# Finding the upper limit of the integral

• Jan 31st 2010, 09:09 AM
racewithferrari
Finding the upper limit of the integral
Find the numbers b such that the average value of f(x) = 2+6x-3x^2 on the interval [0,b] is equal to 3

I am stuck at b^2(3-b)=b and don't know what to do. Or I am doing something wrong ?
Attachment 15131
• Jan 31st 2010, 09:18 AM
Plato
Just solve $b^3-3b^2+b=0.$
• Jan 31st 2010, 09:22 AM
racewithferrari
Oh!!! Thanks I didnt realize that
• Jan 31st 2010, 09:55 AM
racewithferrari
So my answer is coming in decimals,is it ok...?
• Jan 31st 2010, 10:04 AM
Plato
Quote:

Originally Posted by racewithferrari
So my answer is coming in decimals,is it ok...?

I get $b=\frac{3\pm\sqrt{5}}{2}$.