2003数学四.pdf

2003 2 x 1lim[1+ln(1+x)] = . x →0 e2 2 2 ln[1+ln(1+x )] lim[1+ln(1+x)]x = lim e x x →0 x →0 + +x +x 2 ln[1 ln(1 )] 2 ln(1 ) lim lim x →0 x x →0 x 2 = e e e . 1 2∫−1 ( x +x )e−x dx = . 2(1−2e−1 ) 1 1 1 ( x +x )e−x dx = x e−x dx + xe −x dx ∫−1 ∫−1 ∫−1 1 1 1 −x −x −x = ∫−1 x e dx +2∫0 xe dx =−2∫0 xde 1 −x 1 −x −1 = −2(xe 0 − e dx) = 2(1−2e ) . ∫0 a, 0 ≤x ≤1, ⎧ (3) a0 f (x) g (x) ⎨ D 0, ⎩ I ∫∫f (x)g (y −x)dxdy = . D a 2 I ∫∫f (x)g (y −x)dxdy = ∫∫a 2 dxdy D 0≤x ≤1,0≤y −x ≤1 1 x +1 1 2 2