求∫ (1,0)x^2dx∫(1,x) e^-y^2dy

∫(0到1) dx ∫(x到1) x²e^(-y²) dy 需要交换积分次序 x = 0 到 x = 1，y = 1 到 y = x ∫(0到1) dy ∫(0到y) x²e^(-y²) dx = ∫(0到1) [e^(-y²) * x³/3]|(0到y) dy = (1/3)∫(0到1) y³e^(-y&#...

换序得到原式=∫〔0到1〕dy∫〔0到y〕【xx*e^(-yy)】dx =(1/3)∫〔0到1〕【y^3*e^(-yy)】dy =(1/6)∫〔0到1〕【-yy*e^(-yy)】d〔-yy)令u=-yy得到 =(1/6)∫〔0到-1〕【ue^u】du =(1/6)∫〔0到-1〕ude^u 用分部积分法得到 (1/6)【ue^u代入上下限并...

=-∫(0→2)y*ln(1+e^y)dy +∫(0→2)y^2dy 即∫(-2→0)x*ln(1+e^x)dx=-∫(0→2)x*ln(1+e^x)dx +∫(0→2)x^2dx 故∫(-2→2)x*ln(1+e^x)dx =∫(-2→0)x*ln(1+e^x)dx +∫(0→2)x*ln(1+e^x)dx =-∫(0→2)x*ln(1+e^x)dx +∫(0→2)x^...

e^(-y^2)的原函数不能用初等函数表示,要交换积分顺序:先对x积分 D={（x,y）|0≤y≤1,0≤x≤y,}

交换积分次序得 ∫(0,1)dy∫(y,1)e^x^2dx =∫(0,1)e^x^2dx∫(0,x)dy =∫(0,1)e^x^2*xdx =1/2e^x^2[0,1]=e/2-1/2

简单计算一下，答案如图所示

解：6题，原式=2∫(0,1/2)(arcsinx)^2dx/(1-x^2)=(2/3)(arcsinx)^3丨(x=0,1/2)=(1/54)π^3。7题，交换积分顺序，D={(x,y)丨0≤x≤y，0≤y≤2}，∴原式=∫(0,2)dy∫(0,y)e^(-y^2)dx=∫(0,2)ye^(-y^2)dy=(-1/2)e^(-y^2)丨(y=0,2)=(1/2)(...

0≤y≤1，0≤x≤y 所以 原式=∫[0,1]e^(-y^2)dy∫[0,y] x^2dx =∫[0,1]e^(-y^2)dy x^3/3[0,y]=∫[0,1] y^3/3*e^(-y^2)dy =1/6∫[0,1] y^2*e^(-y^2)dy^2 =-1/6∫[0,1] y^2de^(-y^2)=-1/6y^2*e^(-y^2)[0,1]+1/6∫[0,1] e...