IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

KAB

Apply Newton-Leibniz rule which states that

If h(x),p(x),q(x) and f(t) are 4 functions such that

h(x) =_[q(x)}

^{[p(x) ]} f(t).dt

then

d(h(x))/dx = f(p(x))*(d(p(x))/dx) - f(q(x))*(d(q(x))/dx)

So here h(x)=f(x) , p(x)=x , q(x)=0 and f(t)=1/(f(x))²

On applying this rule,

f '(x)=(1/(f(x))²)*1-f(0)*0=1/(f(x))²

Which implies that (f(x))²f '(x) = 1 [Correction]

Or (f(x))³=x+c

Given f(2/3) = 3^{[2 ]}

2

So (3^{[2 ]}

2)³ = 2/3+c

Find c and then find f(72)

I think this is the shortest method!