Calculus · Grade 12 and undergraduate

Motion and Net Change: Integrals in Context

Quick answer

Integrating a rate gives the total change in the quantity: ∫ₐᵇ R′(t) dt = R(b) − R(a). For motion along a line, the integral of velocity is the displacement, the integral of speed |v| is the total distance, and the position is the starting position plus the accumulated change. The same reasoning tracks any amount with a rate in and a rate out, such as water in a tank: the amount at time t is the starting amount plus the integral of the rate in minus the rate out.

What you'll learn

  • Find position from velocity and velocity from acceleration
  • Distinguish displacement from total distance traveled
  • Model an amount with a rate in and a rate out
  • Interpret integrals of rates, with units, in context

Integrating a rate gives a change

The Fundamental Theorem says that integrating a rate of change recovers the total change:

∫abR′(t) dt=R(b)−R(a)\int_a^b R'(t)\,dt = R(b) - R(a)

This is the net change theorem. It turns a rate into an amount: gallons per minute integrated over minutes gives gallons, and meters per second integrated over seconds gives meters.

Knowing where the quantity started, you can find where it is:

s(b)=s(a)+∫abv(t) dtv(b)=v(a)+∫aba(t) dts(b) = s(a) + \int_a^b v(t)\,dt \qquad\qquad v(b) = v(a) + \int_a^b a(t)\,dt

Position is the starting position plus the accumulated change in position. Velocity is found from acceleration the same way.

Displacement and total distance

The integral of velocity is the displacement: the change in position. When the object moves backward, v<0v < 0, and that stretch subtracts. The total distance traveled counts every stretch as positive, so it integrates the speed:

displacement=∫abv(t) dttotal distance=∫ab∣v(t)∣ dt\text{displacement} = \int_a^b v(t)\,dt \qquad\qquad \text{total distance} = \int_a^b \lvert v(t) \rvert\,dt
Velocity v(t) = t² − 4t + 3 on 0 ≤ t ≤ 4 The parabola v = t squared minus 4t plus 3, above the t-axis from 0 to 1, below it from 1 to 3, and above it again from 3 to 4. The area between the curve and the axis is shaded in three pieces, the middle one below the axis in a second shade. Each piece has area 4/3. 1234-1123tv
  • v(t) = t² − 4t + 3
Velocity v(t) = t² − 4t + 3 on 0 ≤ t ≤ 4

Here each shaded piece has area 43\tfrac{4}{3}. The displacement is 43−43+43=43\tfrac{4}{3} - \tfrac{4}{3} + \tfrac{4}{3} = \tfrac{4}{3}, but the total distance is 43+43+43=4\tfrac{4}{3} + \tfrac{4}{3} + \tfrac{4}{3} = 4.

Why distance needs the absolute value

Over a short time Δt\Delta t, the position changes by about v(t) Δtv(t)\,\Delta t: positive going forward, negative going back. Adding those changes gives where the object ends up relative to where it began, so backward steps cancel forward ones.

An odometer never runs backward. It adds the length of every step, which is ∣v(t)∣ Δt\lvert v(t) \rvert\,\Delta t. Displacement asks where you ended up; distance asks how far you went. The two agree only when the object never turns around.

To integrate ∣v∣\lvert v \rvert in practice, find where vv changes sign, integrate vv on each piece, and add the absolute values.

Amounts with a rate in and a rate out

Water flows into a tank at a rate rin(t)r_{\text{in}}(t) and out at a rate rout(t)r_{\text{out}}(t). The amount changes at the net rate W′(t)=rin(t)−rout(t)W'(t) = r_{\text{in}}(t) - r_{\text{out}}(t), so

W(t)=W(0)+∫0t(rin(u)−rout(u)) duW(t) = W(0) + \int_0^t \big(r_{\text{in}}(u) - r_{\text{out}}(u)\big)\,du

The amount rises while more flows in than out and falls when the reverse is true. Its largest value on an interval is at an endpoint or where the net rate changes from positive to negative, so compare the amounts there.

Worked examples

Common mistakes

Practice problems

  1. A particle has velocity v(t)=3t2v(t) = 3t^2 m/s and s(0)=2s(0) = 2 m. Find s(2)s(2).

    Answer

    1010 m

    Full solution

    s(2)=2+∫023t2 dt=2+[t3]02=2+8=10s(2) = 2 + \int_0^2 3t^2\,dt = 2 + \big[t^3\big]_0^2 = 2 + 8 = 10.

  2. For v(t)=t2−2tv(t) = t^2 - 2t on [0,3][0, 3], find the displacement and the total distance traveled.

    Answer

    Displacement 00, total distance 83\tfrac{8}{3}

    Full solution

    An antiderivative is F(t)=t33−t2F(t) = \tfrac{t^3}{3} - t^2, and vv changes sign at t=2t = 2. ∫02v dt=F(2)−F(0)=−43\int_0^2 v\,dt = F(2) - F(0) = -\tfrac{4}{3} and ∫23v dt=F(3)−F(2)=43\int_2^3 v\,dt = F(3) - F(2) = \tfrac{4}{3}.

