EXAMPLE 1 Find the average value of the function f(x) = 3 + x2 on the interval [−3, 3].SOLUTION With a = −3 and b = 3, we havefave= 1b − abaf(x) dx= 13 −  3−3(3 + x2) dx= 16 3−3= .

Sure, I can help with that. Here's the step-by-step solution:

Step 1: Identify the function and the interval. In this case, the function f(x) = 3 + x^2 and the interval is [-3, 3].

Step 2: Apply the formula for the average value of a function on an interval [a, b], which is (1/(b - a)) ∫ from a to b f(x) dx.

Step 3: Substitute the function and the limits of the interval into the formula. This gives us (1/(3 - (-3))) ∫ from -3 to 3 (3 + x^2) dx.

Step 4: Simplify the expression. The denominator becomes 6, and the integral ∫ from -3 to 3 (3 + x^2) dx can be split into two integrals: ∫ from -3 to 3 3 dx + ∫ from -3 to 3 x^2 dx.

Step 5: Evaluate the integrals. The integral of 3 with respect to x from -3 to 3 is 6*3 = 18. The integral of x^2 with respect to x from -3 to 3 is [(3^3)/3 - ((-3)^3)/3] = 18.

Step 6: Add the results of the integrals and divide by 6 to get the average value of the function on the interval. This gives us (18 + 18) / 6 = 6.

So, the average value of the function f(x) = 3 + x^2 on the interval [-3, 3] is 6.