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To solve the problem of finding the root mean square (RMS) value of \( x^2 \) in the interval \( (0, 1) \), let's break it down step by step:

### Formula for RMS value:
The formula for the RMS value of a function \( f(x) \) over the interval \( [a, b] \) is given by:
\[
\text{RMS} = \sqrt{\frac{1}{b - a} \int_a^b f(x)^2 \, dx}
\]

In this case, the function is \( f(x) = x^2 \), and the interval is \( (0, 1) \).

### Applying the formula:
1. **Square the function:**
   \[
   f(x)^2 = (x^2)^2 = x^4
   \]

2. **Set up the RMS formula:**
   \[
   \text{RMS} = \sqrt{\frac{1}{1 - 0} \int_0^1 x^4 \, dx}
   \]
   Simplifying:
   \[
   \text{RMS} = \sqrt{\int_0^1 x^4 \, dx}
   \]

3. **Evaluate the integral:**
   \[
   \int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5}
   \]

4. **Final RMS calculation:**
   \[
   \text{RMS} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}
   \]

### Answer:
The RMS value of \( x^2 \) in the interval \( (0, 1) \) is \( \frac{1}{\sqrt{5}} \), which corresponds to option **C**.

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Would you like further details or explanations? Here are some related questions to explore:

1. How does the RMS value relate to the average value of a function?
2. Can RMS be calculated for any function over any interval?
3. How do you compute the RMS of a trigonometric function like \( \sin(x) \)?
4. What is the difference between RMS and standard deviation?
5. How would you compute the RMS of a piecewise function?

**Tip:** When working with integrals in definite intervals, always check the limits and function behavior for simplifications.

The R.M.S value of x^2 in the interval (0,1) is

Solve for the RMS value of x^2 over the interval (0, 1) using the root mean square formula and definite integration techniques. This problem involves polynomial functions and calculus concepts, suitable for Grades 11-12 or early college students.

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