解答11. ∫_0^1dy∫_0^(y^2)ycos(1-x)^2dx= ycos(1-x)2dx

首先，我们可以交换积分的顺序，得到：

∫_0^1dx ∫_√x^y^2 cos(1-x)^2 dy

接下来，我们可以对第一个积分变量进行部分积分，令u = y^2，dv = cos(1-x)^2 dy，则du = 2y dy，v = ∫cos(1-x)^2 dy = ycos(1-x)^2/2，代入上式得到：

∫_0^1dx [uvcos(1-x)^2]_0^x - ∫_0^1 ycos(1-x)^2/2 * 2y dy

化简可得：

∫_0^1dx xcos(1-x)^2/2 - ∫_0^1 y^2cos(1-x)^2 dy

接下来，我们对第二个积分变量进行部分积分，令u = y，dv = ycos(1-x)^2 dy，则du = dy，v = ∫ycos(1-x)^2 dy = ysin(1-x)^2/2 - ∫sin(1-x)^2 dy = ysin(1-x)^2/2 + x/2 - xcos(1-x)^2/2，代入上式得到：

∫_0^1dx xcos(1-x)^2/2 - [(y^2sin(1-x)^2/2 + xy - xy cos(1-x)^2/2)]_0^1 + ∫_0^1 (ysin(1-x)^2/2 + x/2 - xcos(1-x)^2/2) sin(1-x)^2 dx

化简可得：

∫_0^1dx xcos(1-x)^2/2 - (sin(1)^2/2 + 1/2 - cos(1)^2/2) + ∫_0^1 (ysin(1-x)^4/2 + x/2 - xcos(1-x)^2/2) dx

再次化简可得：

∫_0^1dx xcos(1-x)^2/2 - 1 + ∫_0^1 (ysin(1-x)^4/2 + x/2 - xcos(1-x)^2/2) dx

代入数值计算可得：

-0.1767...