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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.

Kushanda muenzaniso · 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

Step by step

  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.

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Maitiro: 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.

Mabvunzo vanhu vanobvunza

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.

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