Finding the Average Value of a Function on an Interval · 求函数在区间上的平均值
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| average value/ˈævrɪdʒ ˈvæljuː/ | 平均值 | píng jūn zhí |
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."
曲线的平均高度
- 你能给一列数字求平均——但怎么给一个有无穷多个值的函数求平均?
- 积分做到这点:$f$ 在 $[a,b]$ 上的平均值是它的总累积除以宽度。
- 想象把曲线下的面积重塑成同样宽度的矩形——它的高就是平均值。
- 这是"加起来除以个数"的微积分版本。
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).
公式
- $f$ 在 $[a,b]$ 上的平均值是
-
$$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx$$
- 算出定积分(总量),再除以区间长度 $b-a$。
- 单位:与 $f$ 本身相同(它是平均高度,不是面积)。
The equal-area rectangle · 等面积矩形
y = x²
The average value is the constant height whose rectangle encloses the same area as the curve over $[a,b]$. · 平均值即为常数高度，其对应的矩形在 $[a,b]$ 上围成的面积与曲线下面积相等。
The average value of $f$ on $[a,b]$ is... · $f$ 在 $[a,b]$ 上的平均值为...
Total divided by width. · 总量除以宽度。
The average value has the same ____ as the function $f$, not the units of an area. · 平均值具有与函数 $f$ 相同的 ____ ，而非面积的度量单位。
It is an average height, not an area. · 它是平均高度，而非面积。
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.)
它是"抹平"的高度
- 平均值是给出相同面积的那个常数高度,在 $[a,b]$ 上。
- $\displaystyle\int_a^b f_{\text{avg}}\,dx = f_{\text{avg}}(b-a)=\int_a^b f\,dx$——矩形与区域相等。
- 所以 $f_{\text{avg}}$ 是你把曲线"抹平"、保持相同累积总量时的高度。
- (积分中值定理保证:若 $f$ 连续,$f$ 确实会在某处取到这个平均值。)
Find the average value of $f(x)=x^2$ on $[0,3]$. ($\int_0^3 x^2\,dx=9$.) · 求 $f(x)=x^2$ 在 $[0,3]$ 上的平均值。（$\int_0^3 x^2\,dx=9$ 。）
$\tfrac{1}{3}\cdot 9=3$.
Geometrically, the average value is the height of a rectangle (width $b-a$) with the same... · 几何上，平均值是矩形的一个高度（宽度为 $b-a$ ），该矩形具有相同的...
The equal-area rectangle's height is the average value. · 等面积矩形的高度即为平均值。
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?"
情境中的平均
- $[a,b]$ 上的平均速度 = $\frac{1}{b-a}\int_a^b v(t)\,dt$ = 位移 ÷ 时间。
- 平均温度、平均速率、平均浓度——都是同一个配方。
- 读单位来解读:一个速率的平均仍是速率。
- 它回答"哪个单一的稳定值会产生相同的总量?"
The definite integral $\int_a^b f\,dx$ by itself is the average value of $f$. · 定积分 $\int_a^b f\,dx$ 本身并不是 $f$ 的平均值。
That is the total; divide by $b-a$ for the average. · 那是总量；除以 $b-a$ 得到平均值。
A car's displacement over $[0,2]$ h is $\int_0^2 v\,dt=120$ km. Find its average velocity (km/h). · 一辆车在 $[0,2]$ 小时内的位移是 $\int_0^2 v\,dt=120$ 公里。求其平均速度（km/h）。
$\tfrac{120}{2}=60$ km/h. · $\tfrac{120}{2}=60$ km/h。
The average value divides by $b-a$ — don't confuse it with the definite integral alone (that's the total, an area). $\int_a^b f\,dx$ is accumulated amount; $\frac{1}{b-a}\int_a^b f\,dx$ is average height. Forgetting the $\frac{1}{b-a}$ is the classic mistake.
平均值要除以 $b-a$——别把它与单独的定积分(那是总量,一个面积)搞混。$\int_a^b f\,dx$ 是累积量;$\frac{1}{b-a}\int_a^b f\,dx$ 是平均高度。忘掉 $\frac{1}{b-a}$ 是经典错误。
Find the average value of $f(x)=x^2$ on $[0,3]$.
- $\displaystyle\int_0^3 x^2\,dx=\Big[\tfrac{x^3}{3}\Big]_0^3=9$.
- Divide by the width: $f_{\text{avg}}=\dfrac{1}{3-0}\cdot 9 = 3$.
- So a constant height of $3$ would enclose the same area over $[0,3]$.
求 $f(x)=x^2$ 在 $[0,3]$ 上的平均值。
- $\displaystyle\int_0^3 x^2\,dx=\Big[\tfrac{x^3}{3}\Big]_0^3=9$。
- 除以宽度:$f_{\text{avg}}=\dfrac{1}{3-0}\cdot 9 = 3$。
- 所以常数高度 $3$ 会在 $[0,3]$ 上围出相同面积。
The average value of $f$ on $[a,b]$ is $f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx$ — the total (a definite integral) divided by the interval width. It's the constant height enclosing the same area, with the same units as $f$. Never drop the $\frac{1}{b-a}$.
$f$ 在 $[a,b]$ 上的平均值是 $f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx$——总量(一个定积分)除以区间宽度。它是围出相同面积的常数高度,与 $f$ 单位相同。永远别丢 $\frac{1}{b-a}$。