ここで勉強すれば数学検定１級の壁は超えられるか。

[math]f\left( x\right) =\begin{cases}\dfrac {1}{4}a^{2}x-x^{3}\left( 0\leqq x\leqq \dfrac {a}{2}\right) \\ 0\left( x <0,x >\dfrac {a}{2}\right) \end{cases}[/math]

(1)

[math]\int ^{\dfrac {a}{2}}_{0}\left( \dfrac {1}{4}a^{2}x-x^{3}\right) dx=\left[ \dfrac {a^{2}x^{2}}{8}-\dfrac {x^{4}}{4}\right] ^{\dfrac {a}{2}}_{0}[/math]

[math]=\dfrac {a^{2}}{8}\times \left( \dfrac {a}{2}\right) ^{2}-\dfrac {1}{4}\times \left( \dfrac {a}{2}\right) ^{4}=\dfrac {a^{4}}{64}=1[/math]

[math]a=2\sqrt {2}[/math]・・・(1)の答え

（２）

[math]F\left( x\right) =\int ^{x}_{0}\left( 2x-x^{3}\right) dx=x^{2}-\dfrac {x^{4}}{4}[/math]

[math]F\left( x\right) =0\left( x\leq 0\right) [/math]

[math]F\left( x\right) =x^{2}-\dfrac {x^{4}}{4}\left( 0\leqq x\leqq \sqrt {2}\right)[/math]

[math]F\left( x\right) =1\left( x\geqq \sqrt {2}\right)[/math]              ・・・（２）の答え