Update cheatsheet.fr.md

12.temporary_ins/10.electrostatics-vacuum/20.causes-stationary-electric-field/20.overview/cheatsheet.fr.md

@@ -785,9 +785,9 @@ figure à faire
     \int_{\rho=0}^{R} \rho\,(\rho^2+z_M^2)^{\,-3/2}\,d\rho`$    
     <br>
     $`\displaystyle\hspace{1cm}=\dfrac{\dens^{2D}\,z}{2\epsilon_0}  
-    \int_{\rho = 0}^R - \Big(\underbrace{-\dfrac{1}{2}}_{n+1}\Big)\cdot\underbrace{2\rho}_{\color{blue}{u^{\,'}}}\,\underbrace{(\rho^2+z^2)^{-\,3/2}}_{u^n}\,d\rho`$   
+    \int_{\rho = 0}^R - \Big(\underbrace{-\dfrac{1}{2}}_{\color{blue}{n+1}}\Big)\cdot\underbrace{2\rho}_{\color{blue}{u^{\,'}}}\,\underbrace{(\rho^2+z^2)^{-\,3/2}}_{\color{blue}{u^n}}\,d\rho`$   
     <br>
-    $`\displaystyle \hspace{1cm} =  - \dfrac{\dens^{2D}\,z}{2\epsilon_0} \big[(\rho^2+z^2)^{-\,1/2}\big]_0^R`$  
+    $`\displaystyle \hspace{1cm} =  - \dfrac{\dens^{2D}\,z}{2\epsilon_0} \big[\underbrace{(\rho^2+z^2)^{-\,1/2}}_{\color{blue}{u^{n+1}}\big]_0^R`$  
     <br>
     $`\color{blue}{\scriptsize{\text{Le signe moins devient plus}}}`$
     $`\color{blue}{\scriptsize{\text{en inversant les bornes d'intégration}}}`$