maths.free › Real Analysis › Improper integrals

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.

Gumagana halimbawa · integrate 1/x^2 dx from 1 to oo

Calculus · Definite-Integrals

Integrate x^(-2) from 1 to oo

\int_{1}^{\infty} \frac{1}{x^{2}}\, dx

Hakbang-hakbang

  1. \int_{1}^{\infty} \frac{1}{x^{2}}\, dx

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. \int \frac{1}{x^{2}}\, dx = - \frac{1}{x}

    Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).

  3. F(\infty) - F(1) = \left(0\right) - \left(-1\right)

    Fundamental theorem of calculus: plug in the limits.

  4. = 1

    Simplify.

Ipahayag ang sagot
1

Ngayon ikaw Pumili ng isang problema, o i-type o gumuhit ng iyong sarili. Sa bawat hakbang, isang larawan, ang sagot na nakatago hanggang sa hilingin mo.

Panatilihin ang iyong sariling nagtatrabaho

Ang isang libreng account ay nagdaragdag ng mga tala sa bawat aralin, isang tala ng kung ano ang iyong natapos, ang iyong mga nalulutas na problema sa isang lugar, at isang guro na maaari mong tanungin tungkol sa pahinang ito. Ang matematika mismo ay bukas sa lahat, naka-sign in o hindi.

Mag-sign Up Mag-login

Mga simbolong ginagamit dito

I-tap ang anumang simbolo para sa buong kahulugan, isang larawan, at kung ano ang ibig sabihin ng bawat titik sa ito.

Paano: Improper integrals

  1. First find an antiderivative F, then evaluate F(b) − F(a).
  2. Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).
  3. Fundamental theorem of calculus: plug in the limits.
  4. Simplify.

Mga katanungan na tanong ng mga tao

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.

Higit pa sa Real Analysis

Sequences and their limitsConvergence of seriesTaylor approximation and errorFunctions of a complex variableContinuity and differentiability, rigorouslySequences of functions and uniform convergence