Applications of Integration

The definite integral began as a way to find the area under a curve, but that is only the doorway. Once you understand that an integral is a way of adding up infinitely many infinitesimally thin pieces, you can measure almost anything that varies continuously: the area trapped between two curves, the volume of a solid spun on a lathe, the length of a winding path, and the true average of a quantity that never stops changing. This page is where calculus stops being abstract and starts measuring the physical world.

The single idea underneath everything here is slice, approximate, sum, and take the limit. If you can describe one thin slice of your object and add up all the slices, integration does the rest. Master that mental move and every formula below becomes something you can rebuild from scratch rather than memorize.

Learning Objectives

• Compute the area between two curves using vertical and horizontal slices.
• Find volumes of revolution using the disk/washer method and the shell method, and know when each is easier.
• Calculate the arc length of a smooth curve.
• Determine the average value of a function and interpret the Mean Value Theorem for Integrals.
• Recognize the common "slice it up" pattern that unifies all of these applications.

Quick Answer

Every application here comes from integrating a slice. The area between curves $y = f(x)$ (top) and $y = g(x)$ (bottom) is $\int_a^b [f(x) - g(x)]\,dx$. A volume of revolution by disks is $\int_a^b \pi [R(x)]^2\,dx$; by cylindrical shells it is $\int_a^b 2\pi x\,h(x)\,dx$. The arc length of $y = f(x)$ is $\int_a^b \sqrt{1 + [f'(x)]^2}\,dx$. The average value of $f$ on $[a,b]$ is $\frac{1}{b-a}\int_a^b f(x)\,dx$. In each case you build the integrand by describing one representative slice.

Where It Came From

Long before calculus had a name, Archimedes of Syracuse (c. 287–212 BCE) was already doing integration in disguise. Faced with a genuine problem — how do you find the area of a curved region or the volume of a curved solid when geometry only gives you formulas for straight-edged shapes? — he invented the method of exhaustion. He would trap a curved region between inscribed and circumscribed polygons, then squeeze the two estimates together. Using this, he proved that the volume of a sphere is exactly two-thirds the volume of the cylinder that encloses it, a result he was so proud of that he asked for a sphere-and-cylinder to be carved on his tomb.

What Archimedes lacked was a systematic engine. Each result required a fresh, ingenious geometric argument. The need that drove the invention of integral calculus was precisely this: mathematicians and physicists in the 1600s — computing areas, centers of gravity, distances traveled under changing velocity, and the volumes of wine barrels (Kepler literally wrote a treatise on barrel volumes in 1615) — wanted one general procedure instead of a thousand clever tricks. Newton and Leibniz (late 1600s) supplied it: the Fundamental Theorem of Calculus turned Archimedes' exhausting limiting arguments into routine antidifferentiation. The applications on this page are the modern, mechanized versions of the very questions Archimedes struggled with by hand.

Area Between Curves

To find the area of the region enclosed between two curves, slice it into thin vertical strips. Each strip has width $dx$ and height equal to (top curve − bottom curve). Summing the strips gives

$A = \int_a^b [f(x) - g(x)]\,dx$

where $f(x) \ge g(x)$ throughout $[a,b]$. The order matters: always top minus bottom, so the height is never negative.

Worked example. Find the area between $y = x + 2$ and $y = x^2$.

First find where they intersect: $x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0$, so $x = -1$ and $x = 2$. Between these points, test $x = 0$: the line gives $2$, the parabola gives $0$, so the line is on top. Then

$A = \int_{-1}^{2} \big[(x + 2) - x^2\big]\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}.$

At $x = 2$: $\frac{4}{2} + 4 - \frac{8}{3} = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$.

At $x = -1$: $\frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}$.

So $A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} = 4.5$.

When curves are easier to describe as functions of $y$ (for example a sideways-opening parabola), slice horizontally instead: $A = \int_c^d [x_{\text{right}}(y) - x_{\text{left}}(y)]\,dy$.

Volumes of Revolution: Disks and Washers

Spin a region around an axis and it sweeps out a solid. Slice the solid perpendicular to the axis of rotation and each slice is a thin disk (or, if there is a hole, a washer). A disk of radius $R$ and thickness $dx$ has volume $\pi R^2\,dx$, so

$V = \int_a^b \pi [R(x)]^2\,dx.$

If the solid has a hole — an outer radius $R(x)$ and inner radius $r(x)$ — subtract the missing core:

$V = \int_a^b \pi\big([R(x)]^2 - [r(x)]^2\big)\,dx.$

Worked example (disk). Rotate the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ about the $x$-axis. Here $R(x) = \sqrt{x}$, so $R^2 = x$ and

