부분 적분, 순서 반전 및 대칭성
도입됨| English | 의미 |
|---|---|
| integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | 부분 적분 |
| reflection symmetry/rɪˈflekʃn ˈsɪmətri/ | 반사 대칭 |
A decision before an answer
- An integral can become simple after changing its representation. A hard inner integral may turn into an elementary one when you reverse the region’s integration order.
- Your goal: Derive integration-by-parts reductions with valid endpoint limits.
Read the relationship
- Integration by parts comes from the product rule: ∫u dv=uv−∫v du. For definite integrals include the boundary term at both ends. With I_n(x)=∫₁ˣ(ln t)^n dt, choose u=(ln t)^n and dv=dt. For n≥1 the lower boundary vanishes, giving I_n=x(ln x)^n−n I_(n−1). A plus sign would violate the differentiated identity. Keep the same lower limit in a recurrence; changing it changes the constants.
- Reverse a double integral by reconstructing its region.
If I_n(x)=∫₁ˣ(ln t)^n dt for n≥1, which recurrence is correct?
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Integration by parts gives uv minus the remaining integral. The lower boundary is zero for n≥1.
Use the defining rule
- For improper integrals first apply the identity on finite endpoints, then justify the limiting boundary and remaining integral. If ∫₋∞^∞e^(−x²)dx=√π, set u=x and dv=x e^(−x²)dx for ∫x²e^(−x²)dx. Since v=−e^(−x²)/2 and x e^(−x²)→0 at both infinities, the full second moment is √π/2. Evenness gives the positive-half-line moment √π/4. The vanished boundary is part of the argument, not an automatic property of every improper integral.
- Use reflection symmetry while checking removable endpoint behaviour.
After reversing ∫₀¹∫ₓ¹e^(y²)dy dx, what integral remains?
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At each y, x ranges from 0 to y. The inner integral supplies length y.
Check the conditions
- To reverse ∫₀¹∫ₓ¹F(x,y)dy dx, describe the triangle 0≤x≤y≤1, then rewrite it as ∫₀¹∫₀ʸF(x,y)dx dy. Both the outer interval and inner bounds change; swapping symbols alone changes the region. For F=e^(y²), integrating over x first gives ∫₀¹y e^(y²)dy=(e−1)/2. Continuity on this compact triangle makes order reversal valid; singular or conditionally convergent cases require stronger care.
- Use reflection symmetry while checking removable endpoint behaviour.
For I₂(x)=∫₁ˣ(ln t)²dt, the recurrence gives x(ln x)²−2I₁(x), with I₁(x)=x ln x−x+1. Therefore I₂=x[(ln x)²−2ln x+2]−2. The constant −2 ensures I₂(1)=0. Separately, reversing the triangular e^(y²) integral produces a factor y from the inner x-length; dropping that factor would restore the original difficulty and change the value.
For I₂(x)=∫₁ˣ(ln t)²dt, the boundary value I₂(1) is ____.
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The upper and lower integration endpoints coincide, and the explicit recurrence has the same zero boundary value.
Apply the task format
- If an integrable function on [a,b] satisfies f(a+b−x)=−f(x), reflection makes its integral equal to its negative, hence zero. For sin(2mx)/sin x on [0,π] with positive integer m, reflection x↦π−x changes the numerator sign and preserves the denominator. The apparent endpoint singularities are removable: limits are 2m at zero and −2m at π. Check these limits before invoking symmetry; cancellation is not a substitute for integrability.
- Use reflection symmetry while checking removable endpoint behaviour.
Improper integration by parts needs endpoint limits. Reconstruct the region before reversing order. An odd-looking reflected integrand must be integrable, including any removable endpoints.
다음 중 이 사례에 부합하는 답안은 무엇입니까?
유효한 경계점 극한을 포함하는 parts-by-parts 적분 감소 공식 도출
Integration by parts on a finite interval automatically remains valid after replacing its endpoints with infinities.
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The boundary term and remaining integral must have the required limits. Divergent or oscillating boundaries invalidate an automatic replacement.
Keep the distinctions
- integration by parts 分部积分 — An integration identity derived from the product rule with a boundary term.
- reflection symmetry 反射对称 — A relation between integrand values at points mirrored across an interval midpoint.
- Derive integration-by-parts reductions with valid endpoint limits.
- Reverse a double integral by reconstructing its region.
- Use reflection symmetry while checking removable endpoint behaviour.
Match each term with its precise meaning in this lesson.
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Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
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The order follows the stated process; check each stage before the next.