40. Let \( f \) be the function defined by

\[ f(x) = \begin{cases} \frac{1}{16}x^2 + 1 & \text{for } 0 \leq x \leq 4 \\ 3\sqrt{x} - x & \text{for } 4 < x \leq 9 \end{cases} \]

What is the average value of \( f \) on the closed interval \( 0 \leq x \leq 9 \)?






Answer is: option1

\( \frac{65}{54} \)

Solution:

\( \frac{65}{54} \)

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