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Chapter 2: Problem 28

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

Expert verified
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

01

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.
02

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]: $$\text{Avg} = \frac{1}{(a+2) - a}\int_{a}^{a+2} f(x) dx \\ \Rightarrow \text{Avg} = \frac{1}{2}\int_{a}^{a+2} f(x) dx$$
03

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]: $$\frac{1}{2} \int_{a}^{a+2} f(x) dx$$
04

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.

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Most popular questions from this chapter

If each case, give an example of a continuous function \(\mathrm{f}\) satisfying the conditions stated for all real \(\mathrm{x}\), or else explain why there is no such function : (a) \(\int_{0}^{x} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{e}^{x}\) (b) \(\int_{0}^{x^{2}} f(t) d t=1-2^{x^{2}}\). (c) \(\int_{0}^{x} f(t) d t f^{2}(x)-1\).

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