Applications of Integration

Computing areas between curves, volumes of solids of revolution, arc lengths, and other accumulated quantities.

Area between curves: subtract bottom from top over the interval

Area = integral of [f(x) - g(x)] from x = -0.30 to x = 3.30.

Definition

The integral is a measuring machine. The same operation that computes area can compute length, volume, mass, work, probability — anything that can be expressed as an accumulation of infinitely thin slices.

Area between two curves: the area between $f(x)$ and $g(x)$ over $[a, b]$ (where $f \geq g$ ):

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

Displacement and distance: if $v(t)$ is velocity, then:

Displacement (signed): $\int_a^b v(t)\,dt$
Distance traveled (always positive): $\int_a^b |v(t)|\,dt$

Key properties

Every application here follows the same template: identify an infinitesimal contribution, write it as $f(x)\,dx$ , integrate
Displacement and distance traveled coincide exactly when velocity never changes sign
All these formulas reduce to ordinary area/length/volume formulas in the simplest cases (e.g. revolving a rectangle gives a cylinder's volume)
Units work out automatically: integrating a rate (velocity, force, density) over its variable yields the corresponding total quantity

Common mistakes

Confusing displacement with distance traveled: if velocity changes sign, $\int v\,dt$ (signed) underestimates how far the particle actually moved — split at the sign changes and integrate $|v(t)|$
Squaring the difference instead of differencing the squares in the washer method: $(R-r)^2 \neq R^2 - r^2$ except when $r=0$ or $r=R$

Area between two curves

Find the area between $y = x^2$ and $y = x$ on $[0, 1]$ .

Since $x \geq x^2$ on this interval: $A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ .