1. **State the problem:** Evaluate the integrals using the definition of the integral (Theorem 4): - (19) $$\int_{-1}^5 (1 + 3x) \, dx$$ - (20) $$\int_1^4 (x^2 + 2x - 5) \, dx$$ 2. **Recall the definition of the integral (Theorem 4):** The definite integral from $a$ to $b$ of a function $f(x)$ is given by $$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$$ where $\Delta x = \frac{b-a}{n}$ and $x_i^*$ is a sample point in the $i$th subinterval. 3. **Evaluate integral (19):** - Here, $f(x) = 1 + 3x$, $a = -1$, $b = 5$. - Calculate $\Delta x = \frac{5 - (-1)}{n} = \frac{6}{n}$. - Choose right endpoints: $x_i = -1 + i \Delta x = -1 + \frac{6i}{n}$. - Sum: $$\sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n \left(1 + 3\left(-1 + \frac{6i}{n}\right)\right) \frac{6}{n} = \sum_{i=1}^n \left(1 - 3 + \frac{18i}{n}\right) \frac{6}{n} = \sum_{i=1}^n \left(-2 + \frac{18i}{n}\right) \frac{6}{n}$$ - Distribute: $$= \sum_{i=1}^n \left(-\frac{12}{n} + \frac{108i}{n^2}\right) = \sum_{i=1}^n -\frac{12}{n} + \sum_{i=1}^n \frac{108i}{n^2} = -12 + \frac{108}{n^2} \sum_{i=1}^n i$$ - Use formula $\sum_{i=1}^n i = \frac{n(n+1)}{2}$: $$= -12 + \frac{108}{n^2} \cdot \frac{n(n+1)}{2} = -12 + \frac{108(n+1)}{2n} = -12 + \frac{54(n+1)}{n}$$ - Simplify: $$= -12 + 54 + \frac{54}{n} = 42 + \frac{54}{n}$$ - Take limit as $n \to \infty$: $$\lim_{n \to \infty} \left(42 + \frac{54}{n}\right) = 42$$ 4. **Evaluate integral (20):** - Here, $f(x) = x^2 + 2x - 5$, $a=1$, $b=4$. - Calculate $\Delta x = \frac{4-1}{n} = \frac{3}{n}$. - Right endpoints: $x_i = 1 + i \Delta x = 1 + \frac{3i}{n}$. - Sum: $$\sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n \left(\left(1 + \frac{3i}{n}\right)^2 + 2\left(1 + \frac{3i}{n}\right) - 5\right) \frac{3}{n}$$ - Expand inside: $$= \sum_{i=1}^n \left(1 + \frac{6i}{n} + \frac{9i^2}{n^2} + 2 + \frac{6i}{n} - 5\right) \frac{3}{n} = \sum_{i=1}^n \left(-2 + \frac{12i}{n} + \frac{9i^2}{n^2}\right) \frac{3}{n}$$ - Distribute: $$= \sum_{i=1}^n \left(-\frac{6}{n} + \frac{36i}{n^2} + \frac{27i^2}{n^3}\right) = \sum_{i=1}^n -\frac{6}{n} + \sum_{i=1}^n \frac{36i}{n^2} + \sum_{i=1}^n \frac{27i^2}{n^3}$$ - Use formulas: $$\sum_{i=1}^n i = \frac{n(n+1)}{2}, \quad \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$ - Substitute: $$= -6 + \frac{36}{n^2} \cdot \frac{n(n+1)}{2} + \frac{27}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$$ - Simplify each term: $$= -6 + 18 \frac{n+1}{n} + \frac{27}{6} \frac{(n+1)(2n+1)}{n^2} = -6 + 18 \frac{n+1}{n} + 4.5 \frac{(n+1)(2n+1)}{n^2}$$ - Write as: $$= -6 + 18 \left(1 + \frac{1}{n}\right) + 4.5 \left(\frac{2n^2 + 3n + 1}{n^2}\right) = -6 + 18 + \frac{18}{n} + 4.5 \left(2 + \frac{3}{n} + \frac{1}{n^2}\right)$$ - Simplify: $$= 12 + \frac{18}{n} + 9 + \frac{13.5}{n} + \frac{4.5}{n^2} = 21 + \frac{31.5}{n} + \frac{4.5}{n^2}$$ - Take limit as $n \to \infty$: $$\lim_{n \to \infty} \left(21 + \frac{31.5}{n} + \frac{4.5}{n^2}\right) = 21$$ **Final answers:** - (19) $$\int_{-1}^5 (1 + 3x) \, dx = 42$$ - (20) $$\int_1^4 (x^2 + 2x - 5) \, dx = 21$$