I need help Finding the average value of F(x) =3x^2-2 on the interval [0,2]
V venialove Junior Member Joined Mar 30, 2008 Messages 53 Mar 30, 2008 #1 I need help Finding the average value of F(x) =3x^2-2 on the interval [0,2]
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Mar 30, 2008 #2 Re: math (average value) The average value is given by 1b−a∫abf(x)dx\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dxb−a1∫abf(x)dx
Re: math (average value) The average value is given by 1b−a∫abf(x)dx\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dxb−a1∫abf(x)dx
V venialove Junior Member Joined Mar 30, 2008 Messages 53 Mar 30, 2008 #3 Re: math (average value) so do I take [0,2] then do this 1/2-0 which is undefined?
D Deleted member 4993 Guest Mar 30, 2008 #4 Re: math (average value) venialove said: so do I take [0,2] then do this 1/2-0 which is undefined?<<<< why do you think so? Click to expand... 1b−a = 12−0 = 12\displaystyle \frac{1}{b-a}\, = \, \frac{1}{2-0}\, = \,\frac{1}{2}b−a1=2−01=21 do not forget the definite integral part.
Re: math (average value) venialove said: so do I take [0,2] then do this 1/2-0 which is undefined?<<<< why do you think so? Click to expand... 1b−a = 12−0 = 12\displaystyle \frac{1}{b-a}\, = \, \frac{1}{2-0}\, = \,\frac{1}{2}b−a1=2−01=21 do not forget the definite integral part.
