What is #f(x) = int xsqrt(x^2-1) dx# if #f(2) = 3 #?

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Answer 1 · 2016-07-25T19:44:09

#f(x)=1/3(x^2-1)sqrt(x^2-1)+3-sqrt3#..

Let us subst. #x^2-1=t^2 rArr 2xdx=2tdtrArrxdx=tdt#.

Hence, #intxsqrt(x^2-1)dx=intsqrt(t^2)tdt=intt^2dt=t^3/3#

But, #x^2-1=t^2 rArr t=(x^2-1)^(1/2)#

So, #f(x)=intxsqrt(x^2-1)dx=1/3(x^2-1)^(3/2)+C..................(1)#

To determine #C#, we are given the cond. that #f(2)=3#

By #(1)#, then, we have, #1/3*3^(3/2)+C=3 rArr C=3-sqrt3#

Hence, #f(x)=1/3*(x^2-1)^(3/2)+3-sqrt3#, or,

#f(x)=1/3(x^2-1)sqrt(x^2-1)+3-sqrt3#..