j t integral j积分
c t integral c积分
c t-integral c积分
T Integral Test 前卫摇滚
t fuzzy integral fuzzy测度
LAUDA Integral T 工艺过程恒温器
And then, you'll do a one variable integral over t.
那么你要做一个以t为变量的积分。
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。
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。
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。
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。
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。
应用推荐