Find ∫₀³ (2x + 1) dx as the limit of a right-hand sum…

Step 1. Sum. $S_n = h\displaystyle\sum_{r=1}^{n}(2rh + 1) = 2h^2 \times \dfrac{n(n + 1)}{2} + hn = h^2n(n + 1) + hn$.

Step 2. Put $h = \dfrac{3}{n}$. $S_n = \dfrac{9(n + 1)}{n} + 3 = 12 + \dfrac{9}{n}$.

Step 3. $n = 10$. $S_{10} = 12 + 0.9 = 12.9$.

Step 4. The limit. As $n \to \infty$, $\dfrac{9}{n} \to 0$, so the limit is $12$.