Question

求助定积分的一个题 设y'(x)=e^(x-1)^2,y(0)=0,则 ∫[0,1] y(x)dx=?

Answer 1

∫[0,1] y(x)dx=y(x)*x[0,1]-∫[0,1]x y'(x)dx=y(1)-∫[0,1]x e^(x-1)^2dx=y(1)-1/2([e^(x-1)^2)[0,1]
=y(1)-1/2(1-e^2)=y(1)-1/2+0.5e^2)