Question

设f(x,y)为连续函数,交换二次积分I=∫(0,1)x^2dx∫(x,1)(e^(-y^2))dy的积分次序后则I=

Answer 1

I = ∫(0~1) x² dx ∫(x~1) e^(- y²) dy
= ∫(0~1) e^(- y²) dy ∫(0~y) x² dx

之后的计算如下：
= ∫(0~1) e^(- y²) dy • (1/3)[ x³ ] |(0~y)
= (1/3)∫(0~1) e^(- y²) • y³ dy
= (1/3)(- 1/2)∫(0~1) e^(- y²) • y² d(- y²)
= (- 1/6)[ y²e^(- y²) ] |(0~1) + (- 1/6)∫(0~1) e^(- y²) d(- y²)
= (- 1/6)[ e^(- 1)] - (1/6)[ e^(- y²) ] |(0~1)
= - 1/(6e) - (1/6)[ e^(- 1) - 1 ]
= - 1/(6e) - 1/(6e) + 1/6
= 1/6 - 1/(3e)

Answer 2

I=∫(0,1)x^2dx∫(x,1)(e^(-y^2))dy

=∫(0,1)e^(-y^2)dy∫(0,y)x^2dx
=1/3∫(0,1)y^3*e^(-y^2)dy
=-1/6∫(0,1)y^2*d(e^(-y^2))
=-1/6y^2*e^(-y^2)[0,1]+1/3∫(0,1)y*e^(-y^2)dy
=-1/(6e)-1/6e^(-y^2)[0,1]
=-1/(6e)-1/(6e)+1/6
=1/6-1/(3e)