부정적 적분과 기하학적 응용
도입됨| English | 의미 |
|---|---|
| improper integral/ɪmˈprɒpə ˈɪntɪɡrəl/ | 부정적 적분 |
| arc length/ɑːk leŋθ/ | The distance along a curved part of a circle or another curve. |
| principal value/ˈprɪnsɪpl ˈvæljuː/ | 주값 |
Finite volume, infinite surface
- Rotate $y = 1/x$, $x \ge 1$, about the $x$-axis. The horn holds a finite volume of paint but has infinite surface area.
- GRE integration tests improper integrals 反常积分, comparison, and the right geometric formula: area, volume, arc length 弧长 or surface area.
Improper integrals
- Replace an infinite bound or singular point by a limit. A singularity inside the interval: split, and require both pieces to converge.
- $\int_{-1}^{1} \frac{dx}{x}$ diverges: each half diverges, even though symmetric cut-offs cancel (that is only a principal value 主值).
Since $\lim_{\varepsilon \to 0}\left(\int_{-1}^{-\varepsilon} + \int_\varepsilon^1\right)\frac{dx}{x} = 0$, the improper integral $\int_{-1}^1 \frac{dx}{x}$ converges to 0.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Each half must converge on its own.
The $p$-test and comparison
- $\int_1^\infty x^{-p}\,dx$ converges $\iff p > 1$. $\int_0^1 x^{-p}\,dx$ converges $\iff p < 1$.
- Positive integrands: below a convergent one converges; above a divergent one diverges. Keep the inequality pointing the right way.
Converge or diverge?
Apply the p-test at infinity and at zero.
Evaluate $\int_1^\infty x^{-3}\,dx$.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
$[-\tfrac{1}{2x^2}]_1^\infty = \tfrac12$.
Which integral converges?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
For x ≥ 1, 0 < 1/(x² + sin²x) ≤ 1/x², and ∫₁^∞ dx/x² converges, so the comparison test gives convergence. 1/√x and 1/x decay too slowly at infinity, and 1/x² blows up too fast at 0.
Volumes, arc length, surface area
- Disks/washers: $\pi\int (R^2 - r^2)\,dx$. Shells: $2\pi\int (\text{radius})(\text{height})$.
- Arc length: $\int \sqrt{1 + f'^2}\,dx$. Surface of revolution: $2\pi\int f\sqrt{1 + f'^2}\,dx$.
Horn volume: $\pi\int_1^\infty x^{-2}\,dx = \pi$. Surface: $2\pi\int_1^\infty \tfrac1x\sqrt{1 + x^{-4}}\,dx \ge 2\pi\int_1^\infty \tfrac{dx}{x} = \infty$. Comparison settled it without an antiderivative.
Rotate $y = \sqrt x$, $0 \le x \le 4$, about the $x$-axis. The volume is $k\pi$. Find $k$.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
$\pi\int_0^4 x\,dx = 8\pi$.
Match each quantity to its integrand (rotation about the $x$-axis).
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Three different integrals.
Area between curves
- Integrate upper minus lower and split where the order changes; a signed integral is not an area.
- For $\max(f, g)$, find the crossings first: $\max(\sqrt{1-x^2}, x+1)$ on $[-1,1]$ uses the semicircle on $[-1,0]$ and the line on $[0,1]$, total $\tfrac\pi4 + \tfrac32$.
Recap
- Split at interior singularities; a principal value is not convergence.
- $p$-test: $p > 1$ at infinity, $p < 1$ at zero.
- Volume, arc length and surface area are different integrals; one finite says nothing about another.