Accumulation Functions and Net Change Theorem

Accumulation and Net Change

By the end of this lesson, you will be able to interpret the definite integral as a net change or accumulation, and solve applied problems involving rates.

The Net Change Theorem

The total change in a quantity $F$ over $[a, b]$ equals the definite integral of its rate of change $F'$:

$$F(b) - F(a) = \int_a^b F'(x), dx$$

This is the applied version of FTC Part 2.

Examples of (rate, accumulated quantity) pairs:

• (velocity, displacement): $\int_a^b v(t), dt = s(b) - s(a)$
• (flow rate, total volume): $\int_a^b r(t), dt$ = total flow
• (marginal cost, total cost change): $\int_a^b C'(x), dx = C(b) - C(a)$
• (population growth rate, population change): $\int_a^b P'(t), dt = P(b) - P(a)$

Accumulation Functions

An accumulation function $A(x) = \int_a^x f(t), dt$ represents the total amount accumulated from $a$ to $x$.

If $f(t)$ is a rate (gallons per minute, people per hour, etc.), then $A(x)$ is the total quantity accumulated.

Example 1: Water Flow

Water flows into a tank at rate $r(t) = 3t^2 + 2$ gallons per hour for $0 \leq t \leq 4$.

Total water added: $\int_0^4 (3t^2+2), dt = [t^3+2t]_0^4 = 64+8 = 72$ gallons.

Example 2: Net vs. Total Distance

A particle moves with velocity $v(t) = t^2 - 4$ on $[0, 3]$.

Net displacement: $\int_0^3 (t^2-4), dt = \left[\frac{t^3}{3}-4t\right]_0^3 = 9 - 12 = -3$ ft

(Particle ends up 3 feet behind where it started)

Total distance: $v(t) = t^2 - 4 = 0$ at $t = 2$; $v < 0$ for $0 < t < 2$; $v > 0$ for $2 < t < 3$

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

$= \left[4t - \frac{t^3}{3}\right]_0^2 + \left[\frac{t^3}{3}-4t\right]_2^3 = \left(8-\frac{8}{3}\right) + \left(9-12-\frac{8}{3}+8\right) = \frac{16}{3} + \frac{5}{3} = 7$ ft

Example 3: Population

A city's population changes at rate $P'(t)$ people per year. Given data from a table:

t	P'(t)
0	1200
1	1500