Comparison Test for Improper Integrals — Question 3

Section 8-9 · Calculus pages 1585–1586

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Question 3

Use a lower comparison to determine convergence: ∫1∞2+sin⁡xxdx.\int_1^\infty\frac{2+\sin x}{x}\,dx.

Question 3 – Solution

Step 1: Bound the numerator. Since sin⁡x≥−1\sin x\ge-1, 2+sin⁡x≥1.2+\sin x\ge1. For x≥1x\ge1, divide by the positive number xx: 2+sin⁡xx≥1x≥0.\frac{2+\sin x}{x}\ge\frac1x\ge0. Step 2: Test the lower comparison. ∫1∞dxx=limb→∞ln⁡b=∞.\int_1^\infty\frac{dx}{x} =\lim_{b\to\infty}\ln b=\infty. Step 3: Apply direct comparison. The smaller integral diverges, so the original integral diverges. diverges\boxed{\text{diverges}}