I have two expressions, they need to be the same and equal but I can`t see where I am making mistake:

$\int (p_0 dp_1+p_1 dp_0)=\int d(p_1p_0)$

On the other side, when I integrate both sides I have result:


What I have missed here?

  • 1
    $\begingroup$ What are $p_0$ and $p_1$ and what does $d(p_0 p_1)$ mean to you? $\endgroup$ – higgs Apr 10 '18 at 9:39
  • $\begingroup$ $p_0$ and $p_1$ are pressures, how is that connected with that is it equation correct or not? $\endgroup$ – nick_name Apr 10 '18 at 9:40
  • 2
    $\begingroup$ You assumed that $p_0$ and $p_1$ are constants, because you put them out of the integral in the left side of the equation. If they're not constants, then your integration on the left side is wrong. $\endgroup$ – mitchbus Apr 10 '18 at 9:44
  • $\begingroup$ Left side is wrong. For example if $p_0=p_1=p$ then $\int pdp+pdp =p^{2}/2+p^{2}/2+C=p{2}+C$ but you are taking it to be $p^{2}+p^{2}+C$ $\endgroup$ – Kavi Rama Murthy Apr 10 '18 at 9:45
  • $\begingroup$ They are not constants, but isn`t is that like derivation of product $(uv)'=u'v+uv'$? $\endgroup$ – nick_name Apr 10 '18 at 9:46

$\int (p_0 dp_1+p_1 dp_0)=\int d(p_1p_0)\qquad$ : This is correct.

$\int d(p_1p_0)=p_1p_0+C_2\qquad$ : This is correct.

$\int (p_0 dp_1+p_1 dp_0)=p_0p_1+p_1p_0+C_1\qquad$ : This is false.

Because $\quad\int p_0 dp_1 \neq p_0p_1+c_1\quad$ and $\quad\int p_1 dp_0 \neq p_1p_0+c_2$

One cannot integrate separately $\int p_0 dp_1$ and separately $\int p_0 dp_1$. One have to integrate them together, that is $\int (p_0 dp_1+p_1 dp_0)$ in writing : $$\int (p_0 dp_1+p_1 dp_0)=\int d(p_1p_0)=p_1p_0+C_2$$ The term on the right gives the result for the whole term on the left.


One can integrate $\int f(x)dx$ because the function $f(x)$ is function of the same variable than the variable of integration $x$.

One cannot integrate $\int f(x)dt$ because the function $f(x)$ is function of $x$ but not of $t$. More exactly, if $f(x)$ is function of $x$ only, not function of $x$ and $t$, then $\int f(x)dt=f(x)\int dt=f(x)(t+c)$.

One can integrate $\int f(x(t))dt$ because the function of function $f(x(t))$ is function of the same variable than the variable of integration $t$.

In the case $\int p_0dp_1$ it is not specified that $p_0$ is function of $p_1$. So, this is the same case as $\int f(x)dt$ above.


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