It might be an elliptic integral, were you working with ellipse when you got it?
Not an elementary integral. The attachment shows what QuickMathOriginally 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
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
I used the midpoint approximation with 10 subintervals. You could try trapezoid, right, left, Simpson,etc.
This is very close to the actual of 63.12410226.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
Of course, the more intervals the greater the accuracy.
Let me try to explain you.Originally Posted by nertil1
The [I]natural logarithm function[I] is definied as,
.
Note, if you never made up such a function there is was no way to evaluated for example,
cuz it gives,
. 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.