The Net Change Theorem

The Net Change Theorem

The Net Change Theorem is a direct application of the Fundamental Theorem of Calculus to real-world problems:

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

The integral of a rate of change equals the net (total) change in the quantity $F$ over the interval $[a,b]$.

Net Displacement from Velocity

The most common application is motion. When $v(t)$ is velocity and $s(t)$ is position:

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

This gives net displacement—the object's final position relative to its starting point, signed.

Example: For $v(t) = 3t^2 - 2t$ on $[0,2]$: $\int_0^2 (3t^2 - 2t)\,dt = \left[t^3 - t^2\right]_0^2 = 4 \text{ meters}$

The object ends up 4 meters ahead. But this doesn't reveal whether it moved backward at some point.

Displacement vs. Total Distance

Net displacement hides backward motion through cancellation. If you need the total distance traveled (counting all motion as positive), use:

$\text{Total distance} = \int_a^b |v(t)|\,dt$

To compute this, find all zeros of $v(t)$ in $[a,b]$, split the interval at those points, and integrate:

$\int_a^b |v(t)|\,dt = -\int_a^c v(t)\,dt + \int_c^b v(t)\,dt \quad \text{(if } v<0 \text{ on } [a,c], v>0 \text{ on } [c,b]\text{)}$

Example: For $v(t) = t - 1$ on $[0,3]$, where $v(1) = 0$:

• Net displacement: $\int_0^3 (t-1)\,dt = 1.5$ m
• Total distance: $-\int_0^1(t-1)\,dt + \int_1^3(t-1)\,dt = 0.5 + 2 = 2.5$ m

Key Distinction

Always identify whether the problem asks for net change (signed integration) or total accumulation (split at sign changes, integrate absolute value).