# Thread: Help Imporper integral by comparison??

1. ## Help Imporper integral by comparison??

Integral of:
dx / sqrt (x^4 + 4x^3) from upper limit b=infinte and a=1

Integral of:

1 + 2t(cos(t))^2(dt)/ (t)^3 from upper limit infinte an a=lowerlimit=pi

Please I need help using the comparison method to determine if integral converges or diverges.!
THANKS

2. Originally Posted by zangestu888
Integral of:
dx / sqrt (x^4 + 4x^3) from upper limit b=infinte and a=1

Integral of:

1 + 2t(cos(t))^2(dt)/ (t)^3 from upper limit infinte an a=lowerlimit=pi

Please I need help using the comparison method to determine if integral converges or diverges.!
THANKS
Since $\displaystyle x^4 < x^4 + 4x^3$ for $\displaystyle x \ge 1$ then $\displaystyle \frac{1}{x^4 + 4x^3} < \frac{1}{x^4}\,dx$ then

$\displaystyle \int_1^{\infty}\frac{1}{x^4 + 4x^3}\,dx < \int_1^{\infty}\frac{1}{x^4}\,dx < \infty$

For the second problem try $\displaystyle \cos^2 t \le 1$ and see if you can get an inequality like above.

3. OKay thank i will work on it

4. ## ??

so (cost)^2 <=1

am still not quite understanding it, can you give me a hint on how to begin i know i have to find a function that I know is either convergent or divergent 1/t^3 would be my function? am so lost? i wana know how to start if u may.
And for the first question I would like to knnow what hapened to the sqrt( of the function ) the radical u showed me disspeard?

5. Originally Posted by zangestu888
so (cost)^2 <=1

am still not quite understanding it, can you give me a hint on how to begin i know i have to find a function that I know is either convergent or divergent 1/t^3 would be my function? am so lost? i wana know how to start if u may.
And for the first question I would like to knnow what hapened to the sqrt( of the function ) the radical u showed me disspeard?
I totally missed the square root. Hopefully this fixes it.

Since $\displaystyle x^4 < x^4 + 4x^3$ for $\displaystyle x \ge 1$ then $\displaystyle \sqrt{x^4} < \sqrt{x^4 + 4x^3}$ then $\displaystyle \frac{1}{\sqrt{x^4 + 4x^3}} < \frac{1}{x^2}$ then

$\displaystyle \int_1^{\infty}\frac{1}{\sqrt{x^4 + 4x^3}}\,dx < \int_1^{\infty}\frac{1}{x^2}\,dx < \infty$

$\displaystyle \cos^2t < 1\;\;\; \Rightarrow \;\;\;1 + 2 t \cos^2 t < 1 + 2t \;\;\; \Rightarrow\;\;\; \frac{1 + 2 t \cos^2 t }{t^3} < \frac{1 + 2t}{t^3}$