The average value of a function

About this lesson

Calculus AB (AP-aligned) › Integrals and accumulation

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The average value of a function

Integrals and accumulation.

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is y=x2y = x^2 on [0,3][0, 3]; the blue line is y=ay = a. Move aa until the rectangle under the line on [0,3][0, 3] has the same area as the region under the curve, ∫03x2 dx=9\displaystyle\int_0^3 x^2\,dx = 9.

The parabola y = x² on [0, 3] and a horizontal line y = a. At a = 3 the rectangle of width 3 under the line has area 9, the same as under the curve.

Which height aa works?

  1. a=3a = 3
  2. a=4.5a = 4.5

Key idea

The average value is the height of the rectangle with the same area.

The average value of ff on [a,b][a, b] is 1b−a∫abf(x) dx\dfrac{1}{b - a}\displaystyle\int_a^b f(x)\,dx.

Worked example

Average value of f(x)=2x+1f(x) = 2x + 1 on [0,4][0, 4].

∫04(2x+1) dx=20\displaystyle\int_0^4 (2x + 1)\,dx = 20
F(x) = x² + x.
Average =204=5= \tfrac{20}{4} = 5
Divide by the length 4.

Your turn, guided

Your turn: average of f(x)=3x2f(x) = 3x^2 on [0,2][0, 2].

Average =4= 4
∫ = 8; divide by 2.

Practice ★

  1. Find the average value of f(x)=x2f(x) = x^2 on [0,3][0, 3].
  2. Find the average value of f(x)=x2f(x) = x^2 on [0,6][0, 6].
  3. Find the average value of f(x)=x2f(x) = x^2 on [0,2][0, 2].

Watch out!

Practice ★★

  1. v(t)=3t+6v(t) = 3t + 6 m/s. Find the average velocity on [0,8][0, 8], in m/s.
  2. v(t)=3t+1v(t) = 3t + 1 m/s. Find the average velocity on [0,7][0, 7], in m/s.
  1. Find the average value of f(x)=6sin⁡xf(x) = 6\sin x on [0,π][0, \pi]. Give an exact value.

Exit ticket

Three questions to check you have got it.

  1. Find the average value of f(x)=x2f(x) = x^2 on [0,3][0, 3].
  2. Find the average value of f(x)=x2f(x) = x^2 on [0,6][0, 6].
  3. Find the average value of f(x)=x2f(x) = x^2 on [0,2][0, 2].

Summary card

Average value

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

Watch out: divide by the length; do not just average the end values.

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