# 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.

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We will apply the formula:

`avg = 1/(b-a) int_a^b f(x)dx`

Therefore, `a = 0` , `b = b` and `avg = 3` .

We will have:

`3= 1/(b-0) int_0^b (2+6x-3x^2)dx`

`3= 1/(b) int_0^b (2+6x-3x^2)dx `

Remember the formula

`int_a^b u^ndu = (u^(n+1))/(n+1) |_a^b `

So, we will have,

`3= 1/(b) (2x + (6x^2)/(2) - (3x^3)/(3))|_0^b`

Plug-in the limits.

`3 = 1/b(2b + (6b^2)/(2) - (3b^3)/(3))`

`3 = 1/b(2b+3b^2-b^3)`

` `

Distributing the 1/b.

`3 = (2b)/(b) + (3b^2)/(b) - (b^3)/(b)`

`3 = 2 + 3b - b^2`

Making the left side equal to `0` , by subtracting both sides by `3` .

`0 = -1 + 3b - b^2`

We can rewrite that as `-b^2 + 3b -1 = 0` .

Multiply both sides by `-1` to make `-b^2` a positive.

`b^2 - 3b + 1 = 0`

Apply quadratic formula.

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

`a = 1` , `b = -3 ` and `c = 1`

Plug-in the values in the formula:

`x = (-(-3) +- sqrt((-3)^2 - 4(1)(1)))/(2(1))`

Taking note that negative times negative is positive.

`x =(3 +-sqrt(9 - 4))/(2)`

`= (3 +-sqrt(5))/(2)`

Therefore, the values of b are: `(3 + sqrt(5))/(2)` and `(3 - sqrt(5))/(2)` .

That is it! :)