Question

For the integral \[ I = \int_{-1}^{1} \frac{1}{x^2}\, dx \] which of the following statements is TRUE?

• A. $I = 0$
• B. $I = 2$
• C. $I = -2$
• D. The integral does not converge

Hint:

Whenever the integrand has a singularity inside the interval (like \(x=0\) here), check convergence using limits. If it tends to infinity, the improper integral diverges.

Answer 1

The Correct Option is D

Step 1: Analyze the integrand.
The function to integrate is \(\dfrac{1}{x^2}\). Notice that it has a singularity (infinite discontinuity) at \(x=0\). Since the interval \([-1,1]\) includes 0, the integral becomes an improper integral.

Step 2: Break the integral.
\[ I = \int_{-1}^{1} \frac{1}{x^2}\, dx = \int_{-1}^{0} \frac{1}{x^2}\, dx + \int_{0}^{1} \frac{1}{x^2}\, dx \]

Step 3: Compute the antiderivative.
\[ \int \frac{1}{x^2}\, dx = \int x^{-2}\, dx = -\frac{1}{x} \]

Step 4: Evaluate near the singularity.
- For \(\int_{0}^{1} \frac{1}{x^2} dx\): \[ \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} \frac{1}{x^2}\, dx = \lim_{\epsilon \to 0^+} \left[-\frac{1}{x}\right]_{\epsilon}^{1} = \lim_{\epsilon \to 0^+} \left(-1 + \frac{1}{\epsilon}\right) = \infty \] - Similarly, for \(\int_{-1}^{0} \frac{1}{x^2} dx\): \[ \lim_{\epsilon \to 0^+} \int_{-1}^{-\epsilon} \frac{1}{x^2}\, dx = \lim_{\epsilon \to 0^+} \left[-\frac{1}{x}\right]_{-1}^{-\epsilon} = \lim_{\epsilon \to 0^+} \left(\frac{1}{\epsilon} - 1\right) = \infty \]

Step 5: Conclude.
Both integrals diverge to \(+\infty\). Therefore, the overall integral \(\int_{-1}^{1} \frac{1}{x^2}\, dx\) does not converge.
\[ \boxed{\text{The integral does not converge.}} \]