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- Thread starter ISITIEIW
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- Jan 31, 2012

- 253

Thus the integral converges by the p-test.

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- Jan 31, 2012

- 253

- Feb 9, 2012

- 118

You're taking the integral of such function for all $x\ge1$ and within this range, the function is continuous as said above which is important to bound the integrand the way we want. So for example $\dfrac{1}{{{x}^{6}}+2}<\dfrac{1}{{{x}^{6}}}$ holds always for $x\ge1$ and besides $3x^2-2<3x^2+2$ holds always, then actually for all $x\ge1$ you have \(\displaystyle \displaystyle\frac{3{{x}^{3}}-2}{{{x}^{6}}+2}<\frac{3{{x}^{3}}+2}{{{x}^{6}}}= \frac{3}{x^3}+\frac{2}{{{x}^{6}}},\) implying $\displaystyle\int_{1}^{\infty }{\frac{3{{x}^{3}}-2}{{{x}^{6}}+2}\,dx}<\infty $Hey, not too sure about what function i would compare this integral from 1 to infinity of (3x^3 -2)/(x^6 +2) dx. I also have to show that it converges.

- Feb 15, 2012

- 1,967

1. If you are integrating a function $\dfrac{f(x)}{g(x)}$ it is often helpful to find a bounding function:

$\dfrac{h(x)}{k(x)}$ where:

$f(x) \leq h(x)$ for all $x > N$

$g(x) \geq k(x)$ for all $x > M$ and

$h(x),k(x)$ share some common factor so we have some nice cancellation occuring.

Close examination shows this is exactly what is happening in Krizalid's post (caveat: finding the "right" bounding functions can often take some algebraic ingenuity).