Course: MTH 132, Calculus I (Michigan State University) Source: Recitation Activity 13, Section 5.5 Difficulty: Intermediate Prerequisites: FTC Part 2 (evaluating definite integrals), absolute value, basic kinematics concepts (position, velocity).
When a particle moves with a known velocity function, integrating velocity gives displacement (net change in position), while integrating the absolute value of velocity gives total distance travelled. The average value of a function on an interval is the integral divided by the interval length. These three ideas are the main applications of definite integrals in this section.
You now know how to evaluate definite integrals. This section asks: what do those integrals mean in context? The headline application is motion. If you have a velocity function v(t), integration answers two different questions depending on how you set it up. One gives you where the particle ended up relative to where it started (displacement). The other gives you the total ground covered, regardless of direction (total distance). The average value formula rounds out the section as a way to summarise a function's behaviour over an interval with a single number.
You need to be comfortable evaluating polynomial and trigonometric integrals before tackling these problems.
Displacement
The net change in position over a time interval. Computed as s(b) − s(a) = ∫ from a to b of v(t) dt. Displacement can be positive, negative or zero. Think of it as: where did you end up minus where you started? If you walk 5 metres forward and then 5 metres back, your displacement is zero.
Total distance
The total ground covered regardless of direction. Computed as ∫ from a to b of |v(t)| dt. Always non-negative. In the same walk example, your total distance is 10 metres.
Velocity function, v(t)
The rate of change of position with respect to time. Positive v means moving in the positive direction, negative v means moving in the negative direction.
Average value of a function
f_avg = (1/(b − a)) · ∫ from a to b of f(x) dx. In simple terms, it is the constant height of a rectangle over [a, b] that has the same area as the region under f(x). It compresses the entire behaviour of f into one representative number.
Displacement is the straightforward integral of velocity:
s(b) − s(a) = ∫ from a to b of v(t) dt
Just find the antiderivative of v(t), evaluate at the endpoints, and subtract.
Example: v(t) = t² − 8t + 12 on [0, 4]
Antiderivative: t³/3 − 4t² + 12t
s(4) − s(0) = [64/3 − 64 + 48] − [0] = 64/3 − 16 = 16/3 metres
The particle ends up 16/3 metres from where it started (in the positive direction)
Total distance requires integrating |v(t)|. The key step: find where v(t) = 0 (the particle changes direction), then integrate each sub-interval separately, taking absolute values.
Step 1: Set v(t) = 0 and solve
t² − 8t + 12 = 0 → (t − 2)(t − 6) = 0 → t = 2 and t = 6
On [0, 4], only t = 2 is relevant
Step 2: Determine the sign of v(t) on each sub-interval
On [0, 2]: pick t = 1, v(1) = 1 − 8 + 12 = 5 > 0, so v is positive
On [2, 4]: pick t = 3, v(3) = 9 − 24 + 12 = −3 < 0, so v is negative
Step 3: Integrate, flipping the sign where v is negative
Total distance = ∫ from 0 to 2 of (t² − 8t + 12) dt + ∫ from 2 to 4 of −(t² − 8t + 12) dt
First integral: [t³/3 − 4t² + 12t] from 0 to 2 = 8/3 − 16 + 24 = 32/3
Second integral: [−t³/3 + 4t² − 12t] from 2 to 4 = (−64/3 + 64 − 48) − (−8/3 + 16 − 24) = −16/3 − (−32/3) = 16/3
Total distance = 32/3 + 16/3 = 48/3 = 16 metres
Notice: the total distance (16 m) is larger than the displacement (16/3 m), because the particle reversed direction at t = 2 and some of the ground was covered twice.
The average value formula is:
f_avg = (1/(b − a)) · ∫ from a to b of f(x) dx
The procedure: evaluate the definite integral first (you already know how), then divide by the length of the interval.
Example: f(x) = 6x² − 2x + 9 on [2, 4]
We computed ∫ from 2 to 4 of (6x² − 2x + 9) dx = 118 in the definite integrals section
Interval length: 4 − 2 = 2
f_avg = (1/2)(118) = 59
Example: f(x) = 5 sec²(x) on [0, π/4]
We computed ∫ from 0 to π/4 of 5 sec²(x) dx = 5
Interval length: π/4 − 0 = π/4
f_avg = (1/(π/4))(5) = (4/π)(5) = 20/π
The average value has a geometric interpretation: it is the height of a rectangle with base [a, b] whose area equals the area under the curve. The Mean Value Theorem for Integrals guarantees there is at least one point c in [a, b] where f(c) equals this average value.
