The formula for the average value of a function over the interval [a,b] is:

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

So to find b, susbtitute `f(x) = 2+6x-3x^2` , `f_(avg)=3` , `a=0` to the formula above.

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

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

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

`3=1/b[(2b +...

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The formula for the average value of a function over the interval [a,b] is:

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

So to find b, susbtitute `f(x) = 2+6x-3x^2` , `f_(avg)=3` , `a=0` to the formula above.

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

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

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

`3=1/b[(2b + 3b^2-b^3)-0]`

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

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

Express the equation in quadratic form `ax^2+bx+c=0` .

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

Use the quadratic formula to solve for b.

`b=[-b+-sqrt(b^2-4ac)]/(2a) = [-(-3)+-sqrt((-3)^2-4*1*1)]/(2*1) = (3+-sqrt5)/2`

`b_1= 3/2 + sqrt5/2 ` and ` b_2=3/2-sqrt5/2`

**Hence, the values of `b` are `3/2+sqrt5/2` and `3/2-sqrt5/2` .**