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Code Scholars Tutoring · AP Calculus AB

Average value, motion, and net change

Unit 8 · Topics 8.1, 8.2, 8.3. Original AP Calculus AB notes, worked example, and self-check.

Understand the idea

The average value of a function is its integral divided by the interval length; it is different from average rate of change. Integrating velocity gives displacement, while integrating speed gives distance. To recover a quantity from its rate, add the accumulated change to its initial amount. Inflow minus outflow gives net rate, not total throughput. Track units at every stage.

favg=1b−a∫abf(x) dx,s(b)=s(a)+∫abv(t) dtf_{\mathrm{avg}}=\frac1{b-a}\int_a^bf(x)\,dx,\qquad s(b)=s(a)+\int_a^bv(t)\,dt

Worked example

A particle has velocity v(t)=t−1\displaystyle v(t)=t-1 on [0,3]\displaystyle [0,3]. Find displacement and total distance.

Displacement is ∫03(t−1) dt=3/2\displaystyle \int_0^3(t-1)\,dt=3/2. Velocity changes sign at 1\displaystyle 1, so distance is ∫01(1−t) dt+∫13(t−1) dt=1/2+2=5/2\displaystyle \int_0^1(1-t)\,dt+\int_1^3(t-1)\,dt=1/2+2=5/2.

Try it yourself

Find the average value of x2\displaystyle x^2 on [0,3]\displaystyle [0,3].

Show solution and reasoning

13∫03x2 dx=13(9)=3\displaystyle \frac13\int_0^3x^2\,dx=\frac13(9)=3. This differs from the average of endpoint outputs, 9/2\displaystyle 9/2.

Unit 8 MCQNext: Area between curves and changing orientation →