Questions: If the average value of f(x) = x^2 on the closed interval [0, b] is equal to 27, what is the value of b? 9 27 54 sqrt(27) sqrt(54)

Solution Steps

We start by calculating the integral of the function \( f(x) = x^2 \) over the interval \([0, b]\):

\[ \int_{0}^{b} x^2 \, dx = \frac{b^3}{3} \]

Next, we set up the equation for the average value of the function over the interval:

\[ \frac{1}{b} \int_{0}^{b} x^2 \, dx = 27 \]

Substituting the integral we calculated:

\[ \frac{1}{b} \cdot \frac{b^3}{3} = 27 \]

This simplifies to:

\[ \frac{b^2}{3} = 27 \]

To find \( b \), we multiply both sides by 3:

\[ b^2 = 81 \]

Taking the square root of both sides gives us:

\[ b = \pm 9 \]

Since \( b \) represents a length in the context of the problem, we only consider the positive value:

\[ b = 9 \]

Final Answer