题目

计算二重积分∫[1,3]dx∫[x-1,2]e^( y^2) dy

提问时间：2020-10-13

答案

∫(x = 1→3) dx ∫(y = x - 1→2) e^(y²) dy
交换积分次序：dydx → dxdy
x = 1 到 x = 3,y = x - 1 到 y = 2 < => y = 0 到 y = 2,x = 1 到 x = y + 1
= ∫(y = 0→2) e^(y²) dy ∫(x = 1→y + 1) dx
= ∫(y = 0→2) e^(y²) * [x] |(x = 1→y + 1) dy
= ∫(y = 0→2) e^(y²) * [(y + 1) - 1] dy
= (1/2)∫(y = 0→2) e^(y²) d(y²)
= (1/2)[e^(y²)] |(y = 0→2)
= (1/2)[e^(2²) - e^(0)]
= (e⁴ - 1)/2