Single-Variable-Calculus · Unit 15 · Video 2 · Interactive Practice

Net Change and Signed Area: Integrating Velocity

IKey Formulas

Formula	Name	What it measures
$\displaystyle\int_a^b v(t)\,dt = x(b) - x(a)$	Net change (FTC 1 with $x' = v$ )	How far you end up from where you started
$\displaystyle\sum_{i=1}^{n} v(t_i)\,\Delta t \;\to\; \int_a^b v(t)\,dt$ as $\Delta t \to 0$	Sum of speedometer readings	Each term is the distance in one short time step, negative while moving backward
$\displaystyle\int_a^b f(x)\,dx = (\text{area above}) - (\text{area below})$	Signed area	Area above the $x$ -axis counts positive, area below counts negative
$\displaystyle\int_a^b \lvert v(t)\rvert\,dt$	Total distance	The ground covered in either direction, as an odometer reads it

Key Insight: The odometer reads $\int_a^b \lvert v\rvert\,dt$ ; the fundamental theorem gives $\int_a^b v\,dt$ . The two agree only when $v(t) \ge 0$ on the whole interval.

IIVisualization 1 — Speedometer Readings Add Up to the Trip

On the round trip $v(t) = 2t - t^2$ , $0 \le t \le 3$ , the terms $v(t_i)\,\Delta t$ sum toward $x(3) - x(0) = 0$ .

Readings n 3

IIIVisualization 2 — Net Change Versus Total Distance

On the same trip, the signed area from $0$ to $T$ is $x(T) - x(0)$ , while total distance counts the return leg as positive.

Time T 1.00

IVVisualization 3 — The Definite Integral as Signed Area

For $\sin x$ , the value $\int_a^b \sin x\,dx = -\cos b + \cos a$ is the area above the axis minus the area below.

Left end a 0 Right end b 2π

💡 Challenge: find an interval of length $\pi$ on which $\int_a^b \sin x\,dx = 0$ .

VQuiz Questions

Problem 1 · Net Change from a Velocity

Given: A cart on a track has velocity $v(t) = 4 - 2t$ m/s for $0 \le t \le 3$ s. Find its net change in position $x(3) - x(0)$ .

✅ Correct!

\int_0^3 (4 - 2t)\,dt = \big[4t - t^2\big]_0^3 = 12 - 9 = 3

, so the cart ends

3

m ahead of where it started.

❌ Check the antiderivative. An antiderivative of

2t

t^2

, not

2t^2

; with

4t - t^2

the value is

12 - 9 = 3

❌ That is the total distance. The cart moves forward

4

m on

[0, 2]

and back

1

m on

[2, 3]

; the integral of

v

keeps the sign, giving

4 - 1 = 3

❌ That is the velocity at the end.

v(3) = -2

m/s is a single speedometer reading; the net change adds up all of them,

\int_0^3 v(t)\,dt

Show solution

Position is an antiderivative of velocity, so by the fundamental theorem

x(3) - x(0) = \int_0^3 (4 - 2t)\,dt = \big[\,4t - t^2\,\big]_0^3 = (12 - 9) - 0 = 3.

The velocity changes sign at $t = 2$ : the cart goes forward $\int_0^2 (4 - 2t)\,dt = 4$ m and back $\int_2^3 (4 - 2t)\,dt = -1$ m. The net change is $4 + (-1) = 3$ m; the total distance would be $4 + 1 = 5$ m.

Problem 2 · Net and Total on One and a Half Humps

Given: A particle has velocity $v(t) = \sin t$ for $0 \le t \le \tfrac{3\pi}{2}$ . Find the net change in position and the total distance traveled.

Net change $\int_0^{3\pi/2} \sin t\,dt$

Total distance $\int_0^{3\pi/2} \lvert \sin t\rvert\,dt$

✅ Correct! The full hump contributes

+2

and the half dip

-1

, so the net change is

2 - 1 = 1

and the total distance is

2 + 1 = 3

❌ That is the total distance. On

[\pi, \tfrac{3\pi}{2}]

the velocity is negative, so that piece counts as

-1

, not

+1

❌ Sign slip in the evaluation. An antiderivative of

\sin t

-\cos t

, so the value is

-\cos\tfrac{3\pi}{2} + \cos 0 = 0 + 1 = 1

, not

\cos\tfrac{3\pi}{2} - \cos 0

❌ Not quite. Evaluate

-\cos t

at the two endpoints:

-\cos\tfrac{3\pi}{2} - (-\cos 0)

❌ That is the net change. The absolute value flips the dip above the axis, so the half dip adds

+1

instead of

-1

❌ Not quite. Split at

t = \pi

, where

\sin t

changes sign, and add the two areas: one full hump and one half dip.

