Find the average value of the function on the given interval.
\[
f(x)=\sqrt{x+1} ; \quad[0,3]
\]

The average value is $\square$. (Type an integer or a fraction.)

Step 1 ：The problem is to find the average value of the function \(f(x) = \sqrt{x + 1}\) on the interval [0, 3].

Step 2 ：The formula for the average value of a function on an interval [a, b] is \(\frac{1}{b-a} \int_{a}^{b} f(x) dx\).

Step 3 ：In this case, the function is \(f(x) = \sqrt{x + 1}\), and the interval is [0, 3]. So, we need to integrate the function from 0 to 3, and then divide the result by 3 - 0 = 3.

Step 4 ：The integral of \(f(x) = \sqrt{x + 1}\) from 0 to 3 is 14/3.

Step 5 ：Dividing this by 3 gives the average value of the function on the interval [0, 3].

Step 6 ：The average value of the function on the given interval is \(\boxed{\frac{14}{9}}\).