## anonymous one year ago How do you integrate from 0 to x: (t^2)*sqrt(2+5t^4)? I've tried integration by parts where t^2 is dv/dt and sqrt(2+5t^4) is u but it ends up funny and possibly more complex. Integration by substitution also leaves me with a funny number. I'm really stuck, so any ideas would help. Thanks

1. Astrophysics

$\large \int\limits_{0}^{x}t^2 \sqrt{2+5t^4})dt$ correct?

2. anonymous

Yup, that's the integral we need to find

3. Astrophysics

I was thinking fundamental theorem of calculus part 1 but that's more for derivative, hmm.

4. anonymous

try to do it by multiplying and dividing by t and then apply by parts by that way you can easily integrate the radical part

5. anonymous

@divu.mkr I'm not quite sure what you mean by multiplying and dividing by t. Do you mean I should multiply the outside t into the square root?

6. phi

according to wolfram, this is not a simple integral

7. anonymous

What's a simple integral? Is it still a simple integral without the boundaries (0 and x)?

8. phi

it looks ugly either way

9. phi
10. anonymous

Wholey, the integral got symbols I've never seen before! On the other hand, it does make a very nice looking curve.

11. phi

That is what I meant by not simple. I have no idea how to get that answer.

12. Empty

Yeah, I get the feeling that the real question was "Differentiate this with respect to x" haha

13. anonymous

Hmm in that case then, maybe Astrophysics was on the right track... The full question is: calculate lim (x-->infinite) (x^5) $\lim_{x \rightarrow \infty}\frac{ x^5 }{ \int\limits_{0}^{x}(t^2)*\sqrt(2+5t^4) }$

14. anonymous

Opps there's a line between the x^5 and the bottom integral

15. anonymous

@Empty yea, I should've posted the whole question

16. Empty

Ok, yeah you are still asked to do a limit, so almost the same thing. Especially if you use L'Hopital's rule ;) $\lim_{x \rightarrow \infty}\frac{ x^5 }{ \int\limits_{0}^{x}(t^2)*\sqrt{2+5t^4}dt }$ Also, to make the square root work use {} instead of ().

17. anonymous

@Empty Haha and that would have been a much less stressful question.

18. phi

you have and infty/infty so take the derivative of top and bottom

19. anonymous

Oh good old L'Hopital's rule! Thank you so much!

20. Empty

To show that we can do L'Hopital's rule in a more convincing way here, notice: $\large \int\limits_{0}^{x}t^2 \sqrt{2+5t^4})dt \ge \int\limits_{0}^{x}t^2 \sqrt{t^4}dt$ Which is a lot easier to solve and compare to, so since this integral diverges the other one will too in case you weren't sure.

21. anonymous

Do you mean that we can use the right hand function as the denominator in place of the left one since they both diverge to similar degrees? Oh and, was the condition for being able to use L'Hopital's rule that your function is infinity/infinity?

22. phi

no, you keep the original Empty is just showing we have an infinity / infinity (specifically the denominator) so it is valid to use L'Hopital

23. Empty

Yeah that ^ hah

24. phi

I am getting $$\sqrt{5}$$

25. anonymous

Ah right I see (since t^2*sqrt(t^4) diverges)! Yup, the answer is sqrt5 I'm just a bit stuck on trying to find the limit... I've differentiated twice and got to $\frac{ 10x }{ \sqrt{(2+5x^4)^3}*20t^2 }$ But I can't divide by x^4 inside the sqrt as I don't know how to take it out.

26. anonymous

Oh wait hang on, I found my silly mistake (as always). Got it! Thank you guys and everyone else so much, having had to be baffled by trying to integrate the question. Sorry for not posting the whole question earlier!

27. phi

you should only differentiate once you get $\frac{5x^4}{x^2 \sqrt{2+5x^4} }$ divide top and bottom by x^4 $\frac{5}{x^{-2} \sqrt{2+5x^4} }$ bring the x^-2 inside the root $\frac{5}{\sqrt{2x^{-4} +5} }$ now take the limit. the 2/x^4 tends to zero and we get $\frac{5}{\sqrt{5}}= \sqrt{5}$

28. anonymous

Thanks, I eventually got onto the right track. I made the mistake of multiplying the denominator by 1/x^2 after dividing the inside by x^4 instead of multiplying by x^2 outside