Let g(x) be a continuous and differentiable function such that `int_0^2{int_(sqrt2)^(sqrt5/2)[2x^2-3]dx}.g(x)dx=0,` then g(x) = 0 when `x in (0, 2)` has (where [ *] denote greatest integer function)

Answer

To solve the given problem step by step, we start with the integral provided in the question:

\[ \int_0^2 \left( \int_{\sqrt{2}}^{\frac{\sqrt{5}}{2}} (2x^2 - 3) \, dx \right) g(x) \, dx = 0 \]

### Step 1: Evaluate the inner integral

First, we need to evaluate the inner integral:

\[ \int_{\sqrt{2}}^{\frac{\sqrt{5}}{2}} (2x^2 - 3) \, dx \]

To do this, we find the antiderivative of \(2x^2 - 3\):

\[ \int (2x^2 - 3) \, dx = \frac{2}{3}x^3 - 3x + C \]

Now we will evaluate this from \(\sqrt{2}\) to \(\frac{\sqrt{5}}{2}\):

\[ \left[ \frac{2}{3}x^3 - 3x \right]_{\sqrt{2}}^{\frac{\sqrt{5}}{2}} \]

Calculating the upper limit:

\[ \frac{2}{3} \left( \frac{\sqrt{5}}{2} \right)^3 - 3 \left( \frac{\sqrt{5}}{2} \right) = \frac{2}{3} \cdot \frac{5\sqrt{5}}{8} - \frac{3\sqrt{5}}{2} \]

This simplifies to:

\[ \frac{5\sqrt{5}}{12} - \frac{18\sqrt{5}}{12} = \frac{-13\sqrt{5}}{12} \]

Now calculating the lower limit:

\[ \frac{2}{3} (\sqrt{2})^3 - 3(\sqrt{2}) = \frac{2}{3} \cdot 2\sqrt{2} - 3\sqrt{2} = \frac{4\sqrt{2}}{3} - \frac{9\sqrt{2}}{3} = \frac{-5\sqrt{2}}{3} \]

Now, we subtract the lower limit from the upper limit:

\[ \frac{-13\sqrt{5}}{12} - \left( \frac{-5\sqrt{2}}{3} \right) \]

To combine these, we convert \(-\frac{5\sqrt{2}}{3}\) to a fraction with a denominator of 12:

\[ -\frac{5\sqrt{2}}{3} = -\frac{20\sqrt{2}}{12} \]

Thus, we have:

\[ \frac{-13\sqrt{5}}{12} + \frac{20\sqrt{2}}{12} = \frac{-13\sqrt{5} + 20\sqrt{2}}{12} \]

### Step 2: Set the result of the inner integral into the outer integral

Now we substitute this result into the outer integral:

\[ \int_0^2 \left( \frac{-13\sqrt{5} + 20\sqrt{2}}{12} \right) g(x) \, dx = 0 \]

### Step 3: Analyze the integral

Since \(\frac{-13\sqrt{5} + 20\sqrt{2}}{12}\) is a constant, we can factor it out:

\[ \frac{-13\sqrt{5} + 20\sqrt{2}}{12} \int_0^2 g(x) \, dx = 0 \]

For this product to equal zero, either the constant must be zero or the integral must be zero.

### Step 4: Determine the conditions

1. The constant \(\frac{-13\sqrt{5} + 20\sqrt{2}}{12}\) is not zero (as \(-13\sqrt{5} + 20\sqrt{2} \neq 0\)).
2. Therefore, we must have:

\[ \int_0^2 g(x) \, dx = 0 \]

### Step 5: Conclusion about \(g(x)\)

Since \(g(x)\) is continuous and differentiable over the interval \((0, 2)\), the only way for the integral of a continuous function to equal zero over an interval is if the function itself is zero almost everywhere in that interval. Therefore, we conclude that:

\[ g(x) = 0 \quad \text{for all } x \in (0, 2) \]