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Problem: Compute the area between the curve $4-x^{2}$ and the $x$-axis between $[-2,2]$. Do this by taking a limit of a Riemann sum using a partition into $n$ subintervals of equal width and choosing right endpoints.

Step 1 ：Define the function and the interval as \(f(x) = 4-x^{2}\) and \([-2,2]\) respectively.

Step 2 ：Divide the interval \([-2,2]\) into \(n\) subintervals of equal width. The width of each subinterval is \(\Delta x = \frac{2-(-2)}{n} = \frac{4}{n}\).

Step 3 ：The right endpoint of the \(i\)-th subinterval is \(x_{i} = -2 + i\Delta x = -2 + i\frac{4}{n}\).

Step 4 ：Compute the Riemann sum as \(S_{n} = \sum_{i=1}^{n} f(x_{i})\Delta x = \sum_{i=1}^{n} (4-x_{i}^{2})\Delta x = \sum_{i=1}^{n} (4-(-2 + i\frac{4}{n})^{2})\frac{4}{n}\).

Step 5 ：The area between the curve and the \(x\)-axis is given by the limit of the Riemann sum as \(n\) goes to infinity, which is \(\lim_{n\to\infty} S_{n}\).

Step 6 ：Simplify the expression for \(S_{n}\) to get \(S_{n} = 4 - \frac{16}{n^{2}}\sum_{i=1}^{n} i^{2}\).

Step 7 ：Use the formula for the sum of squares \(\sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}\) to substitute into the expression for \(S_{n}\), we get \(S_{n} = 4 - \frac{16}{n^{2}}\cdot\frac{n(n+1)(2n+1)}{6} = 4 - \frac{8(n+1)(2n+1)}{3n}\).

Step 8 ：Take the limit as \(n\) goes to infinity to get the area between the curve and the \(x\)-axis, which is \(\lim_{n\to\infty} S_{n} = 4 - \lim_{n\to\infty} \frac{8(n+1)(2n+1)}{3n} = 4 - \frac{16}{3} = \frac{4}{3}\).

Step 9 ：\(\boxed{\text{So, the area between the curve } 4-x^{2} \text{ and the } x\text{-axis between } [-2,2] \text{ is } \frac{4}{3} \text{ square units.}}\)