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21

Day 21

Average Value And Particle Motion

The Average Value of a FunctionParticle Motion with Integrals: Displacement vs. Total Distance

The Average Value of a Function

The average value of ff over [a,b][a, b] is the height of the rectangle on base [a,b][a, b] whose area equals ∫abf(x) dx\int_a^b f(x)\,dx. The Mean Value Theorem for Integrals guarantees the function actually attains this average somewhere on the interval.

Core Theorem
Average value:
favg=1b−a∫abf(x) dxf_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx
Mean Value Theorem for Integrals: if ff is continuous on [a,b][a, b], there exists c∈(a,b)c \in (a, b) with f(c)=favgf(c) = f_{\text{avg}}.
Step-by-Step SOP
  1. 1

    Integrate f over the interval

    Compute ∫abf(x) dx\int_a^b f(x)\,dx.
  2. 2

    Divide by the width

    Divide by b−ab - a to get favgf_{\text{avg}}.
  3. 3

    If asked, find c

    Solve f(c)=favgf(c) = f_{\text{avg}} and keep the solution(s) inside (a,b)(a, b).

Practice Exercises

Example 01Easy
Find the average value of f(x)=x2f(x) = x^2 on [0,3][0, 3].
NEED A HINT?
Compute 13−0∫03x2 dx\frac{1}{3 - 0}\int_0^3 x^2\,dx.
SHOW DETAILED EXPLANATION

Set up

favg=13∫03x2 dxf_{\text{avg}} = \frac{1}{3}\int_0^3 x^2\,dx.

Integrate

∫03x2 dx=[x33]03=9\int_0^3 x^2\,dx = \Big[\frac{x^3}{3}\Big]_0^3 = 9.

Divide by the width

favg=13(9)=3f_{\text{avg}} = \frac{1}{3}(9) = 3.
Example 02Medium
The average value of f(x)=x2f(x) = x^2 on [0,3][0, 3] is 33. Find the cc guaranteed by the MVT for Integrals.
NEED A HINT?
Set f(c)=3f(c) = 3 and keep the solution in (0,3)(0, 3).
SHOW DETAILED EXPLANATION

Set f(c) equal to the average

c2=3c^2 = 3.

Solve

c=±3c = \pm\sqrt{3}.

Keep the value in the interval

c=3≈1.732∈(0,3)c = \sqrt{3} \approx 1.732 \in (0, 3); reject c=−3c = -\sqrt{3}.
Example 03Medium
Find the average value of f(x)=exf(x) = e^x on [0,ln⁡4][0, \ln 4].
NEED A HINT?
favg=1ln⁡4−0∫0ln⁡4ex dxf_{\text{avg}} = \frac{1}{\ln 4 - 0}\int_0^{\ln 4} e^x\,dx.
SHOW DETAILED EXPLANATION

Integrate

∫0ln⁡4ex dx=[ex]0ln⁡4=4−1=3\int_0^{\ln 4} e^x\,dx = \Big[e^x\Big]_0^{\ln 4} = 4 - 1 = 3.

Divide by the width

favg=3ln⁡4≈2.164f_{\text{avg}} = \frac{3}{\ln 4} \approx 2.164.
Common Pitfalls
  • ⚠
    Forgetting to divide by b minus aThe average value is not just the integral. Students often stop after computing ∫abf(x) dx\int_a^b f(x)\,dx.
  • ⚠
    Average value vs. average rate of changeAverage value of ff uses an integral; average rate of change of ff is f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a} (a slope). Read the question carefully.

Particle Motion with Integrals: Displacement vs. Total Distance

Given a velocity function, integration recovers position. The key AP distinction: the integral of velocity gives net displacement, while the integral of speed ∣v(t)∣|v(t)| gives total distance traveled.

Core Theorem
For a particle with velocity v(t)v(t) on [a,b][a, b]:
Displacement =∫abv(t) dt=s(b)−s(a)= \int_a^b v(t)\,dt = s(b) - s(a)
Total distance =∫ab∣v(t)∣ dt= \int_a^b |v(t)|\,dt
Position =s(a)+∫atv(τ) dτ= s(a) + \int_a^t v(\tau)\,d\tau
Step-by-Step SOP
  1. 1

    Read the question

    Displacement / net change → signed integral. Total distance → integral of ∣v∣|v|.
  2. 2

    For total distance, find where v equals 0

    These are the times the particle changes direction — split the integral there.
  3. 3

    Integrate each subinterval

    Take the absolute value of each piece, then sum.
  4. 4

    For position

    Add the initial position: s(t)=s(a)+∫atv(τ) dτs(t) = s(a) + \int_a^t v(\tau)\,d\tau.

Practice Exercises

Example 01Easy
A particle has velocity v(t)=t2−4v(t) = t^2 - 4 (m/s) for 0≤t≤30 \le t \le 3. Find the displacement.
NEED A HINT?
Displacement is ∫03v(t) dt\int_0^3 v(t)\,dt — no absolute value.
SHOW DETAILED EXPLANATION

Integrate v

∫03(t2−4) dt=[t33−4t]03\int_0^3 (t^2 - 4)\,dt = \Big[\frac{t^3}{3} - 4t\Big]_0^3.

Evaluate

(9−12)−0=−3(9 - 12) - 0 = -3 m.

Interpret

The particle ends 3 m to the left of where it started.
Example 02Hard
For the same v(t)=t2−4v(t) = t^2 - 4 on [0,3][0, 3], find the total distance traveled.
NEED A HINT?
v(t)=0v(t) = 0 at t=2t = 2. Split the integral there and take absolute values.
SHOW DETAILED EXPLANATION

Find the sign change

t2−4=0  ⟹  t=2t^2 - 4 = 0 \implies t = 2. On [0,2)[0, 2), v<0v < 0; on (2,3](2, 3], v>0v > 0.

Integrate each piece

∫02(t2−4) dt=83−8=−163\int_0^2 (t^2 - 4)\,dt = \frac{8}{3} - 8 = -\frac{16}{3}; ∫23(t2−4) dt=−3−(−163)=73\int_2^3 (t^2 - 4)\,dt = -3 - \left(-\frac{16}{3}\right) = \frac{7}{3}.

Add absolute values

Total distance =163+73=233≈7.67= \frac{16}{3} + \frac{7}{3} = \frac{23}{3} \approx 7.67 m.
Example 03Medium
A particle starts at s(0)=5s(0) = 5 with velocity v(t)=3t2v(t) = 3t^2. Find its position at t=2t = 2.
NEED A HINT?
Position =s(0)+∫02v(t) dt= s(0) + \int_0^2 v(t)\,dt.
SHOW DETAILED EXPLANATION

Integrate velocity

∫023t2 dt=[t3]02=8\int_0^2 3t^2\,dt = \Big[t^3\Big]_0^2 = 8.

Add the starting position

s(2)=5+8=13s(2) = 5 + 8 = 13.
Common Pitfalls
  • ⚠
    Using the integral of v for total distance∫v dt\int v\,dt gives displacement, which can be smaller than total distance (or zero) when the particle reverses direction. Always split at v=0v = 0 for distance.
  • ⚠
    Speed vs. velocitySpeed is ∣v(t)∣|v(t)|. The particle is speeding up when vv and aa have the same sign, slowing down when they have opposite signs.
Gary Chang

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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