Show solution

Net change. With the antiderivative $-\cos t$ ,

\int_0^{3\pi/2} \sin t\,dt = -\cos\tfrac{3\pi}{2} - (-\cos 0) = 0 + 1 = 1.

Total distance. Split at $t = \pi$ , where $\sin t$ changes sign:

\int_0^{\pi} \sin t\,dt = -\cos\pi + \cos 0 = 2, \qquad \int_{\pi}^{3\pi/2} \sin t\,dt = -\cos\tfrac{3\pi}{2} + \cos\pi = -1.

\int_0^{3\pi/2} \lvert\sin t\rvert\,dt = 2 + \lvert -1\rvert = 3.

As a check, the net change is $2 + (-1) = 1$ , matching the first computation.

Problem 3 · Two Turnarounds

Given: A car's velocity is $v(t) = t^2 - 4t + 3$ for $0 \le t \le 4$ , so its position satisfies $x(t) - x(0) = \tfrac{t^3}{3} - 2t^2 + 3t$ . Find when the car moves backward and the total distance it travels.

When is the car moving backward?

Total distance traveled on $[0, 4]$

✅ Correct! The three legs move

+\tfrac{4}{3}

-\tfrac{4}{3}

and

+\tfrac{4}{3}

, so the total distance is

3 \cdot \tfrac{4}{3} = 4

while the net change is only

\tfrac{4}{3}

❌ Those are the forward legs.

v(t) = (t - 1)(t - 3)

is positive outside

[1, 3]

and negative between the roots.

❌ Not quite. Factor

v(t) = (t - 1)(t - 3)

and test the sign of

v

between its roots, for example

v(2) = -1

❌ That is the net change.

x(4) - x(0) = \tfrac{4}{3}

lets the backward leg cancel a forward one; total distance adds every leg as positive.

❌ You left out the backward leg. The car also drives

\tfrac{4}{3}

backward on

[1, 3]

, and that ground counts toward total distance.

❌ Not quite. Split at

t = 1

and

t = 3

, find each leg's change in position, and add their absolute values.

Show solution

Sign of $v$ . $v(t) = t^2 - 4t + 3 = (t - 1)(t - 3)$ , which is negative for $1 < t < 3$ and positive on $[0, 1)$ and $(3, 4]$ . The car moves backward on $1 < t < 3$ .

Each leg. With $F(t) = \tfrac{t^3}{3} - 2t^2 + 3t$ , we have $F(0) = 0$ , $F(1) = \tfrac{4}{3}$ , $F(3) = 9 - 18 + 9 = 0$ and $F(4) = \tfrac{64}{3} - 32 + 12 = \tfrac{4}{3}$ .

\int_0^1 v\,dt = \tfrac{4}{3}, \qquad \int_1^3 v\,dt = 0 - \tfrac{4}{3} = -\tfrac{4}{3}, \qquad \int_3^4 v\,dt = \tfrac{4}{3} - 0 = \tfrac{4}{3}.

Total distance.

\int_0^4 \lvert v(t)\rvert\,dt = \tfrac{4}{3} + \tfrac{4}{3} + \tfrac{4}{3} = 4.

The net change is $\tfrac{4}{3} - \tfrac{4}{3} + \tfrac{4}{3} = \tfrac{4}{3} = F(4) - F(0)$ .

Problem 4 · Net Change in a Water Tank

Given: Water flows into a tank at the rate $r(t) = 2 - t$ litres per minute for $0 \le t \le 6$ ; a negative rate means the tank is draining. The tank holds $10$ L at $t = 0$ . How much water does it hold at $t = 6$ ?

✅ Correct! The net change is

\int_0^6 (2 - t)\,dt = 12 - 18 = -6

L, so the tank holds

10 - 6 = 4

❌ That is the net change, not the amount.

\int_0^6 r\,dt = V(6) - V(0) = -6

, so add the starting

10

❌ The draining counts negative.

10 + \int_0^6 \lvert r\rvert\,dt = 10 + 10 = 20

treats the

8

L that drained out as if it had flowed in.

❌ You kept only the filling. The tank gains

2

L on

[0, 2]

but then loses

8

L on

[2, 6]

, where

r(t) < 0

Show solution

The rate $r = V'$ plays the role of velocity and the volume $V$ the role of position, so the fundamental theorem gives the net change:

V(6) - V(0) = \int_0^6 (2 - t)\,dt = \Big[\,2t - \tfrac{t^2}{2}\,\Big]_0^6 = 12 - 18 = -6.

As signed area: the triangle above the axis on $[0, 2]$ has area $\tfrac{1}{2}\cdot 2 \cdot 2 = 2$ and the triangle below on $[2, 6]$ has area $\tfrac{1}{2}\cdot 4 \cdot 4 = 8$ , so the integral is $2 - 8 = -6$ .

V(6) = 10 + (-6) = 4\ \text{L}.

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