Evaluating \lim_{x \to \infty}\frac{1}{x}\int_0^x|\sin(t)|dt

Jenny Branch

Jenny Branch

Answered question

2022-01-29

Evaluating limx1x0x|sin(t)|dt

Answer & Explanation

Darrell Boone

Darrell Boone

Beginner2022-01-30Added 9 answers

Note that |sin(t)| is non-negative, periodic with period π, and that
0π|sin(t)|dt=2
Let f(x) be the largest integer smaller than or equal to xπ. Then it holds that
0f(x)π|sin(t)|dt0x|sin(t)|dt0[f(x)+1]π|sin(t)|dt
This can be written as
2f(x)0x|sin(t)|dt2[f(x)+1]
Dividing by x and noting that limxf(x)x=1π it follows that
2πlimx+1x0x|sin(t)|dt2π

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