$V = \int_0^4 \pi x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi.$

Worked example (washer). Rotate the region between $y = x$ and $y = x^2$ about the $x$-axis. On $[0,1]$ the line $y = x$ is above $y = x^2$, so the outer radius is $R = x$ and the inner radius is $r = x^2$:

$V = \int_0^1 \pi\big(x^2 - x^4\big)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \pi \cdot \frac{2}{15} = \frac{2\pi}{15}.$

Volumes of Revolution: Cylindrical Shells

Sometimes slicing perpendicular to the axis forces you to solve for $x$ in terms of $y$ (or leaves an awkward washer). The shell method slices parallel to the axis instead. Imagine peeling the solid into nested cylindrical shells, like tree rings. A shell at radius $x$, height $h(x)$, thickness $dx$ has volume $2\pi x\,h(x)\,dx$ (circumference × height × thickness), giving

$V = \int_a^b 2\pi x\,h(x)\,dx$

for rotation about the $y$-axis.

Worked example (shell). Rotate the region under $y = x^2$ from $x = 0$ to $x = 2$ about the $y$-axis. Each shell has radius $x$ and height $h(x) = x^2$:

$V = \int_0^2 2\pi x \cdot x^2\,dx = 2\pi \int_0^2 x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^2 = 2\pi \cdot \frac{16}{4} = 8\pi.$

Doing this by disks would require rewriting the boundary as $x = \sqrt{y}$ and integrating in $y$ with a washer — more work. Shells shine when the axis of rotation is the $y$-axis and the region is naturally described by $y = f(x)$.

Rule of thumb: if the slice is perpendicular to the axis, use disks/washers; if parallel, use shells. Pick whichever avoids solving for the other variable.

Arc Length

How long is a curve? Straighten a tiny piece and it is the hypotenuse of a right triangle with legs $dx$ and $dy$: its length is $\sqrt{dx^2 + dy^2} = \sqrt{1 + (dy/dx)^2}\,dx$. Summing gives

$L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx.$

Worked example. Find the arc length of $y = \frac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$.

Then $f'(x) = x^{1/2}$, so $[f'(x)]^2 = x$ and

$L = \int_0^3 \sqrt{1 + x}\,dx = \left[\frac{2}{3}(1 + x)^{3/2}\right]_0^3 = \frac{2}{3}\big(4^{3/2} - 1^{3/2}\big) = \frac{2}{3}(8 - 1) = \frac{14}{3}.$

The integrand $\sqrt{1 + [f'(x)]^2}$ is usually messy, which is why arc-length problems are often set up carefully so the square root simplifies to a perfect square — as it did here.

Average Value of a Function

The average of finitely many numbers is their sum divided by the count. For a continuously varying quantity, "sum" becomes an integral and "count" becomes the length of the interval:

$f_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx.$

Geometrically, $f_{\text{avg}}$ is the height of the rectangle over $[a,b]$ that has the same area as the region under $f$. The Mean Value Theorem for Integrals guarantees that a continuous $f$ actually attains this average value at some point $c$ in $[a,b]$.

Worked example. Find the average value of $f(x) = x^2$ on $[0, 3]$.

$f_{\text{avg}} = \frac{1}{3 - 0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}\cdot \frac{27}{3} = \frac{1}{3}\cdot 9 = 3.$

So on average the function equals $3$; and indeed $f(c) = 3$ when $c = \sqrt{3} \approx 1.73$, a point inside $[0,3]$, exactly as the Mean Value Theorem promises.

Real-World Applications

• Engineering and manufacturing: Volumes of revolution model any object made on a lathe — pistons, funnels, bottles, turbine housings. Engineers compute material volume and mass this way.
• Physics: The average value integral gives root-mean-square voltage in AC circuits and average power over a cycle; the area between a velocity curve and the time axis gives displacement.
• Medicine: Cardiac output and drug concentration studies use "area under the curve" (AUC) of concentration-versus-time graphs — a direct area-between-curves computation.
• Cartography and navigation: Arc length measures the true distance along a curved road, coastline, or cable, not the straight-line shortcut.
• Economics: The area between a demand curve and the price line gives consumer surplus; between supply and price, producer surplus.

Common Mistakes

Mistake 1: Forgetting to check which curve is on top. Students plug curves into $\int (f - g)\,dx$ in the wrong order and get a negative area. Why it is wrong: area is a magnitude; a negative result signals the subtraction is reversed. Correction: test a point inside the interval to see which curve is higher, or if curves cross, split the integral at each crossing and add absolute values.

Mistake 2: Squaring a difference of radii instead of the difference of squares in the washer method. Writing $\pi\int (R - r)^2\,dx$ instead of $\pi\int (R^2 - r^2)\,dx$. Why it is wrong: the washer's area is the big disk minus the small disk, $\pi R^2 - \pi r^2$, and $(R-r)^2 \neq R^2 - r^2$. Correction: always square each radius separately, then subtract.

