Applications of Integration

A structured guide to modeling accumulated change with definite integrals in geometry, motion, average-value problems, and net-change applications.

The Modeling Idea Behind Applications

Definite integrals model totals by adding infinitely many small contributions. The general pattern is

\text{total amount}=\int \text{rate or density}\,d(\text{independent variable}).

A small contribution may be the area of a thin rectangle, the volume of a thin slice, or the change produced by a rate over a short time interval. The most important modeling question is therefore: what quantity does one representative slice contribute?

A reliable setup identifies four features:

The independent variable and its limits.
The size of one small contribution.
The geometric or physical interpretation of the integrand.
The units of the resulting integral.

For example, integrating velocity with respect to time produces distance units, while integrating a cross-sectional area with respect to length produces volume units.

Takeaway: Every application formula comes from adding many small contributions and taking a limit.

Finding Area from Boundary Differences

For vertical slices, the height of a representative rectangle is the upper boundary minus the lower boundary. If $f(x)\ge g(x)$ on $[a,b]$ , then

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

For horizontal slices, use the right boundary minus the left boundary:

A=\int_c^d [R(y)-L(y)]\,dy.

A systematic procedure is:

Sketch the curves and identify the bounded region.
Find the intersection points that determine the limits.
Choose vertical or horizontal slices.
Write the appropriate boundary difference.
Split the interval if the boundary changes.
Evaluate the integral and report square units.

For example, the curves $y=x$ and $y=x^2$ intersect where

x=x^2,

so $x=0$ and $x=1$ . On $[0,1]$ , the line is above the parabola, giving

A=\int_0^1(x-x^2)\,dx =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 =\frac16.

The phrase describes this general boundary-difference method. The order of subtraction is essential: use top minus bottom or right minus left, not the reverse.

Takeaway: Sketch first, determine which boundary is farther in the slicing direction, and split whenever that relationship changes.

Volumes from Cross-Sections

A solid can be divided into thin slices whose cross-sectional area is known. If the cross-sectional area perpendicular to the direction of integration is $A(x)$ , then

V=\int_a^b A(x)\,dx.

For a disk, the cross-section is a solid circle. If its radius is $R(x)$ , then

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

For a washer, the cross-section has an outer radius $R(x)$ and an inner radius $r(x)$ :

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

The radius is a distance from the axis of rotation. It is not automatically the function value, especially when the axis is not a coordinate axis or when the region lies below an axis.

For a sphere of radius $2$ , the upper semicircle is

y=\sqrt{4-x^2}.

A perpendicular slice is a disk with radius $y$ , so

V=\pi\int_{-2}^{2}(4-x^2)\,dx =\frac{32\pi}{3}.

More generally, cross-sections need not be circular. If each slice is a square with side length $s(x)$ , then

V=\int_a^b [s(x)]^2\,dx.

Takeaway: Translate one representative cross-section into its area before writing the volume integral.

Volumes by Rotating Slices

When a vertical rectangle is revolved around the $y$ -axis, it forms a thin cylindrical shell. At position $x$ , the shell has radius $x$ , circumference $2\pi x$ , height $f(x)-g(x)$ , and thickness $dx$ . Therefore,

dV=(2\pi x)(f(x)-g(x))\,dx,

and

V=2\pi\int_a^b x[f(x)-g(x)]\,dx.

For rotation about a vertical line $x=k$ , the radius is $|x-k|$ . For horizontal shells, use horizontal slices and integrate with respect to $y$ .

Consider the region under $y=x$ from $x=0$ to $x=2$ , rotated around the $y$ -axis. Each shell has radius $x$ and height $x$ , so

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

Choose between and shells by comparing the geometry and the algebra. Shells are often simpler when slices parallel to the axis have an easy radius and height, while may be preferable when perpendicular slices have straightforward inner and outer radii.

Takeaway: For shells, identify radius, height, and thickness; then multiply circumference by height before integrating.

Average Values of Functions

For a continuous function $f$ on $[a,b]$ , the is

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

The integral represents the total accumulated output over the interval, and division by $b-a$ converts that total into an average per unit of the independent variable.

This differs from average rate of change:

\text{average rate of change}=\frac{f(b)-f(a)}{b-a}.

The first quantity averages function values; the second is a slope between two points.

For $f(x)=x^2$ on $[0,3]$ ,

f_{\text{avg}}=\frac{1}{3}\int_0^3x^2\,dx =\frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 =3.

Because the function is continuous, there is some $c\in[0,3]$ for which $f(c)=3$ . In this example, $c=\sqrt{3}$ .

Takeaway: is total accumulated output divided by interval length, not a difference quotient.

Accumulation, , and Motion

If $r(t)$ is the rate of change of a quantity, then the accumulated change from $a$ to $x$ is

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

The Fundamental Theorem of Calculus states that

A'(x)=r(x).

The sign of the rate controls the behavior of the :

If $r(x)>0$ , then $A$ is increasing.
If $r(x)<0$ , then $A$ is decreasing.
If the rate changes from positive to negative, the has a local maximum.
If the rate changes from negative to positive, the has a local minimum.

The net-change form is

Q(b)-Q(a)=\int_a^b r(t)\,dt.

If $Q(a)=Q_0$ , then

Q(t)=Q_0+\int_a^t r(u)\,du.

This says final amount equals initial amount plus accumulated .

For motion, position $s(t)$ , velocity $v(t)$ , and acceleration $a(t)$ satisfy

v(t)=s'(t),\qquad a(t)=v'(t)=s''(t).

Thus,

s(b)-s(a)=\int_a^b v(t)\,dt

and

v(b)-v(a)=\int_a^b a(t)\,dt.

The integral of velocity is , while is

\int_a^b|v(t)|\,dt.

For example, if

v(t)=t^2-4t+3=(t-1)(t-3),

on $[0,4]$ , velocity changes sign at $t=1$ and $t=3$ . The is

\int_0^4v(t)\,dt=\frac43,

whereas splitting at the zeros and integrating speed gives $4$ . The difference occurs because the particle reverses direction, causing signed motion to cancel in but not in distance.

Takeaway: Integrate a rate to obtain , and inspect the sign of the rate whenever cancellation or direction changes matter.

Checking an Integral Model

Most errors in applications of integration come from an incorrect model rather than an incorrect antiderivative.

Wrong boundary order: Use top minus bottom for vertical slices and right minus left for horizontal slices.
No interval split: If curves cross or velocity changes sign, split at the relevant points.
Incorrect radius: Measure radius from the axis of rotation; do not automatically use a function value.
Missing squares: Disk and washer areas use $\pi r^2$ , not $\pi r$ .
-distance confusion: Use $\int v(t)\,dt$ for and $\int |v(t)|\,dt$ for .
Average-value confusion: Use $\frac{1}{b-a}\int_a^b f(x)\,dx$ for and $\frac{f(b)-f(a)}{b-a}$ for average rate of change.
Ignored units: Area has square units, volume has cubic units, and an integral of velocity with respect to time has distance units.

A final setup check should ask whether the integrand represents the intended small contribution, whether the limits cover the entire region or interval, and whether the resulting units match the requested quantity.

Takeaway: Verify the geometry, signs, limits, and units before evaluating.