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Now, we are very gentle in this course with respect to knowing integrals, but this is one you have to know.
这门课对积分,的要求并不高，但这个积分你们必须会算。
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So for the reversible process, the work done is the integral under the pressure volume state function, the function of state.
对可逆过程，做的功,是压强体积态函数曲线下，的积分面积。
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We know that charge is equal to the integral of current times the time. And he knows the current, he knows how much time and then he weighs this.
我们知道电量等于电流,对时间的积分，他知道电流，他也知道时间，然后他测量这个。
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The basic idea in solving these equations and integrating is you find one answer, so then when you take enough derivatives, the function does what it's supposed to do.
解决这类方程以及积分的基本思想就是,你求出一个解,然后进行多次求导,求导的结果就满足条件
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Even if the thing that you really want is an integral that would be difficult to evaluate.
即使需要计算的,是一个很难的积分。
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The integral of one over a quantity is the natural log.
这种形式的积分结果,是自然对数。
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And, if you ever get confused, the way to remember these is that you know energy is the integral of a force moving through a distance.
如果你感到很困惑的话，可以这么来记,能量是在沿力的方向进行路径积分。
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This is on the basis of the integral of the current times the time.
这是根据电流对,时间的积分。
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So it's, this is just the integral pdv And it's an ideal gas, isothermal, right.
从一点到二点的,的积分。，from，one，to，two，of，p，dV。,这是理想气体，恒温过程，好的。
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du So, we can also write delta u as integral from 1 to 2 of du.
我们也可以将Δu写成,从1到2的积分。
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You know that much about integrals.
你们应该懂很多积分的。
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And so just like here, w2 now q2 is minus w2, that's integral going from three to four p dV.
就跟这儿一样，现在是q2等于负，等于从第三点到第四点过程的pdv的积分。
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Energy is the integral of force moving through a distance.
能量是在沿力的方向,进行路径积分。
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Cv So, for Cp and Cv, these are often quantities that are measured as a function of temperature, and one could, in fact, calculate this integral.
一般Cp和,都是温度的函数，因此实际上，我们可以将这个积分计算出来。
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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。
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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。
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V2 p1 times V2 minus V1. What that turns out to be, 0 this area right here. It's V1 minus V2 times p1.
第二步做的功是-pdV，积分从V2到，这部分积分出来是多少？应该是。
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So, the work for this process is the integral, pdv or minus the integral, V1, V2, p dV.
此过程中的做功等于负的积分，从v1到v2，外部。
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So, if you're doing physics, x2 work is the integral of force, dx, x1 to x2.
如果你正在干活，功就是力的积分，从x1，到。
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So it's nR integral from V1 to V2 dV over V.
即nR除以V乘以dV从V1到V2的积分。
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That's equivalent to doing the integral, and so, what we end up getting is that the reversible work v2 pdv is equal to minus integral V1, V2, p dV.
这与刚才的积分过程效果相同，最后，我们得到的结论是可逆过程的功,是负的积分，从v1到。
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V2 That's minus p1 times V2 minus V1.
做的功是-p1dV，积分从V1到。
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In other words, your job is to guess a function whose second derivative is a, and this is called integration, which is the opposite of differentiation, and integration is just guessing.
换言之,你的任务是要猜出一个二阶导数为a的函数,这就是积分,和微分恰恰相反,积分就是猜
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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。
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