Mistake 3: Dropping the "1 +" in arc length. Writing $\int \sqrt{[f'(x)]^2}\,dx = \int |f'(x)|\,dx$. Why it is wrong: that would measure only vertical change, ignoring horizontal travel. Correction: the tiny hypotenuse always includes the $dx$ leg, so the integrand is $\sqrt{1 + [f'(x)]^2}$.

Comparison and Connections

The disk and shell methods answer the same question with different slices. Choose based on the geometry, not habit.

Feature	Disk / Washer	Shell
Slice direction	Perpendicular to axis	Parallel to axis
Typical variable	Same as axis direction	Opposite the axis direction
Integrand	$\pi(R^2 - r^2)$	$2\pi x \cdot h(x)$
Best when	Region described in the axis's variable	Region described in the other variable
Handles holes	Naturally (washer)	Naturally (nested rings)

All four applications are the same recursion: average value divides an integral by length; area integrates a height; volume integrates a cross-sectional area; arc length integrates a differential length. They differ only in what one slice contributes.

Practice Questions

Recall

State the formula for the average value of $f$ on $[a,b]$.

Answer: $f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$.

Understanding

Explain in one or two sentences why the washer method uses $R^2 - r^2$ rather than $(R - r)^2$.

Guidance: The cross-section is an annulus whose area is the outer disk minus the inner disk, $\pi R^2 - \pi r^2$. Because squaring is not linear, $(R-r)^2$ would give a completely different (and wrong) area.

Application

Find the volume when the region under $y = \sqrt{x}$ on $[0, 4]$ is rotated about the $y$-axis using shells.

Answer: Shells give $V = \int_0^4 2\pi x \sqrt{x}\,dx = 2\pi\int_0^4 x^{3/2}\,dx = 2\pi\left[\frac{2}{5}x^{5/2}\right]_0^4 = 2\pi \cdot \frac{2}{5}\cdot 32 = \frac{128\pi}{5}$.

Analysis

The region between $y = x$ and $y = x^2$ on $[0,1]$ is rotated about the $x$-axis (washers) and separately about the $y$-axis (shells). Without fully computing both, explain why the two volumes differ.

Guidance: Rotation axis changes the radius of every slice. About the $x$-axis, radii are the $y$-values; about the $y$-axis, radii are the $x$-values. Different radii mean different swept volumes, so the results cannot match. (For the record, about the $x$-axis it is $\frac{2\pi}{15}$; about the $y$-axis, shells give $\int_0^1 2\pi x(x - x^2)\,dx = \frac{\pi}{6}$.)

FAQ

Q: How do I know whether to integrate with respect to $x$ or $y$? Look at how the region is bounded. If the top and bottom boundaries are functions of $x$, slice vertically ($dx$). If left and right boundaries are functions of $y$, slice horizontally ($dy$). Choose the one that keeps a single formula across the whole region.

Q: When should I use shells instead of disks? Use shells when disks would force you to invert your function (solve $y = f(x)$ for $x$) or when rotating about the $y$-axis a region given as $y = f(x)$. If the slice you naturally draw is parallel to the axis, that is a shell.

Q: Why does arc length so often give an ugly integral? Because $\sqrt{1 + [f'(x)]^2}$ rarely has an elementary antiderivative. Textbook problems are engineered so the expression under the root becomes a perfect square. Real-world arc lengths are usually computed numerically.

Q: Is the average value the same as the midpoint value $f(\frac{a+b}{2})$? No. The average value weights every point equally through the integral; the midpoint value is just one sample. They coincide only for linear functions.

Q: Can area between curves ever be zero even if the curves are different? Yes — if the curves cross so that one is above on part of the interval and below on the rest, the signed integral can cancel to zero. That is why you split at crossings and add the absolute areas; the geometric area is never zero for distinct curves.

Quick Revision

• Area between curves: $\int_a^b (\text{top} - \text{bottom})\,dx$; split at intersections.
• Disk/Washer: $V = \int \pi(R^2 - r^2)\,dx$ — slices perpendicular to axis.
• Shell: $V = \int 2\pi x\,h(x)\,dx$ — slices parallel to axis.
• Arc length: $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$.
• Average value: $f_{\text{avg}} = \frac{1}{b-a}\int_a^b f\,dx$; attained at some $c$ (MVT for Integrals).
• Universal move: describe one slice, integrate.

Related Topics

Prerequisites

• The Definite Integral and its meaning (Calculus overview)
• Fundamental Theorem of Calculus and antidifferentiation

Related Topics

• Techniques of Integration (substitution, parts) for evaluating the resulting integrals
• Solids with known cross-sections (a generalization of the disk method)

Next Topics

• Differential Equations, where integration models change over time — see Differential Equations
• Improper integrals and applications to probability