【正确答案】原式=∫_－1^y(y－x)e^xdx+∫_y¹(x－y)e^xdx
=y∫_－1^ye^xdx－∫_－1^yxe^xdx+∫_y¹xe^xdx－y∫_y¹e^ydx
=y(e^y－e^－1)－xe^x∣_－1^y+∫_－1^ye^xdx+xe^x∣_y¹－∫_y¹e^xdx－y(e－e^y)
=ye^y－ye^－1－ye^y—e^－1+e^y－e^－1+e－ye^y－e+e^y－ey+ye^y
=2e^y—y(e+e^－1)－2e^－1．