Maximize the integral!

Solution : Consider the expression, $(f(x)-1)(f(x)+\frac{1}{2})^2$ . Note that, this expression is always negative. Also notice that, this expression is $f(x)^3-\frac{3f(x)}{4}-1/4$ So the integral, $\int_{0}^{1} f(x)-1)(f(x)+\frac{1}{2})^2 = \int_{0}^{1} f(x)^3dx -\frac{1}{4}$ . So answer is atmost $\frac{1}{4}$ . Since this expression will be $0$ when $f(x) = \frac{-1}{2},1$ . Since we can easily find such function, where $\int_{0}^{1} f(x)dx = 0$ . For example take, $f(x) = 1 , \forall x \in [0,1/3]$ and $f(x)=\frac{-1}{2} \forall x \in [1/3,2/3]$ . So we have proved that answer is $\frac{1}{4}$ .

On integrating, we get : $\frac { { (f\left( x \right) ) }^{ 4 } }{ 4 } \quad +\quad c$

On substituting $f(x) = 1$ from given inequality (As f(x) will be maximum when $f(x) = 1$ ), we get : $X=1/4$ .

Therefore, $[100X]{=}{25}$

we can simplify this by assuming that f(x) takes the form

f(x)=b for 0<=x<=a

f(x)=-c for a<x<=1

with 0<=a,b,c<=1

then the integration requirement gives that

ab=(1-a)c

and we wish to maximize

F(a,b,c)=ab^3-(1-a)c^3

from integration requirement we get

c=ab/(1-a)

thus

F(a,b)=ab^3-(1-a)a^3b^3/(1-a)^3

using the usual method of set dF/da=0 and dF/db=0, solving, and finding max we get

F(a,b,c) is maximized at (a,b,c)=(1/3,1,1/2) which gives a maximum area of x=1/4 thus we get [100X]=[100/4]=[25]=25