Is the Integral of ${x}^{2}/({x}^{5}+2)$ Convergent or Divergent?

[karush]

71. Use the Comparison Theorem to determine weather the integral
$$\displaystyle
I=\int\frac{{x}^{2}}{{x}^{5}+2} \, dx$$
is convergent or divergent.

Comparison Theorem Suppose that $f$ and $g$ are continuous with

$f(x) \ge \, g(x) \ge 0 $ for $x\ge a$

(a) if $\displaystyle \int_{a}^{\infty} f(x) \,dx
\text { is convergent then, }
\displaystyle \int_{a}^{\infty} g(x) \,dx
\text { is convergent}$

(b) if $\displaystyle \int_{a}^{\infty} g(x) \,dx
\text { is divergent then, }
\displaystyle \int_{a}^{\infty} f(x) \,dx
\text { is divergent}$

class hasn't started yet so clueless how to do this
looked at some examples but got lost...

the graph converges to 0

 

[I like Serena]

71. Use the Comparison Theorem to determine weather the integral
$$\displaystyle
I=\int\frac{{x}^{2}}{{x}^{5}+2} \, dx$$
is convergent or divergent.

class hasn't started yet so clueless how to do this
looked at some examples but got lost...the graph converges to 0

Hi karush!

Isn't the comparison theorem that we use for instance that:
$$0 \le \frac{{x}^{2}}{{x}^{5}+2} \le \frac{{x}^{2}}{{x}^{5}}$$
(Wondering)

No need for that integral test.

 

[karush]

guess so never have done it...
but well try some more...how would you know what $a$ is

 

[I like Serena]

guess so never have done it...
but well try some more...

how would you know what $a$ is

I think that should be:

Comparison Theorem Suppose that $f$ and $g$ are continuous with

$f(x) \ge \, g(x) \ge {\color{red}0} $ for $x\ge a$

(a) if $\displaystyle \int_{a}^{\infty} f(x) \,dx
\text { is convergent then, }
\displaystyle \int_{a}^{\infty} g(x) \,dx
\text { is convergent}$

(b) if $\displaystyle \int_{a}^{\infty} {\color{red}g}(x) \,dx
\text { is divergent then, }
\displaystyle \int_{a}^{\infty} {\color{red}f}(x) \,dx
\text { is divergent}$So $a$ is the lower boundary of the integrals.
It can be chosen arbitrarily - we can just leave it as is.

 

[karush]

ok i fixed... save it to latex library

does this always have to be an improper integral?

 

[I like Serena]

ok i fixed... save it to latex library

does this always have to be an improper integral?

The concept of convergence or divergence only applies if we're talking about some limit.
For integrals that means they have to be improper, otherwise there's no limit involved.