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26-01-2018
Mathematics

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Use the comparison theorem to determine whether the integral is convergent or divergent integral sin^2x/sqrt(x) from 0 to pi

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LammettHash LammettHash

26-01-2018

[tex]-1\le\sin x\le1\implies0\le\sin^2x\le1\implies0<\dfrac{\sin^2x}{\sqrt x}\le\dfrac1{\sqrt x}[/tex]

Does

[tex]\displaystyle\int_0^\pi\frac{\mathrm dx}{\sqrt x}[/tex]

converge?

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