Math Problem Statement

find the integral of 1/x^2 between[-1,1]

Solution

The integral of 1x2\frac{1}{x^2} over the interval [1,1][-1, 1] is improper because the integrand 1x2\frac{1}{x^2} becomes undefined at x=0x = 0. We need to split the integral into two parts and check whether each part converges:

111x2dx=101x2dx+011x2dx.\int_{-1}^{1} \frac{1}{x^2} \, dx = \int_{-1}^{0} \frac{1}{x^2} \, dx + \int_{0}^{1} \frac{1}{x^2} \, dx.

1. Integral ab1x2dx\int_{a}^{b} \frac{1}{x^2} \, dx:

The antiderivative of 1x2\frac{1}{x^2} is: 1x2dx=1x.\int \frac{1}{x^2} \, dx = -\frac{1}{x}.

2. Applying limits:

For the interval [1,1][-1, 1], splitting the integral around the singularity at x=0x = 0, we write: 111x2dx=limϵ0+(1ϵ1x2dx+ϵ11x2dx).\int_{-1}^{1} \frac{1}{x^2} \, dx = \lim_{\epsilon \to 0^+} \left( \int_{-1}^{-\epsilon} \frac{1}{x^2} \, dx + \int_{\epsilon}^{1} \frac{1}{x^2} \, dx \right).

Evaluating each part:

  • For 1ϵ1x2dx\int_{-1}^{-\epsilon} \frac{1}{x^2} \, dx: 1ϵ1x2dx=[1x]1ϵ=(1ϵ)(11)=1ϵ1.\int_{-1}^{-\epsilon} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_{-1}^{-\epsilon} = \left(-\frac{1}{-\epsilon}\right) - \left(-\frac{1}{-1}\right) = \frac{1}{\epsilon} - 1.

  • For ϵ11x2dx\int_{\epsilon}^{1} \frac{1}{x^2} \, dx: ϵ11x2dx=[1x]ϵ1=(11)(1ϵ)=1+1ϵ.\int_{\epsilon}^{1} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_{\epsilon}^{1} = \left(-\frac{1}{1}\right) - \left(-\frac{1}{\epsilon}\right) = -1 + \frac{1}{\epsilon}.

Adding the results:

limϵ0+(1ϵ1+(1+1ϵ))=limϵ0+(2ϵ2).\lim_{\epsilon \to 0^+} \left( \frac{1}{\epsilon} - 1 + (-1 + \frac{1}{\epsilon}) \right) = \lim_{\epsilon \to 0^+} \left( \frac{2}{\epsilon} - 2 \right).

As ϵ0+\epsilon \to 0^+, the term 2ϵ\frac{2}{\epsilon} \to \infty, so the integral diverges.

Conclusion:

The integral 111x2dx\int_{-1}^{1} \frac{1}{x^2} \, dx does not converge because the function 1x2\frac{1}{x^2} has a singularity at x=0x = 0, causing the integral to diverge to infinity.

Would you like further clarification or visualizations for this?


  1. What is an improper integral?
  2. How can we evaluate improper integrals with infinite limits?
  3. What conditions cause an integral to diverge?
  4. Can an integral diverge on only part of its domain?
  5. What are other examples of integrals with singularities?

Tip: Always check for singularities in the integrand or limits of integration to determine if an integral is improper.

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Math Problem Analysis

Mathematical Concepts

Calculus
Improper Integrals
Definite Integrals
Singularities

Formulas

Integral of 1/x^2: ∫(1/x^2) dx = -1/x
Improper Integral Splitting: ∫(a to b) f(x) dx = ∫(a to c) f(x) dx + ∫(c to b) f(x) dx, where c is the singularity point

Theorems

Divergence of Improper Integrals

Suitable Grade Level

College Calculus or Advanced High School AP Calculus