1. **Problem Statement:** We want to use Riemann sums to show that $$\int_a^b 3x^2 \, dx = b^3 - a^3.$$ 2. **Recall the definition of the definite integral as a limit of Riemann sums:** Divide the interval $[a,b]$ into $n$ equal subintervals of width $$\Delta x = \frac{b-a}{n}.$$ The right endpoint of the $k$-th subinterval is $$x_k = a + k\Delta x.$$ 3. **Set up the Riemann sum for the function $f(x) = 3x^2$:** $$S_n = \sum_{k=1}^n f(x_k) \Delta x = \sum_{k=1}^n 3(x_k)^2 \Delta x = \sum_{k=1}^n 3\left(a + k\frac{b-a}{n}\right)^2 \frac{b-a}{n}.$$ 4. **Expand the square inside the sum:** $$\left(a + k\frac{b-a}{n}\right)^2 = a^2 + 2a k \frac{b-a}{n} + k^2 \frac{(b-a)^2}{n^2}.$$ 5. **Substitute back into the sum:** $$S_n = \sum_{k=1}^n 3 \left(a^2 + 2a k \frac{b-a}{n} + k^2 \frac{(b-a)^2}{n^2}\right) \frac{b-a}{n} = 3 \frac{b-a}{n} \sum_{k=1}^n \left(a^2 + 2a k \frac{b-a}{n} + k^2 \frac{(b-a)^2}{n^2}\right).$$ 6. **Distribute the sum:** $$S_n = 3 \frac{b-a}{n} \left( \sum_{k=1}^n a^2 + \sum_{k=1}^n 2a k \frac{b-a}{n} + \sum_{k=1}^n k^2 \frac{(b-a)^2}{n^2} \right).$$ 7. **Use the formulas for sums:** - $$\sum_{k=1}^n 1 = n,$$ - $$\sum_{k=1}^n k = \frac{n(n+1)}{2},$$ - $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}.$$ 8. **Substitute these into the sum:** $$S_n = 3 \frac{b-a}{n} \left( a^2 n + 2a \frac{b-a}{n} \frac{n(n+1)}{2} + \frac{(b-a)^2}{n^2} \frac{n(n+1)(2n+1)}{6} \right).$$ 9. **Simplify each term:** - First term: $$a^2 n,$$ - Second term: $$2a \frac{b-a}{n} \frac{n(n+1)}{2} = a (b-a)(n+1),$$ - Third term: $$\frac{(b-a)^2}{n^2} \frac{n(n+1)(2n+1)}{6} = \frac{(b-a)^2 (n+1)(2n+1)}{6 n}.$$ 10. **Rewrite $S_n$:** $$S_n = 3 \frac{b-a}{n} \left( a^2 n + a (b-a)(n+1) + \frac{(b-a)^2 (n+1)(2n+1)}{6 n} \right).$$ 11. **Multiply through by $\frac{b-a}{n}$:** $$S_n = 3 (b-a) \left( a^2 + a (b-a) \frac{n+1}{n} + \frac{(b-a)^2}{6} \frac{(n+1)(2n+1)}{n^2} \right).$$ 12. **Take the limit as $n \to \infty$:** - $$\lim_{n \to \infty} \frac{n+1}{n} = 1,$$ - $$\lim_{n \to \infty} \frac{(n+1)(2n+1)}{n^2} = 2.$$ So, $$\lim_{n \to \infty} S_n = 3 (b-a) \left( a^2 + a (b-a) + \frac{(b-a)^2}{6} \times 2 \right) = 3 (b-a) \left( a^2 + a (b-a) + \frac{(b-a)^2}{3} \right).$$ 13. **Expand the terms inside the parentheses:** $$a^2 + a(b-a) + \frac{(b-a)^2}{3} = a^2 + ab - a^2 + \frac{b^2 - 2ab + a^2}{3} = ab + \frac{b^2 - 2ab + a^2}{3}.$$ 14. **Combine like terms:** $$ab + \frac{b^2}{3} - \frac{2ab}{3} + \frac{a^2}{3} = \frac{3ab}{3} + \frac{b^2}{3} - \frac{2ab}{3} + \frac{a^2}{3} = \frac{ab + b^2 + a^2}{3}.$$ 15. **Multiply by $3(b-a)$:** $$3(b-a) \times \frac{ab + b^2 + a^2}{3} = (b-a)(ab + b^2 + a^2).$$ 16. **Expand the product:** $$(b-a)(ab + b^2 + a^2) = b(ab + b^2 + a^2) - a(ab + b^2 + a^2) = b^2 a + b^3 + b a^2 - a^2 b - a b^2 - a^3.$$ 17. **Simplify terms:** Notice that $$b^2 a = a b^2$$ and $$b a^2 = a^2 b,$$ so these cancel out: $$b^3 - a^3.$$ 18. **Conclusion:** We have shown that $$\int_a^b 3x^2 \, dx = \lim_{n \to \infty} S_n = b^3 - a^3,$$ which completes the proof using Riemann sums.