Finding the Average Value of a Function on an Interval

The average height of a curve

• You can average a list of numbers — but how do you average a function, which has infinitely many values?
• Integration does it: the average value 平均值 of $f$ on $[a,b]$ is its total accumulation divided by the width.
• Picture the area under the curve reshaped into a rectangle of the same width — its height is the average.
• It's the calculus version of "add them up and divide by how many."

The formula

• The average value of $f$ on $[a,b]$ is
• $$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx$$
• Compute the definite integral (the total), then divide by the interval length $b-a$.
• Units: same as $f$ itself (it's an average height, not an area).

The rectangle of height f_avg has the same area as the region under f

It's the "leveling out" height

• The average value is the constant height that gives the same area over $[a,b]$.
• $\displaystyle\int_a^b f_{\text{avg}}\,dx = f_{\text{avg}}(b-a)=\int_a^b f\,dx$ — the rectangle and the region match.
• So $f_{\text{avg}}$ is where you'd "level off" the curve to keep the same accumulated total.
• (The Mean Value Theorem for Integrals guarantees $f$ actually hits this average somewhere if $f$ is continuous.)

Averages in context

• Average velocity over $[a,b]$ = $\frac{1}{b-a}\int_a^b v(t)\,dt$ = displacement ÷ time.
• Average temperature, average rate, average concentration — all the same recipe.
• Read the units to interpret: the average of a rate is still a rate.
• It answers "what single steady value would produce the same total?"

Write from an average value of 2

• f(x) = x on [0, 4].
• The average value is wanted.