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I am trying to prove that $$\begin{equation}\int_x^{x+1}\left(\int_0^{v} (u-0)f(u)\textrm{d}u+\int_v^{1} (u-1)f(u)\textrm{d}u\right)\textrm{d}v=\\\int_0^x\int_v^{v+1}f(u)\textrm{d}u\textrm{d}v\end{equation}.$$

With the following sympy code i verified it symbolically for a few cases.

from sympy import *

x,u,v=symbols('x,u,v')
print
f=lambda x:exp(x);
print expand(integrate(integrate((u-0)*f(u), 
(u,0,v))+integrate((u-1)*f(u),(u,v,1)),(v,x,x+1)))
print expand(integrate(integrate(f(u),(u,v,v+1)),(v,0,x)))
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Call $F$ the antiderivative of $f$ such that $F(1)=0$.

Call $\frak{F}$ the antiderivative of $F$ such that $\frak{F}$$(1)=0$.

Now show that both expressions give the same result, i.e., :

$${\frak F}(x+1)-{\frak F}(x)+{\frak F}(0).$$

Method : Write your LHS integral under the form :

$$\begin{equation}\int_x^{x+1}\left(\int_0^1 uf(u)\textrm{d}u-\int_v^{1}f(u)\textrm{d}u\right)\textrm{d}v\end{equation}.$$

Then integrate inside the large parentheses by parts in the first expression, giving

$$\int_{x}^{x+1}([uF(u)]_0^1-\int_0^1 F(u)du + F(v)-F(1)$$

Taking into account $F(1)=0$ and ${\frak F}(1)=0$ :

$$=\int_{x}^{x+1}({\frak F}(0) + F(v))dv$$

I let you find the rest...

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  • $\begingroup$ $\int_0^{v} (u-0)f(u)\textrm{d}u+\int_v^{1} (u-1)f(u)\textrm{d}u=\int_0^{v} uf(u)\textrm{d}u+\int_v^{1} uf(u)\textrm{d}u-\int_v^{1}f(u)\textrm{d}u$ $\endgroup$ – Peter Sheldrick Dec 10 '18 at 10:10
  • $\begingroup$ You are right. Corrected. $\endgroup$ – Jean Marie Dec 10 '18 at 10:12

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