$\Sigma M_{R2} = 0$
$12R_1 = 2400(6)(5) $
$R_1 = 6000 \, \text{lb}$
$\Sigma M_{R1} = 0$
$12R_2 = 2400(6)(7)$
$R_2 = 8400 \, \text{lb}$
$EI \, y'' = 6000x - \frac{1}{2}(2400) \, \langle \, x - 4 \, \rangle^2 + \frac{1}{2}(2400) \, \langle \, x - 10 \, \rangle^2$
$EI \, y'' = 6000x - 1200 \, \langle \, x - 4 \, \rangle^2 + 1200 \, \langle \, x - 10 \, \rangle^2$
$EI y' = 3000x^2 - 400 \, \langle \, x - 4 \, \rangle^3 + 400 \, \langle \, x - 10 \, \rangle^3 + C_1$
$EI \, y = 1000x^3 - 100\, \langle \, x - 4 \, \rangle^4 + 100\, \langle \, x - 10 \, \rangle^4 + C_1x + C_2$
At x = 0, y = 0, therefore C2 = 0
At x = 12 ft, y = 0
$0 = 1000(12^3) - 100(12 - 4)^4 + 100(12 - 10)^4 + 12C_1$
$C_1 = -110\,000 \, \text{lb}\cdot\text{ft}$
Therefore
$EI \, y = 1000x^3 - 100\, \langle \, x - 4 \, \rangle^4 + 100\, \langle \, x - 10 \, \rangle^4 - 110\,000x$
E = 29 × 106 psi
L = 12 ft
At midspan, x = 6 ft
y = -1/360 (12) = -1/30 ft = -2/5 in
Thus,
$EI \, y = 1000x^3 - 100\, \langle \, x - 4 \, \rangle^4 + 100\, \langle \, x - 10 \, \rangle^4 - 110\,000x$
$(29 \times 10^6)I (-\frac{2}{5}) = [ \, 1000(6^3) + 100(2^4) - 110\,000(6) \, ](12^3)$
$I = 66.38 \, \text{ in}^4$ answer
