Grade 12

Applications of the Integral

How slicing a region or solid into infinitely many thin pieces and summing turns area between curves, volume of revolution, and total distance traveled into computable definite integrals.

IntuitionSlicing area and volume into infinitely many thin pieces

A loaf of bread's volume can be found by slicing it into many thin pieces, measuring the area of each slice, multiplying by its thickness, and adding everything up. As the slices get thinner and thinner, this sum becomes an integral: $\int_a^b A(x)\,dx$ , where $A(x)$ is the cross-sectional area at position $x$ . The same slicing idea, applied to the region between two curves instead of a solid, gives area; applied to disks swept out by a rotating curve, it gives the volume of a solid of revolution.

Approximating the area under a curve with

n

rectangular strips. Drag

n

from a few wide strips to many thin ones and watch the sum of rectangle areas converge to the exact area given by the integral.

SchoolArea between two curves, and distance from velocity

Definition: Area between two curves

Let $f$ and $g$ be continuous on $[a,b]$ . The area of the region bounded above and below by the graphs $y=f(x)$ and $y=g(x)$ , and on the sides by $x=a$ and $x=b$ , is $S=\int_a^b |f(x)-g(x)|\,dx$ . When the curves cross inside $(a,b)$ , one solves $f(x)=g(x)$ to find the crossing points, splits $[a,b]$ into pieces where the sign of $f(x)-g(x)$ is constant, drops the absolute value on each piece, and adds the resulting integrals.

S=\int_a^b |f(x)-g(x)|\,dx

The same integral of a signed quantity, applied to velocity instead of a height difference, gives kinematics its two basic formulas. If $v(t)$ is velocity at time $t$ , the displacement (net change in position) is the signed integral of $v$ , while the total distance traveled is the integral of $|v(t)|$ — the two disagree exactly when the object reverses direction, i.e. when $v(t)$ changes sign.

\Delta x = x(t_2)-x(t_1) = \int_{t_1}^{t_2} v(t)\,dt, \qquad s = \int_{t_1}^{t_2} |v(t)|\,dt

UndergraduateTwo theorems: area between curves, and the disk method for volumes

Theorem: Area between two curves as a definite integral

Let $f,g$ be continuous on $[a,b]$ . The region $R$ bounded by $y=f(x)$ , $y=g(x)$ , $x=a$ , and $x=b$ has area $S(R)=\int_a^b|f(x)-g(x)|\,dx$ .

Why is it true?

Approximate the region by thin vertical strips of width $\Delta x$ ; each strip is nearly a rectangle of height $|f(x)-g(x)|$ , and summing these rectangle areas is exactly a Riemann sum, which converges to the definite integral as the strips shrink.

Proof

Partition $[a,b]$ into $n$ subintervals $[x_{i-1},x_i]$ of width $\Delta x=(b-a)/n$ , and pick a sample point $x_i^*$ in each. The strip of the region over $[x_{i-1},x_i]$ has height approximately $|f(x_i^*)-g(x_i^*)|$ , so its area is approximately $|f(x_i^*)-g(x_i^*)|\,\Delta x$ .

Summing over all $n$ strips gives the Riemann sum $\sum_{i=1}^n |f(x_i^*)-g(x_i^*)|\,\Delta x$ , which approximates the true area of $R$ by construction.

As $n\to\infty$ , the error in each strip's approximation goes to $0$ by uniform continuity of $f-g$ on the compact interval $[a,b]$ . Since $|f-g|$ is continuous, the Riemann sum converges to $\int_a^b|f(x)-g(x)|\,dx$ by the definition of the Riemann integral. Hence $S(R)=\int_a^b|f(x)-g(x)|\,dx$ .

When $f(x)\ge g(x)$ throughout $[a,b]$ , this simplifies to $S(R)=\int_a^b[f(x)-g(x)]\,dx$ with no absolute value needed; when the sign of $f-g$ changes at points inside $(a,b)$ , splitting the integral at those points and choosing the correct order of subtraction on each piece gives the practical computation method.

Theorem: Volume of a solid of revolution (disk method)

Let $f$ be continuous and non-negative on $[a,b]$ . Rotating the region under $y=f(x)$ about the $x$ -axis produces a solid whose volume is $V=\pi\int_a^b [f(x)]^2\,dx$ .

Why is it true?

Slicing the solid perpendicular to the $x$ -axis at position $x$ produces a circular disk of radius $f(x)$ and area $\pi[f(x)]^2$ ; stacking these disks and integrating the cross-sectional area over $[a,b]$ gives the total volume, exactly the same slicing argument used for area but with the area of a disk in place of a strip's height.

Proof

Partition $[a,b]$ into $n$ subintervals of width $\Delta x=(b-a)/n$ with sample points $x_i^*$ . The solid restricted to $[x_{i-1},x_i]$ is approximately a cylindrical disk of radius $f(x_i^*)$ and thickness $\Delta x$ , so its volume is approximately $\pi[f(x_i^*)]^2\,\Delta x$ .

Summing these disk volumes over all $n$ subintervals gives the Riemann sum $\sum_{i=1}^n \pi[f(x_i^*)]^2\,\Delta x$ , which approximates the true volume $V$ .

