Let `f:[0,1]to R` be a continuous function then the maximum value of `int_(0)^(1)f(x).x^(2)dx-int_(0)^(1)x.(f(x))^(2)dx` for all such function(s) is:

Answer

To find the maximum value of the expression

\[ I = \int_0^1 f(x) x^2 \, dx - \int_0^1 x (f(x))^2 \, dx \]

for a continuous function \( f: [0, 1] \to \mathbb{R} \), we will use the Cauchy-Schwarz inequality.

### Step 1: Apply Cauchy-Schwarz Inequality

Using the Cauchy-Schwarz inequality, we have:

\[ \left( \int_0^1 f(x) x^2 \, dx \right)^2 \leq \left( \int_0^1 (f(x))^2 \, dx \right) \left( \int_0^1 x^4 \, dx \right) \]

### Step 2: Calculate the Integral of \( x^4 \)

First, we calculate \( \int_0^1 x^4 \, dx \):

\[ \int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5} \]

### Step 3: Substitute into Cauchy-Schwarz Inequality

Substituting this into the inequality gives:

\[ \left( \int_0^1 f(x) x^2 \, dx \right)^2 \leq \left( \int_0^1 (f(x))^2 \, dx \right) \cdot \frac{1}{5} \]

### Step 4: Rearranging the Inequality

Rearranging gives:

\[ \int_0^1 f(x) x^2 \, dx \leq \frac{1}{\sqrt{5}} \sqrt{\int_0^1 (f(x))^2 \, dx} \]

### Step 5: Consider the Expression \( I \)

Now we need to consider the expression for \( I \):

\[ I = \int_0^1 f(x) x^2 \, dx - \int_0^1 x (f(x))^2 \, dx \]

### Step 6: Maximizing \( I \)

To maximize \( I \), we can assume \( f(x) = kx \) for some constant \( k \). Then we have:

\[ \int_0^1 f(x) x^2 \, dx = k \int_0^1 x^3 \, dx = k \cdot \frac{1}{4} \]

And,

\[ \int_0^1 x (f(x))^2 \, dx = \int_0^1 x (k^2 x^2) \, dx = k^2 \int_0^1 x^3 \, dx = k^2 \cdot \frac{1}{4} \]

Thus, substituting these into \( I \):

\[ I = k \cdot \frac{1}{4} - k^2 \cdot \frac{1}{4} = \frac{1}{4} (k - k^2) \]

### Step 7: Find the Maximum of \( \frac{1}{4} (k - k^2) \)

To maximize \( \frac{1}{4} (k - k^2) \), we can differentiate:

\[ \frac{d}{dk} \left( k - k^2 \right) = 1 - 2k \]

Setting this equal to zero gives:

\[ 1 - 2k = 0 \implies k = \frac{1}{2} \]

### Step 8: Calculate the Maximum Value of \( I \)

Substituting \( k = \frac{1}{2} \) back into \( I \):

\[ I = \frac{1}{4} \left( \frac{1}{2} - \left( \frac{1}{2} \right)^2 \right) = \frac{1}{4} \left( \frac{1}{2} - \frac{1}{4} \right) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \]

### Conclusion

Thus, the maximum value of

\[ \int_0^1 f(x) x^2 \, dx - \int_0^1 x (f(x))^2 \, dx \]

is

\[ \boxed{\frac{1}{16}}. \]