Final answer:
The integral \( \int_{1}^{\infty} x^2 dx \) diverges because the function y = x^2 increases without bound as x approaches infinity, leading to an unbounded area under the curve.
Step-by-step explanation:
The integral \( \int_{1}^{\infty} \frac{1}{x^{-2}} dx \) requires evaluation to determine convergence or divergence. Rewriting the integral, we get \( \int_{1}^{\infty} x^2 dx \). This integral represents the sum of the areas under the curve y = x^2 from x = 1 to x \rightarrow \infty. The function y = x^2 is an increasing function, and as x approaches infinity, y also becomes infinitely large, forming an asymptotic behavior.
Since the function y = x^2 increases without bound as x increases, the area under the curve also grows without limit. Consequently, the integral \( \int_{1}^{\infty} x^2 dx \) is indeed divergent, so the statement given in the original question is true. No further justification is needed beyond recognizing the integral of x^2 over the interval from 1 to infinity is unbounded.