The Average Value Audit
A quality engineer samples the temperature (in °C) of a cooling component over an 8-minute window and models it as T(t) = 20 + 16e^(−0.5t) for 0 ≤ t ≤ 8. She must report two numbers to her manager: (1) the average temperature over the window, defined as the mean value of T over [0, 8], and (2) the temperature at the midpoint in time, T(4). A colleague claims these must be equal because "the average of a function is just its value in the middle." The engineer is skeptical and wants a numerical check as well as an exact comparison, and she wants to know which number better represents "typical" operating temperature for warranty purposes.
1. Compute the average value of T over [0, 8] exactly, and compute T(4). Compare them.
Read the worked answer
Average value: (1/8)∫₀⁸ (20 + 16e^(−0.5t)) dt.
Antiderivative: 20t − 32e^(−0.5t). Evaluate 0 to 8: At 8: 160 − 32e^(−4) ≈ 160 − 32(0.01832) = 160 − 0.586 = 159.414. At 0: 0 − 32(1) = −32. Difference = 159.414 − (−32) = 191.414. Divide by 8: ≈ 23.93 °C.
Midpoint value: T(4) = 20 + 16e^(−2) ≈ 20 + 16(0.1353) = 20 + 2.165 = ≈ 22.17 °C.
They are not equal: 23.93 vs. 22.17. The colleague's rule is false for a nonlinear (here, convex decreasing) function.
2. Explain geometrically why the average value exceeds T(4) for this decreasing, convex function, and state which figure better represents typical temperature for a warranty claim.
Read the worked answer
T(t) is decreasing and convex (T′ = −8e^(−0.5t) < 0, T″ = 4e^(−0.5t) > 0). It starts high (36 °C at t = 0) and flattens toward 20 °C. Because the curve spends the early minutes at high values and drops steeply at first, the area-based mean weights those early high temperatures heavily, pulling the average (≈23.93) above the mid-time value (≈22.17).
Geometrically, the mean value is the height of the horizontal line that makes the area under it equal the area under the curve; for this shape that line sits above T(4).
For a warranty claim, the average value (≈23.93 °C) is the better figure: it reflects the time-weighted thermal exposure the part actually experiences, whereas T(4) is a single instantaneous reading and understates cumulative heat load. The colleague's "middle = average" shortcut only holds for linear functions.