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The improper integral ∫_1^∞ (1/x²) dx evaluates to:
AInfinity (diverges, like harmonic series)
B$0$ (vanishes at boundary)
C$2$ (twice the converged value)
D$1$ (converges to a finite value)
Answer & Solution
Correct answer: D. $1$ (converges to a finite value)
∫_1^∞ x^(-2) dx = [-1/x] from 1 to ∞ = 0 − (-1) = 1. Converges. Compare ∫_1^∞ (1/x) dx = ∞ (diverges, harmonic case).
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