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Calculus 2 comparison theorem
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@ganeshie8
Notice that \(x^{1.1} + 5\ge x+ 5\) in \([1, \infty]\)
that means 1/(x^1.1+5) <= 1/x+5
so u can test and see if below integral converges : \(\large \int \limits_1^{\infty} \frac{1}{x+5} dx\)
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and clearly that does NOT converge.
where did u get [1, infinity]?
@ganeshie8
from the problem itself look at the bounds of ur integral
\[ \frac{1}{x^{1.1}+5}<\frac{1}{x^{1.1}} \\ \int \frac{1}{x^{1.1}} \, dx=-\frac{10.}{x^{0.1}}\\ \int_1^{\infty } \frac{1}{x^{1.1}} \, dx=10 \] Your integral is convergent by the comparison theorem @ganeshie8
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In general \[ \int_1^\infty \frac {dx}{x^p} \] is convergent if p>1 and divergent if \( p\le 1\)
Thank you both of you
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