1. ## average value problem

FInd an interval like [0,a] so that 1/x^1/3 is 1

I have that 1/a time int (0 to a) is 1/x^1/3 so then

1/a * 1.5* a^2/3 =1 and I get a=9/4, but this does not work when I plug it back in....what am I doing wrong here?

Then it tells you to find interval like [0,a] for 1/nth root of x.

2. Originally Posted by zhupolongjoe
FInd an interval like [0,a] so that 1/x^1/3 is 1

I have that 1/a time int (0 to a) is 1/x^1/3 so then

1/a * 1.5* a^2/3 =1 and I get a=9/4, but this does not work when I plug it back in....what am I doing wrong here?

Then it tells you to find interval like [0,a] for 1/nth root of x.
Could you state the question as it comes out of the book? What is the integrand supposed to be originally?

3. The questions is "find an interval of form [0,a] s.th. the average value of 1/cubed root of x = 1

4. Originally Posted by zhupolongjoe
The questions is "find an interval of form [0,a] s.th. the average value of 1/cubed root of x = 1
$\displaystyle \Rightarrow\frac{1}{a}\int_0^ax^{-1/3}=1$

$\displaystyle \Rightarrow\frac{1}{a}\left(\frac{3}{2}x^{2/3}\right)\left|\right.^a_0=1$

I think this is right...which is what you have...

$\displaystyle =\frac{1}{a}\left(\frac{3}{2}a^{2/3}-\frac{3}{2}(0)^{2/3}\right)=\frac{3}{2\sqrt[3]{a}}=1\Rightarrow{a}=\frac{27}{16}$