The respective
eccentricities of the hyperbola x
2/ a
2 - y
2/ b
2 = 1 and its conjugate y
2/ b
2 - x
2/ a
2 = 1 are denoted by e and f(e).
Evaluate:
√2
∫ (f(e) + f(f(e))) de
1
I agree with Charlie that
f(e) = sqrt(1+1/(e^2-1)) and f(f(e)) = e
further we can simplify f(e) to
sqrt(e^2/(e^2-1)) = e/sqrt(e^2-1)
so we want
int e + e/sqrt(e^2-1) de so split it into 2 integrals
int e de = e^2/2 from 1 to sqrt(2) equals 2/2 - 1/2 = 1/2
int e/sqrt(e^2-1) de
use substitution with u = e^2-1 thus du =2e de thus we have
int 0.5*u^(-1/2) du =
(1/2)*2*u^(1/2) =
sqrt(x^2-1)
thus from 1 to sqrt(2) we get
1 - 0 = 1
thus the total integral is equal to
(1/2) + 1 = 3/2
which agrees with charlie's numerical calculation
|
|
Posted by Daniel
on 2010-02-19 12:15:26 |
analytical solution
