Find the point where a function attains its own average value
Three steps, the way the exam actually works: work through the lab, write down your own measurements, then answer a 8-point free response. What you recorded goes to the grader with your writing, so a conclusion that does not follow from your own numbers will cost you the point — exactly as it would with a real reader.
Predict before you look
- Write the formula for the average value of f on [a, b], and say in one sentence what the division by (b − a) accomplishes.
- The Mean Value Theorem for integrals guarantees a c where f(c) equals the average value. State the hypothesis it requires.
Nothing to submit here — these are to think through, so the prediction below is an informed one rather than a guess.
Commit to an answer now. It is not graded and being wrong costs nothing — the point is to have something specific to reconcile against once you have the data.
Answer every prediction to unlock the lab. A sentence is enough.
Run the investigation
Predictions first
The procedure and the simulation unlock once you have committed above. Observing before predicting is how a wrong intuition survives a lab intact.
Record what you measured
These are your numbers, not ours. The grader sees them, so your conclusions have to follow from what you actually recorded.
| Accumulated area from 0 to 2.00 | |
|---|---|
| Average value = area ÷ 2 | |
| f(x) readout at x = 1.00 | |
| The x where f(x) equals your average value |
Answer the free response
You have located the point where a function attains its own average value. (a) Report the accumulated area and the average value, showing the division. (b) Report the x you found, and verify algebraically that c = √(4/3) by solving c² equal to the average value. (c) Your c is not the midpoint of [0, 2]. Explain why, referring to the shape of the function, and state for what class of functions c WOULD be the midpoint. (d) State the Mean Value Theorem for integrals and its hypothesis, and explain why your measurement is an instance of it rather than a coincidence.
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