概率分布 概率密度问题∫（上面是2 下面是0）（Ax+1）dx 为什么等于2A+2∫2xe^（（-x）^2）dx 为什么

∫（上面是2 下面是0）(Ax+1)dx = [0.5Ax^2 + x] （上面是2 下面是0）= (0.5A*4 + 2) - 0 = 2A+2
让 t = -x^2
所以 x = √ (-t) => dx/dt = - 1/ [2 √ (-t) ] => dx = - dt / [2 √ (-t) ]
∫2xe^（-x^2）dx
= ∫2 [√ (-t)] e^(t){- dt / [2 √ (-t) ] }
= -∫2 [√ (-t)] e^(t) {dt / [2 √ (-t) ] }
= -∫e^(t) dt
= -∫e^(-x^2) d (-x^2)
= -e^t
= - e^ (-x^2)