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Improper integrals
Integrals to infinity or across a singularity, defined as limits.
∫₁^∞ 1/x² dx is a limit of ∫₁^b as b → ∞, and it converges to 1; the same integral of 1/x diverges. The boundary between the two, p = 1, is the same boundary as for the p-series. The integral test is exactly this correspondence.
Үйлдэл · integrate 1/x^2 dx from 1 to oo
Алхам алхмаар
- \int_{1}^{\infty} \frac{1}{x^{2}}\, dx
First find an antiderivative F, then evaluate F(b) − F(a).
- \int \frac{1}{x^{2}}\, dx = - \frac{1}{x}
Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).
- F(\infty) - F(1) = \left(0\right) - \left(-1\right)
Fundamental theorem of calculus: plug in the limits.
- = 1
Simplify.
Хариулт
Одоо та Даалгавар сонго, эсвэл бич, эсвэл зур. Бүх алхам, зураг, хариулт нь асуулт асуух хүртэл нуугдаж байдаг.
Үнэгүй бүртгэл нь хичээл бүрийн тэмдэглэл, төгссөн хичээл, шийдсэн даалгаврын тэмдэглэл, энэ хуудсан дээр асууж болох багшийг нэгтгэж өгдөг. Математик нь бүртгэлтэй, бүртгэлтэйгүй хэн бүхэнд нээлттэй.
Бүртгүүлэх НэвтрэхЭнд ашигласан тэмдэгтүүд
Бүх тодорхойлолт, зураг, үсгийн утгыг харахын тулд ямар ч тэмдэгтийг дарна уу.
Яаж: Improper integrals
- First find an antiderivative F, then evaluate F(b) − F(a).
- Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).
- Fundamental theorem of calculus: plug in the limits.
- Simplify.
Хүмүүс асуудаг асуултууд
What is the ε, δ definition actually saying?
That you can make the output as close to the limit as anyone demands (within ε) by keeping the input close enough (within δ). It replaces "approaches" with a challenge-and-response that can be checked.
Why does the harmonic series diverge when its terms go to zero?
Because the terms shrink too slowly: group them as 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on: infinitely many halves.
Бүх зүйл Real Analysis
Sequences and their limitsConvergence of seriesTaylor approximation and errorFunctions of a complex variableContinuity and differentiability, rigorouslySequences of functions and uniform convergence