Mathfolis

Average Value

Unit 8 · Applications of Integration

Average Value of a Function

The average value of f on [a, b] is the height of a rectangle with base (b−a) having the same area as the region under f. The Mean Value Theorem for Integrals guarantees f equals its average value at some c in [a, b].

Average Value

favg=1b−a∫abf(x) dxf_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx

Motion with Integrals

Displacement
∫t1t2v(t) dt\int_{t_1}^{t_2} v(t)\,dt
Total distance (always ≥ 0)
∫t1t2∣v(t)∣ dt\int_{t_1}^{t_2} |v(t)|\,dt
Position from acceleration
v(t)=v(t0)+∫t0ta(τ) dτv(t) = v(t_0) + \int_{t_0}^t a(\tau)\,d\tau
AP Tip: For total distance, find zeros of v(t) to split the interval, then add the absolute values of each sub-integral.
Type 1

Computing the Average Value

Apply the average value formula and find cc where f(c)=favgf(c) = f_{avg} (MVT for Integrals).

Example 1
Find the average value of f(x) = 3x² − 2x + 1 on [0, 2].
Example 2
Find the average value of f(x) = cos x on [0, π]. Find cc where f(c)=favgf(c) = f_{avg}.

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Type 2

Displacement vs. Total Distance Traveled

Displacement = ∫v dt (signed). Total distance = ∫|v| dt — split at zeros of v and sum absolute values.

Example 3
v(t) = t² − 4t + 3 for t ∈ [0, 4]. Find (a) displacement, (b) total distance.

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Type 3

Position and Velocity from Acceleration

Integrate a(t) to get v(t), then integrate v(t) to get s(t). Apply initial conditions at each step.

Example 4
a(t) = 6t − 2, v(0) = −3, s(0) = 1. Find s(3).

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