Fundamental Theorem of Calculus help needed:

Find the numbers "b" such that the average value of f(x)=2 + 7x - x^3 on the interval [0,b] = 3

Question

Fundamental Theorem of Calculus help needed:

Find the numbers "b" such that the average value of f(x)=2 + 7x - x^3 on the interval [0,b] = 3

phi · Answer

The idea is that a "box" from x=0 to b, with a height of 3 has the same area as the area under the curve from x=0 to b

phi · Answer

The area under the curve is found by integrating
$$ \int_0^b 2 + 7x - x^3 $$
and you want this area to be the same area as the box with width (b-0)*3 or 3b
$$ 3b= \int_0^b 2 + 7x - x^3 $$

The first step is integrate the right hand side. 
then solve for b

anonymous · Answer

So the integration would be 2x + 7x^2/2 - x^4/4

phi · Answer

yes, now evaluate between the bounds 0 to b
you get
2b + (7/2)b^2 - (¼) b^4 = 3b

anonymous · Answer

Okay Okay

Then aI can subtract 2b to get 3.5b^2 - b^4/4 = 1b

Where am I left then?

phi · Answer

you get b=0 
and 
0.25 b^3 -3.5b + 1 =0

this is difficult to solve by hand, but wolfram gives the values.

anonymous · Answer

Oh of course you do. Thank you I will look there.