∫_{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

Show them Gauss’s derivation:

1 + 2 + ... + n-1 + n n + n-1 + ... + 2 + 1

——————————————————- n+1 + n+1 + ... + n+1 + n+1= n(n+1) Thus Σ_{i=1}^{n}i = \frac{n(n+1)}{2} 5.4) Indefinite integral: ∫ f(x)dx = anti-derivative of f. I.e, ∫ f(x)dx = F (x) where F′(x) = f(x) Ex: 1.) ∫ x^{3}dx = 2.) ∫ [3x^{4} + 2 + \frac{3}{x} + cos(x)]dx = 5.5) Integration by substitution If F (x) = ln|x^{2} − x|, then by chain rule F′(x) = \frac{1}{x^{2}−x}(2x − 1) = \frac{2x−1}{x^{2}−x} Thus ∫ \frac{2x−1}{x^{2}−x}dx = How do you recognize derivative when chain rule involved? u -substition ∫ \frac{2x−1}{x^{2}−x}dx = ∫ \frac{du}{u} = ∫ \frac{1}{u}du = ln|u| + C = ln|x^{2} − x| + C Let u = x^{2} − x, then du = 2x − 1