積分

求める期待値を$E$とする。$E={\displaystyle \frac{\pi }{2}}{\displaystyle {\int }_0^{\sqrt{2}}{x}^2\mathrm{d}x-2{\displaystyle {\int }_0^{\frac{\pi }{4}}{\displaystyle \frac{t\sin t}{{\cos }^{3}t}}\mathrm{d}t(\cos t={\displaystyle \frac{1}{x}})}}$

$={\displaystyle \frac{\sqrt{2}}{3}}\pi -2{\displaystyle {\int }_0^{\frac{\pi }{4}}t{\left({\displaystyle \frac{1}{2{\cos }^{2}t}}\right)}^{\prime }\mathrm{d}t}$

$={\displaystyle \frac{\sqrt{2}}{3}}\pi -2{\left[{\displaystyle \frac{t}{2{\cos }^{2}t}}\right]}_0^{\frac{\pi }{4}}+{\displaystyle {\int }_0^{\frac{\pi }{4}}{\displaystyle \frac{\mathrm{d}t}{{\cos }^{2}t}}}$

$={\displaystyle \frac{\sqrt{2}}{3}}\pi -{\displaystyle \frac{\pi }{2}}+{[\tan t]}_0^{\frac{\pi }{4}}$

$={\displaystyle \frac{\sqrt{2}}{3}}\pi -{\displaystyle \frac{\pi }{2}}+1$

より求める答は $$ $1+({\displaystyle \frac{\sqrt{2}}{3}}-{\displaystyle \frac{1}{2}})\pi (\fallingdotseq 0.91)$