Question

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

Answer 1

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

Answer 2

Yup, that's the integral we need to find

Answer 3

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

Answer 4

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

Answer 5

@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?

Answer 6

according to wolfram, this is not a simple integral

Answer 7

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

Answer 8

it looks ugly either way

Answer 9

without limits
http://www.wolframalpha.com/input/?i=integral+%28t^2%29*sqrt%282%2B5t^4%29+dt

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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 ().

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

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?

Answer 22

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

Answer 23

Yeah that ^ hah

Answer 24

I am getting \( \sqrt{5} \)

Answer 25

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.

Answer 26

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!

Answer 27

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
\[ \frac{5}{\sqrt{5}}= \sqrt{5}\]

Answer 28

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