The area as a function of b is given by
[tex] a(b)=\int\limits^b_0 {(2x^2+6)} \, dx =\frac{2}{3}b^3+6b [/tex]
You want to find b such that a(b) = 36.
... (2/3)b^3 +6b = 36
... b^3 +9b -54 = 0 . . . . multiply by 3/2 to clear fractions
A graphing calculator shows this equation to have one real solution at b=3.
The value of b is 3.
