To find the root mean square (RMS) value of the function $y = 9x^2 + 11$ between $x = 1$ and $x = 4$, we follow these steps:

Step-by-Step Solution

1. Definition of RMS Value:

   The RMS value of a function $y = f(x)$ over the interval $[a, b]$ is given by: $\text{RMS} = \sqrt{\frac{1}{b - a} \int_{a}^{b} [f(x)]^2 \, dx}.$

   For the given function $y = 9x^2 + 11$, the interval is $[1, 4]$, so: $\text{RMS} = \sqrt{\frac{1}{4 - 1} \int_{1}^{4} (9x^2 + 11)^2 \, dx}.$

2. Expand the Function $(9x^2 + 11)^2$:

   Let's expand $(9x^2 + 11)^2$: $(9x^2 + 11)^2 = 81x^4 + 2 \cdot 9x^2 \cdot 11 + 11^2 = 81x^4 + 198x^2 + 121.$

3. Set Up the Integral:

   Substitute the expanded function into the RMS formula: $\text{RMS} = \sqrt{\frac{1}{3} \int_{1}^{4} (81x^4 + 198x^2 + 121) \, dx}.$

4. Integrate with Respect to $x$:

   Now, let's compute the integral term-by-term: $\int (81x^4 + 198x^2 + 121) \, dx = \int 81x^4 \, dx + \int 198x^2 \, dx + \int 121 \, dx.$

   Evaluate each integral separately:

   $\int 81x^4 \, dx = 81 \cdot \frac{x^5}{5} = \frac{81}{5} x^5,$ $\int 198x^2 \, dx = 198 \cdot \frac{x^3}{3} = 66x^3,$ $\int 121 \, dx = 121x.$

   So, the integral becomes: $\int_{1}^{4} (81x^4 + 198x^2 + 121) \, dx = \left[ \frac{81}{5} x^5 + 66x^3 + 121x \right]_{1}^{4}.$

5. Compute the Definite Integral:

   Evaluate at the bounds $x = 4$ and $x = 1$:

   For $x = 4$: $\frac{81}{5} \cdot 4^5 + 66 \cdot 4^3 + 121 \cdot 4 = \frac{81}{5} \cdot 1024 + 66 \cdot 64 + 484.$

   Calculate these terms: $\frac{81}{5} \cdot 1024 = \frac{82944}{5} = 16588.8,$ $66 \cdot 64 = 4224,$ $121 \cdot 4 = 484.$

   Sum them up: $16588.8 + 4224 + 484 = 21396.8.$

   For $x = 1$: $\frac{81}{5} \cdot 1^5 + 66 \cdot 1^3 + 121 \cdot 1 = \frac{81}{5} + 66 + 121.$

   Calculate these terms: $\frac{81}{5} = 16.2,$ $66 + 121 = 187.$

   Sum them up: $16.2 + 187 = 203.2.$

   Now, compute the definite integral: $\int_{1}^{4} (81x^4 + 198x^2 + 121) \, dx = 21396.8 - 203.2 = 21193.6.$

6. Compute the RMS Value:

   Substitute the value of the integral into the RMS formula: $\text{RMS} = \sqrt{\frac{1}{3} \cdot 21193.6} = \sqrt{7064.53}.$

   Calculate the square root: $\text{RMS} \approx 84.02.$

Final Answer

The root mean square value of $y = 9x^2 + 11$ between $x = 1$ and $x = 4$ is approximately 84.02 (rounded to two decimal places).

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Would you like more details, or do you have any questions?

Here are some related questions to explore further:

1. How do you find the RMS value of a trigonometric function over a given interval?
2. What is the physical significance of the RMS value in engineering?
3. How does the RMS value change if the function is shifted vertically or horizontally?
4. What are some practical applications of RMS values in data analysis?
5. Can RMS values be used to compare the variability of different functions over the same interval?

Tip: The RMS value is particularly useful when analyzing alternating currents in electrical engineering.