# Math Help - average value

1. ## average value

Hi everyone,

Could someone please tell me if this is correct?

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.

int. o to b 2+6x-3x^2

2x+6x^2/2-3x^3/3 from 0 to b=
2x+3x^2-x^3 from 0 to b=
2b+3b^2-b^3=3
b(2+3b-b^2)=3
2+3b-b^2=3
-1+3b-b^2=0
-b^2+3b-1=0
b=3
b=.382
b=2.618

Therefore, b=2.618

Thank you very much

2. Originally Posted by chocolatelover
Hi everyone,

Could someone please tell me if this is correct?

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.

int. o to b 2+6x-3x^2

2x+6x^2/2-3x^3/3 from 0 to b=
2x+3x^2-x^3 from 0 to b=
2b+3b^2-b^3=3
b(2+3b-b^2)=3
2+3b-b^2=3
I see two things wrong up to this point:
You haven't divided by b (the size of the interval) and if
$(x - a)(x - b) = c$ then we do not require that x - a = c or x - b = c. (We only do this when c = 0.)

Let me write this out to the point you have:
$3 = \frac{1}{b - 0} \int_0^b(2 + 6x - 3x^2)~dx$

$3 = \frac{1}{b}(2b + 3b^2 - b^3)$

$3b = 2b + 3b^2 - b^3$

Now solve for b.

-Dan

3. Thank you very much

I got 3=1/b(2b+3b^2-b^3)
3=2b/b+3b^2/b-b^3/b
0=-1+3b-b^2
b=.382, 2.618

Which one would I use? They both can't be right, can they?

Thank you very much

4. Originally Posted by chocolatelover
I got 3=1/b(2b+3b^2-b^3)
3=2b/b+3b^2/b-b^3/b
0=-1+3b-b^2
b=.382, 2.618

Which one would I use? They both can't be right, can they?

Thank you
Why the doubled post 3 hours apart??

Why can't there be two solutions? As a simple example look at the average value of sin(x) over the interval 0 to $2\pi$. Now look at the average of sin(x) over the interval 0 to $4\pi$. Both are 0.

-Dan