average value

[tandoorichicken]

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?

[jamesrc]

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.)

[st3dent]

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

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