Update cheatsheet.fr.md

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

@@ -785,11 +785,11 @@ figure à faire
     \int_{\rho=0}^{R} \rho\,(\rho^2+z_M^2)^{\,-3/2}\,d\rho`$    
     <br>
     $`\displaystyle\hspace{1cm}=\dfrac{\dens^{2D}\,z}{2\epsilon_0}`$   
-    $`\int_{\rho = 0}^R - \Big(\underbrace{-\dfrac{1}{2}_{n+1}\Big)}\cdot\underbrace{2\rho}_{u^{\,'}}\,\underbrace{(\rho^2+z^2)^{-\,3/2}}_{u^n}\,d\rho`$   
+    $`\int_{\rho = 0}^R - \Big(\underbrace{-\dfrac{1}{2}}_{n+1}\Big)\cdot\underbrace{2\rho}_{u^{\,'}}\,\underbrace{(\rho^2+z^2)^{-\,3/2}}_{u^n}\,d\rho`$   
     <br>
     $`\displaystyle \hspace{1cm} =  - \dfrac{\dens^{2D}\,z}{2\epsilon_0} \big[(\rho^2+z^2)^{-\,1/2}\big]_0^R`$  
     <br>
-    $`\color{blue}{\scriptsize{\text{Le signe moins devient plus}`$
+    $`\color{blue}{\scriptsize{\text{Le signe moins devient plus}}}`$
     $`\color{blue}{\scriptsize{\text{en inversant les bornes d'intégration}}}`$   
     <br>
     $`\displaystyle \hspace{1cm} = +\dfrac{\dens^{2D}\,z}{2\epsilon_0} \big[(\rho^2+z^2)^{-\,1/2}\big]_R^0`$