Determine the value of a real constant c, given that:

     y
   ∫(ln p)*(1+p)-1 dp = g(y)
    1

and, g(e^c) + g(e^-c) = c³/12

where ln x denotes the natural logarithm of x.

from the fundamental theorem of calculus we have that

g'(x)=Ln(x)/(x+1)  (1)

now if we differentiate g(e^x)+g(e^-x)=x^3/12 in terms of x we get

e^x * g'(e^x) - e^-x * g'(e^-x)=x^2/4  (2)

now from (1) we have

g'(e^x)=x/(e^x + 1)  and g'(e^-x)=-x/(e^-x + 1)

substituting these into (2) we have

x*e^x/(e^x+1)+x*e^-x/(e^-x+1)=x^2/4

now we have a factor of x on both sides so right off we have a solution of x=0 factoring the x we are left with

e^x/(e^x+1) + e^-x/(e^-x+1)=x/4

now when we simplify the left side we end up with 1 thus

x/4=1   x=4 is our other solution

thus the 2 values for c are c=0 and c=4

 

Posted by Daniel on 2008-01-09 09:44:22