本题需交换积分次序来做
∫[0→∞]dx∫[x→2x] e^(-y²)dy
=∫[0→∞]dy∫[y/2→y] e^(-y²)dx
=∫[0→∞] (y-y/2)e^(-y²) dy
=(1/2)∫[0→∞] ye^(-y²) dy
=(1/4)∫[0→∞] e^(-y²) d(y²)
=-(1/4)e^(-y²) |[0→∞]
=1/4
本题需交换积分次序来做
∫[0→∞]dx∫[x→2x] e^(-y²)dy
=∫[0→∞]dy∫[y/2→y] e^(-y²)dx
=∫[0→∞] (y-y/2)e^(-y²) dy
=(1/2)∫[0→∞] ye^(-y²) dy
=(1/4)∫[0→∞] e^(-y²) d(y²)
=-(1/4)e^(-y²) |[0→∞]
=1/4