Areas and Distances for MATH 141
Exam Relevance for MATH 141
Riemann sum concepts in MATH 141 are setup for integrals. Rarely tested standalone.
This skill appears on:
Understanding Areas and Distances
When you know how fast something is moving at every moment (its velocity function), you can use integrals to find how far it traveled. This connects the geometric idea of "area under a curve" to the physical idea of "distance traveled."
The Key Insight
If you're driving at a constant 60 mph for 2 hours, you travel $60 \times 2 = 120$ miles. That's just rate × time.
But what if your speed keeps changing? That's where integrals come in — they add up all those tiny "rate × time" pieces when the rate isn't constant.
Displacement vs. Total Distance
This is the most important distinction in this topic:
| Concept | What it measures | Formula | Can be negative? |
|---|---|---|---|
| Displacement | Net change in position (where you end up minus where you started) | $\int_a^b v(t) \, dt$ | Yes |
| Total Distance | How far you actually traveled (odometer reading) | $\int_a^b |v(t)| \, dt$ | No |
Why the difference?
- If you walk 5 meters forward, then 3 meters backward:
- Displacement = 5 - 3 = 2 meters (you're 2m from where you started)
- Total Distance = 5 + 3 = 8 meters (your legs walked 8m total)
When velocity is negative (moving backward), regular integration subtracts that distance. To get total distance, we need the absolute value.
The Formulas
Displacement (Net Change in Position)
$$\text{Displacement} = \int_a^b v(t) \, dt$$
Just integrate the velocity function directly.
Total Distance Traveled
$$\text{Total Distance} = \int_a^b |v(t)| \, dt$$
Take the absolute value of velocity before integrating. In practice, this means:
- Find where $v(t) = 0$ (when the object changes direction)
- Split the integral at those points
- Take the absolute value of each piece
- Add them all up
A particle moves with velocity $v(t) = 3t^2 - 6t$ m/s. Find the displacement from $t = 0$ to $t = 3$.
$$\text{Displacement} = \int_0^3 (3t^2 - 6t) \, dt$$
$$= \left[ t^3 - 3t^2 \right]_0^3$$
$$= (27 - 27) - (0 - 0) = 0$$
$$\boxed{0 \text{ meters}}$$
The particle ends up exactly where it started! (It went forward, then came back.)
Using the same velocity $v(t) = 3t^2 - 6t$ m/s, find the total distance traveled from $t = 0$ to $t = 3$.
Step 1: Find where velocity is zero
$$3t^2 - 6t = 0$$ $$3t(t - 2) = 0$$ $$t = 0 \text{ or } t = 2$$
The particle changes direction at $t = 2$.
Step 2: Determine the sign of velocity in each interval
- For $0 < t < 2$: test $t = 1$ → $v(1) = 3 - 6 = -3 < 0$ (moving backward)
- For $2 < t < 3$: test $t = 2.5$ → $v(2.5) = 18.75 - 15 = 3.75 > 0$ (moving forward)
Step 3: Split the integral and take absolute values
$$\text{Total Distance} = \left| \int_0^2 (3t^2 - 6t) \, dt \right| + \left| \int_2^3 (3t^2 - 6t) \, dt \right|$$
First integral: $$\int_0^2 (3t^2 - 6t) \, dt = \left[ t^3 - 3t^2 \right]_0^2 = (8 - 12) - 0 = -4$$ $$\left| -4 \right| = 4$$
Second integral: $$\int_2^3 (3t^2 - 6t) \, dt = \left[ t^3 - 3t^2 \right]_2^3 = (27 - 27) - (8 - 12) = 0 - (-4) = 4$$ $$\left| 4 \right| = 4$$
Step 4: Add them up
$$\text{Total Distance} = 4 + 4 = \boxed{8 \text{ meters}}$$
Find the area under $f(x) = x^2$ from $x = 0$ to $x = 3$.
$$\text{Area} = \int_0^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^3 = \frac{27}{3} - 0 = \boxed{9 \text{ square units}}$$
A car's velocity is $v(t) = 4t - t^2$ ft/s for $0 \leq t \leq 4$. How far does the car travel in total?
Step 1: Find where $v(t) = 0$
$$4t - t^2 = 0$$ $$t(4 - t) = 0$$ $$t = 0 \text{ or } t = 4$$
The velocity is zero only at the endpoints — no direction change in between!
Step 2: Check the sign
At $t = 2$: $v(2) = 8 - 4 = 4 > 0$
Velocity is positive throughout $(0, 4)$, so displacement = total distance.
