∫_{2}^{6}(−\frac{1}{2}t + 4)dt = = lim_{n→∞}Σ_{i=1}^{n}f(t_{i})Δt Δt = \frac{6−2}{n} = \frac{4}{n} t_{i} = 2 + iΔt = 2 + \frac{4i}{n} (using right-hand endpoints) ∫_{2}^{6}(−\frac{1}{2}t + 4)dt = lim_{n→∞}Σ_{i=1}^{n}f(2 + \frac{4i}{n})(\frac{4}{n}) = lim_{n→∞}Σ_{i=1}^{n}[−\frac{1}{2}(2 + \frac{4i}{n}) + 4](\frac{4}{n}) = lim_{n→∞}Σ_{i=1}^{n}[−1 − \frac{2i}{n} + 4](\frac{4}{n}) = lim_{n→∞}Σ_{i=1}^{n}[3 − \frac{2i}{n}](\frac{4}{n}) = lim_{n→∞}Σ_{i=1}^{n}[\frac{12}{n} − \frac{8i}{n^{2}} ] = lim_{n→∞}(Σ_{i=1}^{n} \frac{12}{n} − Σ_{i=1}^{n} \frac{8i}{n^{2}} ) = lim_{n→∞}(12 − \frac{8}{n^{2}} Σ_{i=1}^{n}i) = lim_{n→∞}(12 − \frac{8}{n^{2}} \frac{n(n+1)}{2}) = lim_{n→∞}(12 − \frac{4n^{2}+4n}{n^{2}}) = lim_{n→∞}(12 − 4 − \frac{4}{n}) = 8