1. **Problem statement:** Evaluate the definite integral $$\int_1^2 (5x^6 + 2x - 1) \, dx$$ and the indefinite integral $$\int 5x^3 \, dx$$. 2. **Formula and rules:** - For definite integrals, use the Fundamental Theorem of Calculus: $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $$F'(x) = f(x)$$. - For indefinite integrals, use the power rule: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ for $$n \neq -1$$. 3. **Evaluate the definite integral:** - Integrate term-by-term: $$\int (5x^6 + 2x - 1) \, dx = 5 \int x^6 \, dx + 2 \int x \, dx - \int 1 \, dx$$ - Using the power rule: $$5 \cdot \frac{x^{7}}{7} + 2 \cdot \frac{x^{2}}{2} - x + C = \frac{5}{7}x^{7} + x^{2} - x + C$$ - Apply limits from 1 to 2: $$\left(\frac{5}{7} \cdot 2^{7} + 2^{2} - 2\right) - \left(\frac{5}{7} \cdot 1^{7} + 1^{2} - 1\right)$$ - Calculate values: $$2^{7} = 128$$ So, $$\left(\frac{5}{7} \cdot 128 + 4 - 2\right) - \left(\frac{5}{7} + 1 - 1\right) = \left(\frac{640}{7} + 2\right) - \frac{5}{7} = \frac{640}{7} + 2 - \frac{5}{7} = \frac{635}{7} + 2 = \frac{635}{7} + \frac{14}{7} = \frac{649}{7}$$ 4. **Evaluate the indefinite integral:** - $$\int 5x^{3} \, dx = 5 \int x^{3} \, dx = 5 \cdot \frac{x^{4}}{4} + C = \frac{5}{4} x^{4} + C$$ **Final answers:** - $$\int_1^2 (5x^6 + 2x - 1) \, dx = \frac{649}{7}$$ - $$\int 5x^{3} \, dx = \frac{5}{4} x^{4} + C$$