Going around in a cycle the integral of dq over t is less than or equal to zero.

对一个循环过程作d q除以t的积分，小于等于零。

youdao

That just becomes the integral from, well, I guess t goes from zero to one, actually.

事实上，这就变为t从0到1的积分。

youdao

So, that should be, sorry, that's the same thing as the derivative with respect to t of the total amount of smoke in the region, which is given by the triple integral of u.

那么，那将是。。。，抱歉，这和区域中,烟雾总量关于t的导数一样，由u的三重积分给出。

youdao

The ideal gas constant doesn't change, temperature doesn't change, and so v we just have minus nRT integral V1, V2, dV over V.

理想气体常数不变，温度也不变，因此,是负的nRT，积分从v1到v2，dv除以。

麻省理工公开课 - 热力学与动力学课程节选

The reason for inexact doesn't mean it's a crummy measurement, t means that it's path dependent, and so the value of this integral depends on how you get from one to two.

这是因为它是,与积分路径有关的，因此这里的积分值，取决于从一端到二端的,具体路径。

麻省理工公开课 - 热力学与动力学课程节选

Going around in a cycle the integral of dq over T is less than or equal to zero.

对一个循环过程作dq除以T的积分,小于等于零。

麻省理工公开课 - 热力学与动力学课程节选

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