I'm given that the definite integration of some function is equal to $0$.
Now I've to find maximum value of definite integration of cube of same function between same bounds.
How to do it? Please explain.
I'm writing it in mathematical form.
$$\int^1 _0 f(x)=0:-1\le f(x)\le1$$
Find the maximum value of $$\int^1_0 [f(x)]^3$$
Seriously I have no clue to do the question. I tried to draw the graph(experimental) where I kept areas in 1st, 4th quadrants equal so that the integration is zero. Now I'm trying but draw the graph of cubic of same graph. And maximize the area in the first quadrant. I'm thinking about to do it using some graphical calculator and have some visualization of the problem but obviously I have no clue how to do that even.
Thanks for solving the question or giving it time.
Best Answer
Let $f: [0,1] \to [-1,1]$ be any integrable function satisfies the constraint $\int_0^1 f(x) = 0$. For any $x \in [0,1]$,
$$f(x) \in [-1,1]\quad\implies\quad (1 - f(x))\left(f(x) + \frac12\right)^2 \ge 0$$ Integrating RHS gives us
$$\int_0^1 \left[\frac14 + \frac34 f(x) - f(x)^3\right] dx \ge 0 \implies \int_0^1 f(x)^3 dx \le \frac14 $$ Since this value $\frac14$ is achievable by the function
$$f(x) = \begin{cases}1, & x \in \left[0,\frac13\right]\\ -\frac12, & x \in (\frac13,1]\end{cases}$$ The maximum value of the integral $\int_0^1 f(x)^3 dx$ is $\frac14$.