can someone help me do this

the intergral, square root of 1 + 9x^4 dx

i can't seem to figure it out

thanks

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- Jul 29th 2006, 07:11 PMnertil1integral
can someone help me do this

the intergral, square root of 1 + 9x^4 dx

i can't seem to figure it out

thanks - Jul 29th 2006, 08:02 PMThePerfectHacker
It might be an elliptic integral, were you working with ellipse when you got it?

- Jul 29th 2006, 10:14 PMCaptainBlackQuote:

Originally Posted by**nertil1**

has to say about it, together with some explanatory text from

functions.wolfram.com - though how the -ve modulus is supposed to

be interpreted I don't know :(

RonL - Jul 31st 2006, 11:59 AMnertil1
No, I was actually trying to find the length of the function f(X) = x^3 from 1 to 4. The formula that was in my calc book said to do this I have to take the derivative of the function first then add 1 and then take the square root of the whole thing. then integrate it

- Jul 31st 2006, 12:40 PMtopsquarkQuote:

Originally Posted by**nertil1**

-Dan - Jul 31st 2006, 01:13 PMThePerfectHackerQuote:

Originally Posted by**nertil1**

*approximate*the integration. That can be done easily, if that is what you want tell us and we solve it for you. - Jul 31st 2006, 01:59 PMgalactus
I used the midpoint approximation with 10 subintervals. You could try trapezoid, right, left, Simpson,etc.

$\displaystyle \frac{4-1}{10}=\frac{3}{10}$

Code:

x $\displaystyle \sqrt{1+9x^{4}}$

------ ----------------

1.15 4.091584

1.45 6.38629

1.75 9.241762

2.05 12.647097

2.35 16.597652

2.65 21.091220

2.95 26.126645

3.25 31.703275

3.55 37.820723

3.85 44.478743

--------------

(.3)(210.184977)=**63.055493**

Of course, the more intervals the greater the accuracy. - Jul 31st 2006, 05:27 PMc_323_h
i think there's supposed to be a $\displaystyle x^3$ somewhere so you can use $\displaystyle u=1+9x^4$ or maybe not

- Jul 31st 2006, 06:25 PMnertil1
is there a way to evaluate the integral to get the precise answer or can you only estimate it?

- Jul 31st 2006, 07:10 PMThePerfectHackerQuote:

Originally Posted by**nertil1**

The [I]natural logarithm function[I] is definied as,

$\displaystyle \ln x=\int_1^x \frac{dx}{x}$ $\displaystyle ,x>0$.

Note, if you never made up such a function there is was no way to evaluated for example,

$\displaystyle \int_0^1 \frac{1}{x+1} dx$ cuz it gives,

$\displaystyle \ln 2$. So by introducting the natural logaithm function we can give precise values of certain integral.

However, this antiderivative cannot be expressed in regular function (like the natural logarithm so we introduce it) unless we introduce another class of function called "elliptic integral" (not to be confused with elliptic curves and elliptic functions). They enable us to express this value.

In reality, all you use anyways is an approximation so there is no harm in using the trapezodial rule. - Jul 31st 2006, 07:12 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

The problem in most arc length problems is that the functions produce integral of non-elementray functions. When the most basic ones like the one here.