Benedict Irwin edited main.tex  over 9 years ago

Commit id: 9c5e07fa4865aa7ecd98172e9949ac67022edbf0

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\end{matrix}  , \;\; a_1\le r \le a_2  \\  I_r(0.5)=\int_0^{I_r(0.5)} I_r(\beta)=\int_0^{I_r(\beta)}  r - \beta x \; \mathrm{dx} = \Bigg\{ \begin{matrix} \Bigg\{ \\ \\ \Bigg\{ \end{matrix}  \begin{matrix}  I_{a_2}^>(\beta)+2\sqrt{r-a_2}+\frac{\beta^*\delta}{2}(r-a_2)\\  I_{a_2}^>(\beta)-2\sqrt{r-a_2}+\frac{\delta^2}{4}(r-a_2)\\  I_{a_2}^<(\beta)+4\sqrt{r-a_2}+\frac{\beta^*\delta}{2}(r-a_2)\\  I_{a_2}^<(\beta)-4\sqrt{r-a_2}+\frac{\beta^*\delta^2}{4}(r-a_2) I_{a_2}^>(\beta)+\frac{1}{\beta}\sqrt{r-a_2}+\frac{\beta^*\delta}{4\beta}(r-a_2)\\  I_{a_2}^>(\beta)-\frac{1}{\beta}\sqrt{r-a_2}+\frac{\delta^2}{8\beta}(r-a_2)\\  I_{a_2}^<(\beta)+\frac{2}{\beta}\sqrt{r-a_2}+\frac{\beta^*\delta}{4\beta}(r-a_2)\\  I_{a_2}^<(\beta)-\frac{2}{\beta}\sqrt{r-a_2}+\frac{\beta^*\delta^2}{8\beta}(r-a_2)  \end{matrix}  ,\;\; a_2\le r \le a_3 \\