Comparison Test for Integrals

Real Analysis · Axiom Academy

LESSON Comparison Test for Integrals Decide whether an improper integral converges — without ever computing its value — by trapping it between functions you already understand. An improper integral runs to infinity. We define it as a limit of ordinary integrals, and it converges when that limit is a finite number: converges when the limit is finite, diverges when it is infinite The animation sweeps out the area under 1/x^2 from 1 onward. The region is infinitely wide, yet the running total climbs toward a finite ceiling of 1 . That is convergence made visible — an unbounded domain enclosing a bounded area. Faster-than- 1/x decay (larger p ) wins; 1/x itself and anything decaying slower loses. 2. Direct Comparison: Trapped Under a Convergent Curve Suppose for all . If the larger function g encloses a finite area, the smaller f — sitting entirely beneath it — cannot enclose more. So f converges too. If converges , then converges. (smaller than convergent → convergent) If diverges , then diverges. (bigger than divergent → divergent) Watch both areas fill at once. The blue area for f is forever capped by the orange area for g , and g 's total settles at a finite value — so the readout for f can never run away. For every we have , so flipping the inequality: Since converges ( p = 2 > 1 ), the smaller integral converges . (Its exact value happens to be , but we never needed that to decide.) For we have , hence , and since e^ t is increasing, :

This is the written version of the interactive lesson above. See the full Real Analysis course.