<p>Can someone help me? Thank you in advance.</p>
<p>Find the maximum of the integral from 0 to 1 of f(x) cubed, given the constraints
|f(x)| <= 1, integral from 0 to 1 of f(x) dx =0.</p>
<p>I believe I have found the answer. Please forgive me if I am wrong. From the problem, it should be clear that the graph of f(x) is positive on some intervals and negative on the others and that the difference of the positive function area and the negative function area is zero, as given. For simplicity, let’s make f(x) positive only on the first interval and negative on the other interval. Remember that we are only focused on perserving the function’s area. To futher simplify the problem, let’s make f(x) constant on both intervals. This may seem stupid at first - but it really isn’t. Recall that the area under a function equals the average function value times the length of the interval. Now the integral of f(x) (and hence the integral of [f(x)]^3) is reduced to the difference of the areas of rectangles. Let the average function value on the “positive” interval = a and the absolute value of the average function value on the “negative” interval = b and x be where f(x) = 0. Then, by the area condition, ax = (1-x)b. Upon solving for b, we have b=ax/(1-x). Let the function g(x) be the values of the integral of [f(x)]^3 on the interval [0,1] dependent upon the location of the zero of f. This is the function we need to maximize on the right interval. We now have g(x)=x(a^3)-(1-x)[ax/(1-x)]^3. This is, as said before, the difference of two rectangles’ areas. The “right” interval to maximize g(x) on is [0,.5] because, on the interval (.5,1) assuming a may be 1, b becomes greater than 1, which would break one of the conditions. Using differential calculus, we find that the maximum of g(x) on [0,.5] happens at x=1/3 g(x)=(a^3)/4. Noting that the maximum value of f is 1, the long anticipated answer is - drum roll please - 1/4.</p>
<p>I hope that helped. Please continue if you have any more questions or if I have made any mistakes. Please respond upon receiving this.</p>