| English | 意味 |
|---|---|
| average value/ˈævrɪdʒ ˈvæljuː/ | 平均値 |
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).
等面積の長方形
y = x²
平均値とは、$[a,b]$上の曲線と同じ面積を持つ長方形の一定の高さのことである。
$f$の$[a,b]$における平均値は...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
総量を幅で割る。
平均値は関数$f$と同じ____を持ち、面積の単位ではない。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
それは面積ではなく平均の高さである。
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.)
$f(x)=x^2$の$[0,3]$における平均値を求めよ。（$\int_0^3 x^2\,dx=9$。）
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$\tfrac{1}{3}\cdot 9=3$.
幾何学的に、平均値とは、幅が$b-a$で、同じ...を持つ長方形の高さである。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
等面積の長方形の高さが平均値である。
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?"
$\int_a^b f\,dx$の定積分自体は、$f$の平均値ではない。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
それは総量です；平均にするには$b-a$で割る。
ある車が$[0,2]$時間での移動距離は$\int_0^2 v\,dt=120$kmである。その平均速度（km/h）を求めよ。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$\tfrac{120}{2}=60$ 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.
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 $[0,3]$.
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 $\frac{1}{b-a}$.