Average value

  • #1
Hmmm.... got this one wrong

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

Okay, so another average value problem right?

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

Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
476
1
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:

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

(Simple change in variable to solve from here.)
 
  • #3
21
0
Simple use the following to find the Average Value of a function:

1\b-a[tex] \int_a^b{f(x)dx} [/tex]
 
Last edited:

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