ここで勉強すれば数学検定１級の壁は超えられるか。

平面上の領域D｛（ｘ，ｙ)| ０≦ｘ≦１，０≦ｙ≦１ }のとき、

$\int \int _{D}\left| x-y\right| ^{-\dfrac {2}{3}}dxdy$ の２重積分を解く。

ｘ≧ｙの場合(A)

$\int \int \left( x-y\right) ^{-\dfrac {2}{3}}dydx =\int ^{1}_{0}dx\int ^{x}_{0}\left( x-y\right) ^{-\dfrac {2}{3}}dy =\int ^{1}_{0}dx\left[ -3\left( x-y\right) ^{\dfrac {1}{3}}\right] ^{y=x}_{ｙ＝0}$

$=\int ^{1}_{0}3x^{\dfrac {1}{3}}dx=3\left[ \dfrac {3}{4}x^{\dfrac {4}{3}}\right] ^{1}_{0}=\dfrac {9}{4}$

ｙ≧ｘの場合(B)

$=\int \int \left( y-x\right) ^{-\dfrac {2}{3}}dxdy=\int ^{1}_{0}dy\int ^{y}_{0}\left( y-x\right) ^{-\dfrac {2}{3}}dx$

$=\int ^{1}_{0}dy\left[ -3\left( y-x\right) ^{\dfrac {1}{3}}\right] ^{x=y}_{x=0}=\int ^{1}_{0}3y^{\dfrac {1}{3}}dy$

$=3\left[ \dfrac {3}{4}y^{\dfrac {4}{3}}\right] ^{1}_{0}=\dfrac {9}{4}$

$\int \int _{D}\left| x-y\right| ^{-\dfrac {2}{3}}dxdy=\dfrac {9}{4}+\dfrac {9}{4}=\dfrac {9}{2}$

$\dfrac {9}{2}$・・・答え