# Average value

Hmmm.... got this one wrong

What is the average value of $$y = x^{2}\sqrt{x^3+1}$$ on the interval [0,2] ?

Okay, so another average value problem right?

$$f(b) - f(a) = f'(c)(b-a)$$
$$f(2) - f(0) = 2f'(c)$$
$$12 = 2f'(c)$$
So then the average value is 6 right?

## Answers and Replies

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jamesrc
Science Advisor
Gold Member
Your problem is that you're trying to find the average value of a function using the mean value theorem. Try using the definition of the average value of a function:

$$\bar y = \frac{\int_a^b{f(x)dx}}{b-a} = \frac 1 2 \int_0^2{x^2\sqrt{x^3+1}dx}$$

(Simple change in variable to solve from here.)

Simple use the following to find the Average Value of a function:

1\b-a$$\int_a^b{f(x)dx}$$

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