ここで勉強すれば数学検定１級の壁は超えられるか。

$D=\left\{ \left( x,y\right) 0\leqq y\leqq x\right\}$　で

$\int \int _{D}\left( 2+x+y\right) ^{-3}dxdy$を計算します。

$\int \int _{D}\left( 2+x+y\right) ^{-3}dxdy$

$=\lim _{p\rightarrow \infty }\int ^{p}_{0}dx\int ^{x}_{0}\left( 2+x+y\right) ^{-3}dy$

$=\lim _{p\rightarrow \infty }\int ^{p}_{0}\left[ -\dfrac {1}{2}\left( 2+x+y\right) ^{-2}\right] ^{x}_{0}dx$

$=\lim _{p\rightarrow \infty }\int ^{p}_{0}\left( \dfrac {1}{2}\left( 2+x\right) ^{-2}-\dfrac {1}{2}\left( 2+2x\right) ^{-2}\right) dx$　

$\lim _{p\rightarrow \infty }\left[ -\dfrac {1}{2}\left( 2+x\right) ^{-1}+\dfrac {1}{8}\left( 1+x\right) ^{-1}\right] ^{p}_{0}$

$=\lim _{p\rightarrow \infty }\left( -\dfrac {1}{2\left( 2+p\right) }+\dfrac {1}{8\left( 1+p\right) }+\dfrac {1}{4}-\dfrac {1}{8}\right)$

$=0+0+\dfrac {1}{8}=\dfrac {1}{8}$・・・答え