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^21)) and f(f(e)) = e
further we can simplify f(e) to
sqrt(e^2/(e^21)) = e/sqrt(e^21)
so we want
int e + e/sqrt(e^21) 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^21) de
use substitution with u = e^21 thus du =2e de thus we have
int 0.5*u^(1/2) du =
(1/2)*2*u^(1/2) =
sqrt(x^21)
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 20100219 12:15:26 