\text{Displacement} = s(b) - s(a) = \int_a^b v(t)\,dt\text{Total distance} = \int_a^b |v(t)|\,dtf_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dxTotal distance is what your car's odometer reads: it only goes up, regardless of whether you drive north or south. Displacement is the straight-line difference between your starting point and ending point on a GPS. The average value formula is how meteorologists report the "average temperature" over a day when temperature varies continuously.
Students frequently compute ∫ v(t) dt (displacement) when the problem asks for total distance. If the velocity changes sign on the interval, these two quantities are different. Always check which one the problem is asking for.
When computing total distance, students sometimes take the absolute value of the final answer instead of integrating |v(t)|. That gives the absolute value of the displacement, not the total distance. You must split the integral at the zeros of v(t) and handle each piece separately.
Students sometimes forget to find where v(t) = 0 before splitting. If you skip this, you cannot correctly set up the absolute value integral.
In the average value formula, students occasionally forget the 1/(b − a) factor. Without it, you just have the integral, not the average.
⚠️ The distinction between total distance and displacement is one of the most commonly tested concepts in this chapter. Expect a problem that gives v(t) and asks for both, or one that specifically asks for total distance to see whether you handle the absolute value correctly.
⚠️ Average value problems are quick marks if you know the formula. They often appear as standalone questions worth 5 to 8 points.
⚠️ Professors often give a velocity function that is a factorable quadratic, so you can find the zeros cleanly. Practise factoring under time pressure.
⚠️ Watch units. If v(t) is in m/s and t is in seconds, displacement and total distance are in metres. The average value inherits the units of f, not the units of f times x.
True or False: Displacement can be negative.
True or False: Total distance can be negative.
Fill in the blank: To find total distance, you integrate ______ rather than v(t).
True or False: If v(t) ≥ 0 on the entire interval, then total distance equals displacement.
Fill in the blank: The average value of f(x) = 10 on [0, 5] is ______.
Q: A particle has velocity v(t) = 3t² − 12t on [0, 5]. Find the displacement.
A: Antiderivative: t³ − 6t². Evaluate: (125 − 150) − (0) = −25. The particle ends up 25 units behind where it started.
Q: For the same v(t) = 3t² − 12t on [0, 5], find the total distance.
A: Set v(t) = 0: 3t(t − 4) = 0, so t = 0 and t = 4. On [0, 4], v(t) ≤ 0 (check t = 1: 3 − 12 = −9). On [4, 5], v(t) ≥ 0 (check t = 5: 75 − 60 = 15). Total distance = |∫ from 0 to 4 of v(t) dt| + ∫ from 4 to 5 of v(t) dt. First piece: [t³ − 6t²] from 0 to 4 = 64 − 96 = −32, absolute value = 32. Second piece: [t³ − 6t²] from 4 to 5 = (125 − 150) − (64 − 96) = −25 − (−32) = 7. Total distance = 32 + 7 = 39.
Q: Find the average value of f(x) = x³ on [0, 2].
A: f_avg = (1/(2 − 0)) · ∫ from 0 to 2 of x³ dx = (1/2) · [x⁴/4] from 0 to 2 = (1/2)(16/4) = (1/2)(4) = 2.
Q: A car's velocity is v(t) = 60 − 2t km/h for t in [0, 40]. When does it stop, and what is the total distance from t = 0 to t = 40?
A: v(t) = 0 when 60 − 2t = 0, so t = 30. On [0, 30], v > 0. On [30, 40], v < 0. Total distance = ∫ from 0 to 30 of (60 − 2t) dt + ∫ from 30 to 40 of (2t − 60) dt = [60t − t²] from 0 to 30 + [t² − 60t] from 30 to 40 = (1800 − 900) + [(1600 − 2400) − (900 − 1800)] = 900 + (−800 + 900) = 900 + 100 = 1000 km.
Q: The average value of f on [1, 5] is 8. What is ∫ from 1 to 5 of f(x) dx?
A: Rearrange the formula: ∫ = f_avg · (b − a) = 8 · 4 = 32.
This material connects to kinematics in physics: displacement and total distance are the same concepts you use in mechanics, and the same integral setup carries into higher-dimensional motion (vector calculus). The average value formula reappears in probability as the expected value of a continuous random variable. It also connects to the Mean Value Theorem for Integrals, which guarantees the existence of a point where the function equals its average.
Total distance, displacement, net change theorem, velocity integral, absolute value of velocity, average value of a function, mean value theorem for integrals, motion problems calculus, odometer vs GPS, particle motion, MTH 132, Calculus I, Michigan State University, MSU, Section 5.5, Recitation 13, kinematics, signed area, distance vs displacement