Finding whether the following integral converges or diverges?

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Answer 1 · 2018-05-19T01:06:50

The integral converges

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This is the graph of the function $y = \frac{1}{\sqrt{x^{5} + 2}}$ $y = frac(1)(sqrt(x^(5) + 2))$ .

Clearly, from $x = 1$ $x = 1$ onwards, the area under the curve converges to a finite value.

Answer 2 · 2018-05-19T01:20:13

Obviously

$\int_{1}^{\infty} \frac{1}{\sqrt{x^{5} + 2}} d x > 0$ $int_1^oo 1/sqrt(x^5+2)dx >0$

For $x > 1$ $x>1$ we have

$x^{4} < x^{5} + 2 \Rightarrow \frac{1}{x^{2}} > \frac{1}{\sqrt{x^{5} + 2}}$ $x^4 < x^5+2 implies 1/x^2>1/sqrt(x^5+2)$

Thus

$\int_{1}^{\infty} \frac{1}{\sqrt{x^{5} + 2}} d x < \int_{1}^{\infty} \frac{d x}{x^{2}}$ $int_1^oo 1/sqrt(x^5+2)dx < int_1^oo dx/x^2$

The latter integral is

$lim_{Lto oo} int_1^L dx/x^2 = lim_{L to oo} (1-1/L) = 1$

and thus we have

$0 < int_1^oo 1/sqrt(x^5+2)dx < 1$

and thus the integral converges.