Minimum value of integral

Question

If $$A = \int_0^1 x^2f(x)\,\mathrm dx $$ and $$B = \int_0^1 x(f(x))^2\,\mathrm dx,$$ then find the minimum value of $(B-A)$.

Edit: Also deduce $f(x)$.

Answer 1

Here is an elementary way to do it. Write $y=f(x)$, then $$B-A = \int _0^1 (x(y^2-xy+{x^2\over 4}) - {x^3\over 4})\; dx$$

$$= \int _0^1 (x\cdot \underbrace{(y-{x\over 2})^2 }_{\geq 0} - {x^3\over 4})\; dx$$

$$\geq \int _0^1 ( - {x^3\over 4})\; dx =-{1\over 16}$$

and the minimu is attaned at $f(x)={x\over 2}$

Answer 2

In order to minimize the integrand, we want to minize the function that is being integrated.

So in our case we want to find a $f:[0,1]\to\Bbb R$ such that the integrand $xf(x)^2-x^2f(x)$ is as small as possible for every $x\in[0,1]$.

For every $x\in[0,1]$, define $F_x:\Bbb R\to\Bbb R: y\mapsto xy^2-x^2y$. We want to minimize $F_x$ for every $x\in[0,1]$. Note that $F_x'(y)=2xy-x^2$. So $F_x'(y)=0\iff y=\frac x2$. Mini-exercise for you: Check that this $y=\frac x2$ minimizes every $F_x$.

Thus, we conclude that $f(x)=\frac x2$ minimizes our integral. For this $f$, we have

$$B-A=-\int_0^1 \frac{x^3}4\,\mathrm dx = -\frac1{16}.$$

Answer 3

$$B-A = \int_{0}^{1} xf^2-x^2f dx = \int_{0}^{1} L(x,f) dx,$$ $$L(x,f) = xf^2-x^2f.$$ $$\frac{\partial L}{\partial f} = 0 \hspace{0.2in}\text{(Euler-Lagrange equation)}$$ $$\implies 2xf(x)-x^2 = 0 \implies f(x) = \frac{1}{2}x.$$ So, $$B-A = \int_{0}^{1} \frac{1}{4}x^3 - \frac{1}{2}x^3 dx = -\frac{1}{16}.$$