Integration Applications in Motion

In calculus, integration is a powerful tool for analyzing motion. If you know an object's velocity function $v(t)$, you can use integrals to determine both where the object ends up (displacement) and how much ground it covered to get there (total distance). Another common application is finding the average value of a continuous function over a specific interval.

Displacement vs. Total Distance

When an object moves along a straight line, its velocity $v(t)$ can be positive (moving forward) or negative (moving backward).

Displacement is the net change in position. It is calculated by simply integrating the velocity function over the given time interval $[a, b]$:

$\text{Displacement} = \int_{a}^{b} v(t) dt$

Total Distance is the total amount of ground covered, regardless of direction. To find it, you must integrate the speed, which is the absolute value of velocity:

$\text{Total Distance} = \int_{a}^{b} |v(t)| dt$

To calculate total distance, you need to find where $v(t) = 0$ to identify when the object changes direction, and then split the integral into intervals where $v(t)$ is entirely positive or negative.

Example: Finding Displacement and Total Distance

Problem: Given the velocity function $v(t) = t^2 - 4t + 3$, find the displacement and total distance traveled from $t = 0$ to $t = 4$.

1. Displacement: Simply integrate $v(t)$ from 0 to 4: $\text{Displacement} = \int_{0}^{4} (t^2 - 4t + 3) dt$ $= \left[ \frac{1}{3}t^3 - 2t^2 + 3t \right]_{0}^{4}$ $= \left( \frac{64}{3} - 32 + 12 \right) - 0 = \frac{64}{3} - 20 = \frac{4}{3}$ The net displacement is $\frac{4}{3}$.

2. Total Distance: First, find where $v(t) = 0$: $t^2 - 4t + 3 = (t-1)(t-3) = 0$ The object stops at $t = 1$ and $t = 3$. We split the integral into three parts: $[0, 1]$, $[1, 3]$, and $[3, 4]$.

• Interval $[0, 1]$ (Moving forward): $\int_{0}^{1} (t^2 - 4t + 3) dt = \left[ \frac{1}{3}t^3 - 2t^2 + 3t \right]_{0}^{1} = \frac{1}{3} - 2 + 3 = \frac{4}{3}$
• Interval $[1, 3]$ (Moving backward): $\int_{1}^{3} (t^2 - 4t + 3) dt = \left[ \frac{1}{3}t^3 - 2t^2 + 3t \right]_{1}^{3} = (9 - 18 + 9) - \left(\frac{4}{3}\right) = -\frac{4}{3}$ The distance here is $\left|-\frac{4}{3}\right| = \frac{4}{3}$.
• Interval $[3, 4]$ (Moving forward): $\int_{3}^{4} (t^2 - 4t + 3) dt = \left( \frac{64}{3} - 20 \right) - 0 = \frac{4}{3}$

Add the absolute values of these distances: $\text{Total Distance} = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = \frac{12}{3} = 4$

Average Value of a Function

Integrals also allow us to find the average (mean) value of a continuous function $f(x)$ over a closed interval $[a, b]$. The formula is:

$f_{\text{avg}} = \frac{1}{b - a} \int_{a}^{b} f(x) dx$

Geometrically, this represents the height of a rectangle with width $(b-a)$ that has the exact same area as the region under the curve $f(x)$ on that interval.

Example: Finding the Average Value

Problem: Find the average value of $f(x) = \sin x$ on the interval $[0, \pi]$.

Using the formula: $f_{\text{avg}} = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin x dx$ $= \frac{1}{\pi} \left[ -\cos x \right]_{0}^{\pi}$ $= \frac{1}{\pi} (-\cos(\pi) - (-\cos(0)))$ $= \frac{1}{\pi} (-(-1) + 1) = \frac{2}{\pi}$

The average value of $\sin x$ on $[0, \pi]$ is $\frac{2}{\pi}$.