x$

Divide $[0,1]$ into $n$ equal pieces. Then $\Delta x=1/n$. Using right endpoints, $x_i^*=i/n$, so

$$R_n=\sum_{i=1}^{n}\frac{i}{n}\frac{1}{n} =\frac{1}{n^2}\sum_{i=1}^{n}i.$$

Using $\sum_{i=1}^n i=n(n+1)/2$,

$$R_n=\frac{n(n+1)}{2n^2} =\frac12+\frac{1}{2n}.$$

Therefore

$$\int_0^1x\,dx=\lim_{n\to\infty}R_n=\frac12.$$

The right sums overestimate because $f(x)=x$ is increasing. Left sums underestimate. Both squeeze toward the same value.

Sign and geometric area

The integral is signed accumulation. Contributions above the horizontal axis are positive; those below are negative.

For $f(x)=x$ on $[-1,1]$,

$$\int_{-1}^{1}x\,dx=0$$

because the negative and positive contributions cancel. The geometric area is not zero:

$$\int_{-1}^{1}|x|\,dx=1.$$

Units expose meaning

If $v(t)$ is in meters per second and $dt$ is in seconds, then $v(t)dt$ is in meters. If density $\rho(x)$ is kilograms per meter and $dx$ is meters, then $\rho(x)dx$ is kilograms.

Treating $dx$ as a reminder of the input variable and its units prevents many modeling errors.

The accumulation function

Define

$$A(x)=\int_a^x f(t)\,dt.$$

If the upper endpoint increases by a small $h$, the new accumulated amount is approximately a thin rectangle:

$$A(x+h)-A(x)\approx f(x)h.$$

Dividing by $h$ suggests $A'(x)\approx f(x)$. The limiting statement is exact.

Worked application: displacement versus distance

Let velocity be $v(t)=3t^2-6t$ on $0\le t\le3$. An antiderivative is $V(t)=t^3-3t^2$, so displacement is

$$\int_0^3v(t)\,dt=V(3)-V(0)=0.$$

Zero displacement does not mean no motion. Since $v(t)=3t(t-2)$ changes sign at $t=2$, total distance is

$$-\int_0^2v(t)\,dt+\int_2^3v(t)\,dt =4+4=8.$$

The sign split is a conceptual step, not merely algebra.

Definite and indefinite integrals

The definite integral $\int_a^b f(x)dx$ is a number. The notation

$$\int f(x)\,dx=F(x)+C$$

denotes a family of antiderivatives. The constant $C$ is required because differentiation erases constants. In $F(b)-F(a)$, the same constant cancels.

Check your understanding

1. If $f$ is negative everywhere on $[a,b]$, what sign must $\int_a^b f(x)dx$ have?
2. A flow rate is measured in gallons per minute. What are the units of its integral with respect to minutes?
3. Why can a definite integral be zero even when the integrand is not zero?

Show the reasoning

1. Negative: every sufficiently small contribution $f(x_i^*)\Delta x_i$ is negative.
2. Gallons, because $(\text{gallons/minute})(\text{minutes})=\text{gallons}$.
3. Positive and negative contributions can cancel, as they do for an odd function over a symmetric interval.