As $n\to\infty$ , continuity of $f$ on $[a,b]$ makes the approximation error vanish, and since $x\mapsto \pi[f(x)]^2$ is continuous, the Riemann sum converges by definition to $\pi\int_a^b[f(x)]^2\,dx$ . Hence $V=\pi\int_a^b[f(x)]^2\,dx$ .

As a consistency check, take $f(x)=\dfrac{r}{h}x$ on $[0,h]$ , whose rotation about the $x$ -axis is exactly a cone of base radius $r$ and height $h$ : $V=\pi\int_0^h \left(\dfrac{r}{h}x\right)^2 dx = \pi\dfrac{r^2}{h^2}\cdot\dfrac{h^3}{3} = \dfrac{1}{3}\pi r^2 h$ , exactly the classical cone volume formula.

UndergraduateReal-world applications: area, volume, and motion

Once area and volume are written as integrals, computing them becomes an antiderivative exercise rather than a geometric puzzle. Architects use the area-between-curves formula for irregular floor plans, engineers use the disk method to compute the volume of machined parts (bottles, nozzles, turbine blades) that have circular cross-sections but no elementary geometric shape, and physicists use the velocity integral to track a moving object's total distance even when its direction keeps reversing.

Example: Area between a parabola and a line

Find the area of the region bounded by the parabola $y=x^2$ and the line $y=x+2$ .

Solution

First find where the curves cross by solving $x^2=x+2$ , i.e. $x^2-x-2=0$ , which factors as $(x-2)(x+1)=0$ , giving $x=-1$ and $x=2$ .

On the interval $(-1,2)$ , test a point such as $x=0$ : the line gives $y=2$ and the parabola gives $y=0$ , so the line lies above the parabola there, meaning $f(x)-g(x)=(x+2)-x^2\ge 0$ on $[-1,2]$ and no absolute value splitting is needed.

The area is $S=\int_{-1}^{2}\big[(x+2)-x^2\big]\,dx=\left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^{2}$ .

At $x=2$ : $\dfrac{4}{2}+4-\dfrac{8}{3}=2+4-\dfrac{8}{3}=\dfrac{10}{3}$ . At $x=-1$ : $\dfrac{1}{2}-2+\dfrac{1}{3}=-\dfrac{7}{6}$ . Subtracting, $S=\dfrac{10}{3}-\left(-\dfrac{7}{6}\right)=\dfrac{20}{6}+\dfrac{7}{6}=\dfrac{27}{6}=\dfrac{9}{2}$ .

Example: Volume of a paraboloid-shaped water tank

A water tank is shaped like the solid obtained by rotating the curve $y=\sqrt{x}$ , for $0\le x\le 4$ (measured in meters), about the $x$ -axis. Find the volume of the tank.

Solution

By the disk method, $V=\pi\int_0^4 \big(\sqrt{x}\big)^2\,dx=\pi\int_0^4 x\,dx$ .

Computing the antiderivative, $\pi\int_0^4 x\,dx=\pi\left[\dfrac{x^2}{2}\right]_0^4=\pi\left(\dfrac{16}{2}-0\right)=8\pi$ .

So the tank holds $V=8\pi$ cubic meters, approximately $25.1$ m $^3$ — note that the radius of the tank grows like $\sqrt{x}$ , so it is much wider near $x=4$ than near $x=0$ , matching the shape of a shallow, flared basin rather than a straight-sided cylinder.

Common mistake. Two classic mistakes. (1) Forgetting to find the intersection points before integrating. To compute the area between

y=f(x)

and

y=g(x)

, one must first solve

f(x)=g(x)

to know the limits of integration and which curve is on top; skipping this step and guessing the bounds (or the sign of

f-g

) gives a wrong or even negative area. (2) Confusing displacement with total distance. If

v(t)

changes sign on

[t_1,t_2]

, then

\int_{t_1}^{t_2}v(t)\,dt

gives only the net displacement, which can be much smaller than the total distance traveled

\int_{t_1}^{t_2}|v(t)|\,dt

— for instance a ball thrown up and caught back at the same height has zero displacement but a nonzero distance traveled (up and back down).

Research frontier as of 2026

As of 2026, the slicing idea behind area and volume integrals extends into computational geometry and machine learning. Numerical integration (quadrature) methods approximate integrals of functions with no closed-form antiderivative, essential for computing volumes of complex CAD models and for evaluating high-dimensional integrals that appear in Bayesian statistics and physics simulations; when the dimension is very high, deterministic quadrature grids become infeasible and Monte Carlo integration (randomly sampling points and averaging) is used instead, with quasi-Monte Carlo methods (using low-discrepancy sequences rather than pure randomness) an active area of research for reducing the number of samples needed. A related open question in computational topology is efficiently computing volumes of high-dimensional polytopes, which is provably hard in general (no known polynomial-time exact algorithm), driving research into randomized approximation algorithms with provable error bounds.

What is the area of the region bounded by $y=x^2$ and $y=x+2$ ?

Rotating $y=\sqrt{x}$ on $[0,4]$ about the $x$ -axis gives a solid of what volume?

A ball is thrown up and caught at the same height it was thrown from. What can be said about its displacement and total distance?

To find the area between $y=f(x)$ and $y=g(x)$ , what must be done first?

References

Michael Spivak (2008). Calculus
James Stewart (2015). Calculus: Early Transcendentals