| English | 의미 |
|---|---|
| average value/ˈævrɪdʒ ˈvæljuː/ | 평균값 |
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).
등적사각형
y = x²
평균값은 $[a,b]$ 구간의 곡선과 동일한 면적을 가지는 사각형의 상수 높이입니다.
$f$의 $[a,b]$에서의 평균값은...
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
총량을 너비로 나옴.
평균값은 함수 $f$와 같은 ____를 가지며, 면적 단위가 아닙니다.
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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.)
$f(x)=x^2$의 $[0,3]$에서의 평균값을 구하시오. ($\int_0^3 x^2\,dx=9$.)
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
$\tfrac{1}{3}\cdot 9=3$.
기하학적으로, 평균값은 너비가 $b-a$인 사각형의 높이로서 ...와 동일한 ...입니다.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
등적사각형의 높이가 평균값입니다.
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?"
정적분 $\int_a^b f\,dx$ 자체는 $f$의 평균값이 아닙니다.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
那是总量; 平均值为$b-a$除之。
$[0,2]$ 시간 동안 자동차의 변위는 $\int_0^2 v\,dt=120$ km입니다. 평균 속도(km/h)를 구하시오.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
$\tfrac{120}{2}=60$ 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 $\frac{1}{b-a}$.