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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- Oct 25th 2009, 06:47 PMjlmills5improper 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? - Oct 25th 2009, 07:06 PMKrizalid
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.