Question

Is it true that the average value of an integrable function over an interval of length 2 is half the function's integral over the interval?

Short answer

Answer: Yes, the average value of an integrable function over an interval of length 2 is equal to half of the function's integral over the same interval.

Step-by-step solution

Remember the formula for average value of a function over an interval.

The average value of a function f(x) over an interval [a, b] is given by the formula: $$\text{Avg} = \frac{1}{b - a}\int_{a}^{b} f(x) dx$$ In our case, the length of the interval is 2, so b - a = 2.

Calculate the average value for the function over an interval of length 2.

Let's calculate the average value of the function using the formula from step 1, given an interval [a, a+2]: $$\text{Avg} = \frac{1}{(a+2) - a}\int_{a}^{a+2} f(x) dx \\ \Rightarrow \text{Avg} = \frac{1}{2}\int_{a}^{a+2} f(x) dx$$

Calculate half of the function's integral over an interval of length 2.

Now let's calculate half of the integral of the function over the same interval [a, a+2]: $$\frac{1}{2} \int_{a}^{a+2} f(x) dx$$

Compare the results from step 2 and step 3.

Looking at the results from steps 2 and 3, we can see that they are the same: $$\text{Avg} = \frac{1}{2}\int_{a}^{a+2} f(x) dx = \frac{1}{2} \int_{a}^{a+2} f(x) dx$$ So, yes, the average value of an integrable function over an interval of length 2 is half the function's integral over the interval.