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## Re: moment of inertia

$I_{\text{at }x = 2} = \Sigma \left[ {\displaystyle \int_{x_1}^{x_2}} (2 - x)^2 \, dA \right]$

$I_{\text{at }x = 2} = {\displaystyle \int_{\sqrt{2}}^{1.634}} (2 - x)^2(x^3 - 2x) \, dx + {\displaystyle \int_{1.634}^2} (2 - x)^2 [ \, (6 - x) - 2x \, ] \, dx$

$I_{\text{at }x = 2} = 0.0351 ~ \text{unit}^4$