反常積分と幾何学的応用
導入済み| English | 意味 |
|---|---|
| improper integral/ɪmˈprɒpə ˈɪntɪɡrəl/ | 不整積分 |
| principal value/ˈprɪnsɪpl ˈvæljuː/ | 主値 |
A decision before an answer
- A curve can enclose a finite area over an infinite interval while forming a surface of infinite area when rotated.
- Your goal: Define improper integrals by limits at each singular boundary.
Read the relationship
- An infinite integration endpoint is replaced by a finite bound and a limit. At a singular point inside the interval, split the integral and require both one-sided integrals to converge separately. Symmetric cancellation can define a Cauchy principal value but does not prove convergence of the ordinary improper integral. For 1/x across zero, the two sides diverge even though symmetric cutoffs cancel.
- Select area, volume and arc-length formulas from the geometry.
For which p does the integral of x^(−p) from zero to 1 converge?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The lower-bound power limit is finite when 1−p>0; p=1 gives logarithmic divergence.
Use the defining rule
- The integral of x^(−p) from 1 to infinity converges exactly when p>1. From zero to 1 it converges exactly when p<1. The same exponent behaves differently at the two boundaries. For positive integrands, comparison transfers convergence from a larger integrable function or divergence from a smaller nonintegrable one; keep the inequality direction correct. For an integral of a maximum or minimum, solve branch crossings and check which expression dominates each interval. For max(√(1−x²),x+1) on [−1,1], the interior crossing is zero; use the semicircle on [−1,0] and the line on [0,1]. The area is π/4+3/2, not the integral of either branch over the entire interval.
- Separate convergence of integrals from signed cancellation.
Rotating the region under y=x² from x=0 to x=1 around the x-axis gives which volume?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Disk radius is x², so V=π integral 0..1 x⁴ dx=π/5.
Check the conditions
- For rotation around the x-axis, disks or washers integrate π(R²−r²) dx. Cylindrical shells use 2π times radius times height and integrate in the matching variable. Select the method by the geometry and verify nonnegative radii. Area between curves integrates upper minus lower, splitting where their order changes. A signed integral is not automatically geometric area.
- Separate convergence of integrals from signed cancellation.
Rotate y=1/x for x≥1 around the x-axis. The disk volume is π times the integral of 1/x², so V=π. The lateral surface integrand is 2π(1/x)sqrt(1+1/x⁴), at least 2π/x; its improper integral diverges. Finite volume therefore does not imply finite surface area. This comparison avoids trying to find an unnecessary antiderivative.
The integral from 1 to infinity of 1/x² dx equals ____.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The finite-bound integral is 1−1/b and its limit as b grows is 1.
Apply the task format
- For a differentiable plane curve y=f(x), arc length is the integral of sqrt(1+(f′(x))²) dx. A surface formed by rotating a nonnegative f around the x-axis has area integral 2πf sqrt(1+(f′)²) dx. These are different quantities from volume. If an interval is unbounded, the geometric formula still needs a convergence test.
- Separate convergence of integrals from signed cancellation.
Never count a principal value as a convergent improper integral, or infer one geometric quantity's finiteness from another.
このケースに当てはまる答えはどれか？
各特異境界での極限により improper 積分を定義する
The ordinary improper integral of 1/x from −1 to 1 converges because the integrand is odd.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Each side diverges at zero. The symmetric principal value is zero, which is a different limit.
What is ∫₋₁¹ max(√(1−x²),x+1) dx?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The semicircle dominates on [−1,0], giving quarter-circle area π/4. On [0,1] the line integrates to 3/2. The crossing at −1 is an endpoint, not another interior split.
Keep the distinctions
- improper integral 反常积分 — An integral defined by limits at infinite or singular boundaries.
- principal value 主值 — A limit using prescribed symmetric cancellation that may exist when an ordinary improper integral diverges.
- Define improper integrals by limits at each singular boundary.
- Select area, volume and arc-length formulas from the geometry.
- Separate convergence of integrals from signed cancellation.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.