\( \left. \begin{array} { l } { V = \int _ { R } ^ { + \infty } E d r } \\ { V = \frac { Q } { 4 \pi \varepsilon _ { 0 } } \lim _ { T \rightarrow \infty } \int _ { R } ^ { T } \frac { 1 } { r ^ { 2 } } d r } \\ { V = \frac { Q } { 4 \pi \varepsilon _ { 0 } } \lim _ { T \rightarrow + \infty } - \frac { 1 } { r } | _ { R } ^ { T } } \\ { V = \frac { Q } { 4 \pi \varepsilon _ { 0 } } \lim _ { T \rightarrow + \infty } ( - \frac { 1 } { T } - - \frac { 1 } { R } ) } \\ { V = \frac { Q } { 4 \pi \varepsilon _ { 0 } } \lim _ { T \rightarrow + \infty } ( - \frac { 1 } { T } + \frac { 1 } { R } ) } \\ { V = \frac { Q } { 4 \pi \varepsilon R } } \end{array} \right. \)

Let's solve the integral step by step to find the expression for \( V \). 1. **Start with the integral**: \[ V = \frac{Q}{4 \pi \varepsilon_0} \lim_{T \rightarrow +\infty} \int_{R}^{T} \frac{1}{r^2} \, dr \] 2. **Evaluate the integral**: The integral \( \int \frac{1}{r^2} \, dr \) can be computed as follows: \[ \int \frac{1}{r^2} \, dr = -\frac{1}{r} + C \] Therefore, we can evaluate the definite integral from \( R \) to \( T \): \[ \int_{R}^{T} \frac{1}{r^2} \, dr = \left[-\frac{1}{r}\right]_{R}^{T} = -\frac{1}{T} - \left(-\frac{1}{R}\right) = -\frac{1}{T} + \frac{1}{R} \] 3. **Substituting back into the expression for \( V \)**: Now substituting the result of the integral back into the expression for \( V \): \[ V = \frac{Q}{4 \pi \varepsilon_0} \lim_{T \rightarrow +\infty} \left(-\frac{1}{T} + \frac{1}{R}\right) \] 4. **Taking the limit**: As \( T \) approaches infinity, \( -\frac{1}{T} \) approaches 0: \[ V = \frac{Q}{4 \pi \varepsilon_0} \left(0 + \frac{1}{R}\right) = \frac{Q}{4 \pi \varepsilon_0 R} \] Thus, the final expression for \( V \) is: \[ V = \frac{Q}{4 \pi \varepsilon_0 R} \] This completes the evaluation of the potential \( V \).