1. **State the problem:** We want to show that the sum of the areas of the upper approximating rectangles under the curve $f(x) = 5x^2$ on the interval $[0,2]$ approaches $\frac{40}{3}$ as the number of rectangles $n$ approaches infinity. 2. **Define the sum of rectangles:** Each rectangle has width $\frac{2}{n}$ and height given by the function value at the right endpoint of each subinterval: $5\left(\frac{2k}{n}\right)^2$ for $k=1,2,\ldots,n$. 3. **Write the sum $R_n$:** $$ R_n = \sum_{k=1}^n \text{width} \times \text{height} = \sum_{k=1}^n \frac{2}{n} \times 5\left(\frac{2k}{n}\right)^2 = \sum_{k=1}^n \frac{2}{n} \times 5 \times \frac{4k^2}{n^2} = \sum_{k=1}^n \frac{40}{n^3} k^2. $$ 4. **Factor out constants:** $$ R_n = \frac{40}{n^3} \sum_{k=1}^n k^2. $$ 5. **Use the formula for sum of squares:** $$ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. $$ 6. **Substitute the sum into $R_n$:** $$ R_n = \frac{40}{n^3} \times \frac{n(n+1)(2n+1)}{6} = \frac{40}{6} \times \frac{(n+1)(2n+1)}{n^2} = \frac{20}{3} \times \frac{(n+1)(2n+1)}{n^2}. $$ 7. **Take the limit as $n \to \infty$:** $$ \lim_{n \to \infty} R_n = \frac{20}{3} \lim_{n \to \infty} \frac{(n+1)(2n+1)}{n^2} = \frac{20}{3} \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) \left(2 + \frac{1}{n}\right). $$ 8. **Evaluate the limit:** $$ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) = 1, \quad \lim_{n \to \infty} \left(2 + \frac{1}{n}\right) = 2. $$ 9. **Final result:** $$ \lim_{n \to \infty} R_n = \frac{20}{3} \times 1 \times 2 = \frac{40}{3}. $$ Thus, the sum of the areas of the upper approximating rectangles approaches $\frac{40}{3}$ as $n$ becomes very large.