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Thread: improper integral

  1. #1
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    improper integral

    int 4/[sqrt(x)(x+6)] dx from 0 to infinity

    0 and infinity are both problems...solution says to break it up as limits from 0 to 1 and 1 to infinity....why? how is this decided? Is a convenient number just decided upon?
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  2. #2
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    Krizalid's Avatar
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    it can be any number as long as $\displaystyle x\in[0,\infty)$ and besides is usual and convenient.

    we have $\displaystyle \int_{0}^{\infty }{\frac{dx}{(x+6)\sqrt{x}}}=\int_{0}^{1}{\frac{dx} {(x+6)\sqrt{x}}}+\int_{1}^{\infty }{\frac{dx}{(x+6)\sqrt{x}}},$ so in order to get convergence here, these pieces must converge.

    the first one does by simple limit comparison test with $\displaystyle \int_0^1\frac{dx}{\sqrt x},$ and for the second one, for each $\displaystyle x\ge1$ is $\displaystyle \frac{1}{(x+6)\sqrt{x}}<\frac{1}{x\sqrt{x}}=\frac{ 1}{x^{3/2}},$ thus these pieces converge and the original integral does.
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