    The displacement is 00: the particle ends where it started. The distance is 43+43=83\tfrac{4}{3} + \tfrac{4}{3} = \tfrac{8}{3}.

  3. A particle has acceleration a(t)=6ta(t) = 6t and v(0)=−3v(0) = -3. Find v(2)v(2).

    Answer

    99

    Full solution

    v(2)=−3+∫026t dt=−3+[3t2]02=−3+12=9v(2) = -3 + \int_0^2 6t\,dt = -3 + \big[3t^2\big]_0^2 = -3 + 12 = 9.

  4. For v(t)=sin⁡tv(t) = \sin t on [0,2π][0, 2\pi], find the displacement and the total distance.

    Answer

    Displacement 00, total distance 44

    Full solution

    ∫0πsin⁡t dt=2\int_0^{\pi} \sin t\,dt = 2 and ∫π2πsin⁡t dt=−2\int_{\pi}^{2\pi} \sin t\,dt = -2. The sum is 00; the absolute values add to 44.

  5. Water leaks from a tank at 4e−0.1t4e^{-0.1t} liters per hour. How much leaks out in the first 1010 hours?

    Answer

    40(1−e−1)≈25.340(1 - e^{-1}) \approx 25.3 liters

    Full solution

    ∫0104e−0.1t dt=[−40e−0.1t]010=−40e−1+40\int_0^{10} 4e^{-0.1t}\,dt = \big[-40e^{-0.1t}\big]_0^{10} = -40e^{-1} + 40.

  6. r(t)r(t) is the rate, in liters per minute, at which water flows into a pool, with tt in minutes. What are the units of ∫030r(t) dt\int_0^{30} r(t)\,dt, and what does it mean?

    Answer

    Liters: the amount of water that flows into the pool in the first 3030 minutes.

    Full solution

    The integral adds up (liters per minute) × (minutes), which gives liters.

  7. A town of 10 00010\,000 people grows at 200e0.05t200e^{0.05t} people per year. Find the population after 1010 years.

    Answer

    About 12 59512\,595

    Full solution

    ∫010200e0.05t dt=[4000e0.05t]010=4000(e0.5−1)≈2595\int_0^{10} 200e^{0.05t}\,dt = \big[4000e^{0.05t}\big]_0^{10} = 4000(e^{0.5} - 1) \approx 2595. Add this to 10 00010\,000.

  8. A tank holds 5050 gallons at t=0t = 0. Water flows in at 88 gallons per minute and drains at tt gallons per minute, for 0≤t≤120 \le t \le 12. Find the largest amount of water in the tank.

    Answer

    8282 gallons, at t=8t = 8

    Full solution

    W(t)=50+8t−t22W(t) = 50 + 8t - \tfrac{t^2}{2}. The net rate 8−t8 - t turns negative at t=8t = 8. Compare W(0)=50W(0) = 50, W(8)=82W(8) = 82 and W(12)=74W(12) = 74.

  9. A particle has velocity v(t)=2t−6v(t) = 2t - 6 and s(0)=5s(0) = 5. For t≥0t \ge 0, where is it farthest to the left?

    Answer

    At s=−4s = -4, when t=3t = 3

    Full solution

    The particle moves left while v<0v < 0, until t=3t = 3, and right after that. s(3)=5+∫03(2t−6) dt=5+(9−18)=−4s(3) = 5 + \int_0^3 (2t - 6)\,dt = 5 + (9 - 18) = -4.

  10. A student finds the total distance for v(t)=t−2v(t) = t - 2 on [0,5][0, 5] as ∫05(t−2) dt=2.5\int_0^5 (t - 2)\,dt = 2.5. What went wrong?

    Hint

    What is the sign of vv for t<2t < 2?

    Answer

    The backward motion before t=2t = 2 was subtracted instead of added. The total distance is 6.56.5.

    Full solution

    ∫02(t−2) dt=−2\int_0^2 (t - 2)\,dt = -2 and ∫25(t−2) dt=4.5\int_2^5 (t - 2)\,dt = 4.5. The displacement is 2.52.5, but the distance is 2+4.5=6.52 + 4.5 = 6.5.

Frequently asked questions

What is the difference between displacement and total distance?

Displacement is the net change in position, the integral of v, where backward motion subtracts. Total distance is the integral of |v|, where every movement adds.

How do I find position from velocity?

Add the accumulated change to the starting position: s(b) = s(a) + ∫ from a to b of v(t) dt.

How do I compute total distance traveled?

Find where v changes sign, integrate v on each piece, and add the absolute values of the results.

What are the units of the integral of a rate?

The rate's units times the units of the variable. Gallons per minute integrated over minutes gives gallons.

When is the amount in a tank largest?

At an endpoint or where the rate in minus the rate out changes from positive to negative. Compare the amounts at those candidates.

What to learn next

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