Answer :
Answer:
b = (5±√17)/4 ≈ {0.2192, 2.2808}
Step-by-step explanation:
The average on the interval is found from the integral ...
[tex]\frac{1}{b}\int\limits^b_0 {(4+10x-6x^2)} \, dx =\frac{1}{b}(4b+5b^2-2b^3)\\\\=-2b^2+5b+4[/tex]
We want the value of this integral to be 5, so we want to find b such that ...
5 = -2b^2 +5b +4
2b^2 -5b +1 = 0
b^2 -5/2b = -1/2
b^2 -5/2b +25/16 = -1/2 +25/16
(b -5/4)^2 = 17/16
b = (5±√17)/4
The average value of f(x) is 5 on the intervals ...
[0, (5-√17)/4] and [0, (5+√17)/4]