10 Applications of Integration

Building an Integral Model

A definite integral is useful whenever a total can be divided into thin pieces whose contributions vary continuously. The central modeling pattern is

For each problem:

• Identify the total quantity being measured.

• Choose a thin piece, such as a rectangle, slice, ring, shell, rod segment, or liquid layer.

• Express the piece's contribution in terms of the variable.

• Determine the bounds from endpoints, intersections, or physical dimensions.

• Integrate and interpret the result with appropriate units.

This approach connects geometry and physical applications: the integral adds many small contributions into one total.

Area Between Curves

For vertical slices, the height of each slice is the upper function minus the lower function. Thus, if $f(x)$ is above $g(x)$ on $[a,b]$, the Area between curves is

For example, the area bounded above by $f(x)=x+4$, below by $g(x)=3-\frac{x}{2}$, and between $x=1$ and $x=4$ is

If the curves cross, find the intersection points and split the integral wherever the identity of the upper and lower curves changes. For horizontal slices, use right minus left:

Takeaway: sketch first, identify the boundaries, and use the slice direction that gives the simplest description.

Volumes by Slicing

A solid can be built by accumulating cross-sectional areas. If the cross-sectional area perpendicular to the chosen variable is $A(x)$, then

The cross sections may be squares, triangles, semicircles, or other shapes. For a base under $y=4-x^2$ and above the $x$-axis on $[-2,2]$, with square cross sections perpendicular to the $x$-axis, the side length is $s(x)=4-x^2$. Therefore,

The essential step is not the final antiderivative; it is expressing the cross-sectional area correctly.

Disks and Washers

When a region is revolved around an axis, its slices often form disks or washers. The Disk method applies when each cross section has no hole. For a region under $y=f(x)$ revolved around the $x$-axis,

For example, revolving the region under $y=\sqrt{x}$ from $x=1$ to $x=4$ gives

The Washer method applies when each cross section has a hole. If $R(x)$ is the outer radius and $r(x)$ is the inner radius, then

Always measure radii as distances from the axis of rotation. For rotation about a horizontal line $y=k$, a radius may be $|f(x)-k|$.

Volumes of Revolution

A vertical rectangle revolved around a vertical axis forms a thin shell. The shell's approximate volume is circumference times height times thickness. The Cylindrical shell formula is

For rotation about the $y$-axis, a region under $y=f(x)$ produces shells with radius $x$ and height $f(x)$:

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

Choose disks, washers, or shells according to which method produces the simplest correct setup. The axis of rotation and the direction of the slices determine the radius, height, and integration variable.

Average Value

For a continuous function on $[a,b]$, the Average value of a function is

The integral gives the accumulated total, and division by the interval length gives the amount per unit of interval. For $f(x)=x^2$ on $[0,3]$,

Geometrically, the average value is the height of a rectangle with width $b-a$ and the same area as the region under the graph:

Takeaway: average value is accumulated value divided by the length of the interval.

Mass, Work, and Pumping

The same small-piece model applies to quantities that vary in space.

Mass: If a rod has linear density $\rho(x)$ on $[a,b]$, then a segment contributes $dm=\rho(x)\,dx$, so

For a circular object with radial density $\rho(r)$, a thin ring has area approximately $2\pi r\,dr$, giving

Work: If force varies with position as $F(x)$, work over $[a,b]$ is

For a spring with $F(x)=kx$, stretching from $x=a$ to $x=b$ requires

Pumping liquid: A layer with volume $dV$ contributes weight times lifting distance. With weight density $\gamma$, cross-sectional area $A(y)$, and lifting distance $D(y)$,

The lifting distance must run from the layer's position to the destination, not merely describe the layer's thickness.

Hydrostatic Force

Pressure increases with depth in a stationary liquid. At depth $h$, pressure is

where $\gamma$ is weight density. A horizontal strip with width $w(y)$ and thickness $dy$ has area approximately $w(y)\,dy$, so its force contribution is

Consequently, the Hydrostatic force on a surface is

To set up the integral, describe the depth and width of a typical strip, choose bounds covering the entire surface, and verify that the resulting units are force units.

A General Problem-Solving Checklist

A reliable setup can be organized into seven questions:

1. What total quantity is required: area, volume, mass, work, force, or an average?

2. What thin piece makes the geometry or physics easiest to describe?

3. What is the contribution of that piece?

4. Which variable gives the simplest slice or layer description?

5. What are the correct bounds?

6. Do signs, radii, and boundary orders make sense?

7. Do the units match the requested quantity?

Common checks include using top minus bottom for vertical area slices, right minus left for horizontal area slices, outer radius squared minus inner radius squared for washers, and distance to the axis for a radius. A correct integral is the main modeling achievement; evaluation comes afterward.