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

Try it

A particle has velocity $v(t) = t^2 - 4$ for $0 \leq t \leq 3$ . Find the distance traveled.

Solution

$v(t) = 0$ when $t = 2$ . On $[0,2]$ , $v < 0$ ; on $[2,3]$ , $v > 0$ .

Distance $= \int_0^2 (4-t^2)\,dt + \int_2^3 (t^2-4)\,dt$

On $[0,2]$ : $\left[4t - \frac{t^3}{3}\right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}$ .

On $[2,3]$ : $\left[\frac{t^3}{3} - 4t\right]_2^3 = (9-12) - \left(\frac{8}{3}-8\right) = -3 + \frac{16}{3} = \frac{7}{3}$ .

Total distance $= \frac{16}{3} + \frac{7}{3} = \frac{23}{3}$ .

Related concepts

Needs first

Fundamental Theorem of Calculus Integrals

Integration by Parts Techniques of Integration

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