integral

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July 29th 2006, 07:11 PM
nertil1

integral

can someone help me do this
the intergral, square root of 1 + 9x^4 dx
i can't seem to figure it out
thanks
July 29th 2006, 08:02 PM
ThePerfectHacker

It might be an elliptic integral, were you working with ellipse when you got it?
July 29th 2006, 10:14 PM
CaptainBlack

1 Attachment(s)

Quote:

Originally Posted by nertil1

can someone help me do this
the intergral, square root of 1 + 9x^4 dx
i can't seem to figure it out
thanks

Not an elementary integral. The attachment shows what QuickMath
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
July 31st 2006, 11:59 AM
nertil1

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
July 31st 2006, 12:40 PM
topsquark

Quote:

Originally Posted by nertil1

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

Actually $s = \int_a^b \, dx \sqrt{1 + \left ( f'(x) \right ) ^2}$ . You applied it right, but explained it wrong.

-Dan
July 31st 2006, 01:13 PM
ThePerfectHacker

Quote:

Originally Posted by nertil1

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

Ah! Then the book wanted you to approximate the integration. That can be done easily, if that is what you want tell us and we solve it for you.
July 31st 2006, 01:59 PM
galactus

I used the midpoint approximation with 10 subintervals. You could try trapezoid, right, left, Simpson,etc.

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

Code:

x ------ ---------------- 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

This is very close to the actual of 63.12410226.

Of course, the more intervals the greater the accuracy.
July 31st 2006, 05:27 PM
c_323_h

i think there's supposed to be a $x^3$ somewhere so you can use $u=1+9x^4$ or maybe not
July 31st 2006, 06:25 PM
nertil1

is there a way to evaluate the integral to get the precise answer or can you only estimate it?
July 31st 2006, 07:10 PM
ThePerfectHacker

Quote:

Originally Posted by nertil1

is there a way to evaluate the integral to get the precise answer or can you only estimate it?

Let me try to explain you.

The [I]natural logarithm function[I] is definied as,
$\ln x=\int_1^x \frac{dx}{x}$ $,x>0$ .

Note, if you never made up such a function there is was no way to evaluated for example,
$\int_0^1 \frac{1}{x+1} dx$ cuz it gives,
$\ln 2$ . 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.
July 31st 2006, 07:12 PM
ThePerfectHacker

Quote:

Originally Posted by c_323_h

i think there's supposed to be a $x^3$ somewhere so you can use $u=1+9x^4$ or maybe not

If there were then substitute would be the trick.

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.