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Solution assignment 03 Improper integrals

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Assignment 3

Doest the following integral converge:

\displaystyle\int_{1}^{\infty}{\frac{1}{\sqrt{x}}}dx

Solution

We calculate:

\displaystyle\int_{1}^{\infty}{\frac{1}{\sqrt{x}}}dx=\lim_{a\to\infty}\int_{1}^{a}{\frac{1}{\sqrt{x}}}dx=

=\displaystyle\lim_{a\to\infty}[2\sqrt{x}]_{1}^{a}=\infty

Thus the integral diverges.

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