Step 3: Integrate
$$\text{Total Distance} = \int_0^4 (4t - t^2) \, dt = \left[ 2t^2 - \frac{t^3}{3} \right]_0^4$$
$$= \left( 32 - \frac{64}{3} \right) - 0 = \frac{96 - 64}{3} = \frac{32}{3}$$
$$\boxed{\frac{32}{3} \text{ feet} \approx 10.67 \text{ feet}}$$
A particle's velocity graph shows:
- $v(t) = 2$ m/s for $0 \leq t \leq 3$
- $v(t) = -1$ m/s for $3 < t \leq 5$
Find (a) displacement and (b) total distance from $t = 0$ to $t = 5$.
(a) Displacement:
$$\int_0^5 v(t) \, dt = \int_0^3 2 \, dt + \int_3^5 (-1) \, dt$$ $$= 2(3) + (-1)(2) = 6 - 2 = \boxed{4 \text{ meters}}$$
(b) Total Distance:
$$\int_0^5 |v(t)| \, dt = \int_0^3 2 \, dt + \int_3^5 1 \, dt$$ $$= 2(3) + 1(2) = 6 + 2 = \boxed{8 \text{ meters}}$$
True or False: If a particle's displacement over an interval is zero, it must have stayed still the entire time.
False. Zero displacement means the particle ended where it started — but it could have moved around and returned.
For example, if $v(t) = \sin(t)$ from $t = 0$ to $t = 2\pi$: $$\int_0^{2\pi} \sin(t) \, dt = [-\cos(t)]_0^{2\pi} = -1 - (-1) = 0$$
The displacement is zero, but the particle definitely moved (it oscillated back and forth).
True or False: Total distance traveled is always greater than or equal to the magnitude of displacement.
True. Total distance counts every meter traveled. Displacement only measures the net result.
Think of it this way: if you take a detour, your odometer (total distance) increases, but your "as the crow flies" distance (displacement) stays the same.
Mathematically: $\int_a^b |v(t)| \, dt \geq \left| \int_a^b v(t) \, dt \right|$
This is always true because absolute value inside the integral ≥ absolute value of the integral.
True or False: To find total distance traveled when velocity changes sign, you can just integrate $|v(t)|$ directly without splitting the integral.
True in theory, but practically misleading. While $\int_a^b |v(t)| \, dt$ is the correct formula, you typically can't evaluate $|v(t)|$ as a single expression. You need to:
- Find where $v(t) = 0$
- Split the integral at those points
- Remove the absolute value by using $-v(t)$ where $v(t) < 0$
So while the formula is correct, the method requires splitting the integral in practice.
Common Mistakes and Misunderstandings
❌ Mistake: Using displacement when asked for total distance
Wrong: "The total distance is $\int_0^3 v(t) \, dt = 0$ meters"
Why it's wrong: When velocity is sometimes negative, the regular integral gives displacement (net change), not total distance. Parts of the motion cancel out.
Correct: Find where $v(t) = 0$, split the integral, take absolute values of each piece, then add.
❌ Mistake: Forgetting to check for direction changes
Wrong: Assuming velocity is always positive and just integrating directly.
Why it's wrong: If the object reverses direction, you'll miss distance traveled in the "backward" phase.
Correct: Always solve $v(t) = 0$ first to find potential direction changes within your interval.
❌ Mistake: Confusing area "under" a curve with signed area
Wrong: Saying the area under $f(x) = x^2 - 4$ from $x = 0$ to $x = 3$ is just $\int_0^3 (x^2 - 4) \, dx$
Why it's wrong: When the function dips below the x-axis, the integral gives negative area there. If you want the actual geometric area, you need absolute values.
Correct: For geometric area, use $\int_a^b |f(x)| \, dx$ and split at zeros just like with distance problems.
Displacement (Net Change in Position)
The net change in position from time a to time b. Can be positive, negative, or zero. Integrate the velocity function directly.
Variables:
- $v(t)$:
- velocity as a function of time
- $a$:
- starting time
- $b$:
- ending time
Total Distance Traveled
The actual distance traveled (like an odometer reading). Always positive. Find where v(t) = 0, split the integral at those points, and take absolute values of each piece.
Variables:
- $|v(t)|$:
- absolute value of velocity
- $a$:
- starting time
- $b$:
- ending time
Area Under a Curve
The signed area between the curve y = f(x) and the x-axis. Regions below the x-axis contribute negative area. For geometric (unsigned) area, use absolute value and split at zeros.
Variables:
- $f(x)$:
- the function defining the curve
- $a$:
- left boundary
- $b$:
- right boundary
This skill is taught in the following courses. Create an account to access practice exercises and full course materials.