Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{-1} \frac{1}{x^{2}} d x $$

Short answer

The given improper integral converges and its value is 1.

Step-by-step solution

Investigate Convergence or Divergence

To determine whether the integral converges or diverges, a limit substitution is made. Here, let's let b be a real number such that -∞ < b < -1. So, we first consider the integral \(\int_{b}^{-1} \frac{1}{x^{2}} dx\).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is used to evaluate the integral, thus we have \(-\[ \frac{1} {x} \]_{b}^{-1} = -(-1 - - \frac{1} {b}) = 1 - \frac{1} {b}\).

Taking limit as b approaches -∞

Now, the limit is taken as \(b\) approaches \(-∞\), which gives us \(\lim_{b \to -\infty} (1 - \frac{1} {b})\). Given that the limit of a sum is equal to the sum of the limits (and the fact that the limit of a constant is that constant), this limit simplifies to \(1 - 0 = 1\).

Concretizing the Result

Since the limit is finite, the improper integral is convergent and its value equals the established limit, which in this case is 1.