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.

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.