Application Properties of Integrals

Integration is not only an operation inside pure mathematics. It is the bridge between abstract theory and the physical world. The average temperature over a day, the probability that a random variable falls in a given interval, the total energy burned by a circuit — these look like different problems, but all of them reduce to the same thing: integrating a function over an interval.

This page takes three classic settings and shows what the definite integral does in each of them: the average value of a continuously varying quantity, the probability carried by a density curve, and the energy released by a varying power. Each section moves from the general formula to a fully worked example, so that you can watch the whole chain from “identify the integrand” to “read off the number”.

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1. Average value

In daily life we constantly need the typical level of a quantity that varies continuously: the average temperature of a day, the average heart rate during a lecture, the average speed of a car over a trip. Because the quantity changes continuously, a plain arithmetic mean of a few sample readings is not the right tool. The correct notion is the average value of a continuous function.

Intuition. Think of $\int_a^b f(x)\,dx$ as the “sum” of infinitely many values of $f$ over the interval. Dividing that sum by the length $b - a$ of the interval is exactly what you do to an ordinary finite list of numbers to get their mean. The integral is the continuous analogue of the arithmetic average.

Example: average temperature over a day.

The temperature in a certain town during one day, for $0 \leq t \leq 24$ hours, is approximately

$T(t) = 20 + 5\sin\!\left(\frac{\pi t}{12}\right)$

in degrees Celsius. Find the average temperature over the 24 hours.

Solution: substitute into the average value formula:

$T_{\text{avg}} = \frac{1}{24 - 0}\int_0^{24} \left[20 + 5\sin\!\left(\frac{\pi t}{12}\right)\right] dt$

Split the integral:

$T_{\text{avg}} = \frac{1}{24}\left[\int_0^{24} 20\, dt + \int_0^{24} 5\sin\!\left(\frac{\pi t}{12}\right) dt\right]$

First part:

$\int_0^{24} 20\, dt = 20 \times 24 = 480$

Second part: let $u = \dfrac{\pi t}{12}$, so $du = \dfrac{\pi}{12}\,dt$, that is $dt = \dfrac{12}{\pi}\,du$. The limits become $u = 0$ and $u = 2\pi$.

$\int_0^{24} 5\sin\!\left(\frac{\pi t}{12}\right) dt = 5 \cdot \frac{12}{\pi}\int_0^{2\pi} \sin u\, du = \frac{60}{\pi}\Big[-\cos u\Big]_0^{2\pi} = \frac{60}{\pi}(-1+1) = 0$

The sine term contributes nothing, which matches intuition: the extra warmth of the afternoon is exactly cancelled by the cooling of the night. Therefore

$T_{\text{avg}} = \frac{1}{24}(480 + 0) = 20 \text{ °C}$

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2. Probability density

In probability theory a continuous random variable cannot take an exact value with positive probability — the probability of any single point is zero. What matters instead is the probability that the variable lands in an interval, and that is precisely what the integral of a probability density function (PDF) computes.

Intuition. Picture $f(x)$ as a curve. The probability that $X$ falls in $[a, b]$ is the area under that curve between $a$ and $b$. Since the total area is 1 — the “certain event” — every other area is a fraction of the whole.

Example: the exponential distribution.

The lifetime $X$ of an electronic component, measured in thousands of hours, follows an exponential distribution with parameter $\lambda = 0.5$, whose density is

$f(x) = \begin{cases} 0.5\, e^{-0.5x}, & x \ge 0 \\ 0, & x < 0 \end{cases}$

Find the probability that the component survives longer than 2 thousand hours, that is $P(X > 2)$.

Solution:

$P(X > 2) = \int_2^{+\infty} 0.5\, e^{-0.5x}\, dx$

Since $\displaystyle\int 0.5\, e^{-0.5x}\, dx = -e^{-0.5x} + C$, substitute the limits:

$P(X > 2) = \left[-e^{-0.5x}\right]_2^{+\infty} = 0 - \left(-e^{-1}\right) = e^{-1} \approx 0.368$

So the component has about a 36.8 percent chance of lasting more than 2000 hours. The memoryless property of the exponential distribution makes exactly this kind of computation routine in reliability engineering.

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3. Energy

In physics and engineering, power is the rate at which energy is transferred, that is $P = \frac{dE}{dt}$. The reverse direction is therefore an integral: knowing the power as a function of time, the total energy consumed over an interval is the integral of the power.

The same idea works well beyond circuits: in mechanics the integral of force over displacement is work, and in thermodynamics the integral of heat flow over time is total heat. Each is “a rate, accumulated over an interval”.

Example: energy dissipated in a resistor.

The instantaneous power dissipated in a resistor is $P(t) = 100\sin^2(t)$ watts. Find the total energy dissipated between $t = 0$ and $t = \pi$ seconds.

Solution:

$E = \int_{0}^{\pi} 100\sin^2(t)\, dt$

Use the half-angle formula $\sin^2(t) = \dfrac{1 - \cos(2t)}{2}$:

$E = 100\int_0^{\pi} \frac{1 - \cos(2t)}{2}\, dt = 50\int_0^{\pi} \left[1 - \cos(2t)\right] dt$

$= 50\left[t - \frac{\sin(2t)}{2}\right]_0^{\pi} = 50\left[\pi - 0 - (0 - 0)\right] = 50\pi \approx 157.1 \text{ J}$

There is a quick sanity check available here. The waveform $\sin^2(t)$ never goes negative — a resistor only consumes energy, it never generates it — and its average over a half period is exactly half the peak, namely 50 W. So the total energy is 50 W times $\pi$ seconds, which is $50\pi$ joules.

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At a glance

The common essence. All three express the same idea: accumulating by summing. Whether you are averaging infinitely many instantaneous values, adding up infinitely many infinitesimal probabilities, or collecting infinitely many instantaneous power readings, the integral turns local information into a global result.

Working advice: whenever you meet a concrete problem, ask two questions first — what is the integrand, and what is the interval? Once those are identified, the correct physical or statistical meaning usually follows on its own.

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Exercises

Exercise 1

Let $f(x) = x^2$ on $[0, 3]$. Find the average value of $f$ on this interval, and explain why it is not the arithmetic mean of the endpoint values $f(0)$ and $f(3)$.

Exercise 2

The lifetime $X$ of a component follows the exponential distribution with density $f(x) = 0.5\, e^{-0.5x}$ for $x \geq 0$ and $f(x) = 0$ otherwise. Find the probability that the component fails within 1 thousand hours, that is $P(X \leq 1)$.

Exercise 3

The instantaneous power supplied to a device is $P(t) = 200\sin(2t)$ watts for $0 \leq t \leq \frac{\pi}{2}$. Find the total energy supplied over that interval.

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Summary

The definite integral is the universal tool for turning a rate into a total. Averages, probabilities and energies look unrelated on the surface, but each one is an accumulation over an interval, and each one is found by the same two steps: identify the integrand, then identify the limits.

Symbols Used in This Article