Let's solve each problem step-by-step: --- ### **1. \( f(x) = 2 \) on \([3, 6]\)** #### **Step 1: Use the Fundamental Theorem of Calculus** The area under \( f(x) \) from \( a \) to \( b \) is: \[ \int_a^b f(x) \, dx \] Here, \( f(x) = 2 \), \( a = 3 \), and \( b = 6 \). \[ \int_3^6 2 \, dx = [2x]_3^6 = 2(6) - 2(3) = 12 - 6 = \boxed{6} \] #### **Step 2: Verify Using Geometry** The graph is a horizontal line at \( y=2 \). The area is a rectangle: \[ \text{Area} = \text{height} \times \text{width} = 2 \times (6-3) = 2 \times 3 = \boxed{6} \] --- ### **2. \( f(x) = 2x + 9 \) on \([-1, 4]\)** #### **Step 1: Use the Fundamental Theorem of Calculus** \[ \int_{-1}^4 (2x + 9) \, dx \] Find the antiderivative: \[ \int (2x + 9) dx = x^2 + 9x + C \] Evaluate from \(-1\) to \(4\): \[ [x^2 + 9x]_{-1}^4 = \left[ (4)^2 + 9(4) \right] - \left[ (-1)^2 + 9(-1) \right] \\ = (16 + 36) - (1 - 9) \\ = 52 - (-8) \\ = 52 + 8 = \boxed{60} \] --- ### **3. \( f(x) = 5x^3 \) on \([1, 3]\)** #### **Step 1: Use the Fundamental Theorem of Calculus** \[ \int_1^3 5x^3 dx \] Find the antiderivative: \[ \int 5x^3 dx = \frac{5}{4}x^4 + C \] Evaluate from 1 to 3: \[ \left[ \frac{5}{4}x^4 \right]_1^3 = \frac{5}{4}(3^4) - \frac{5}{4}(1^4) = \frac{5}{4}(81) - \frac{5}{4}(1) = \frac{405}{4} - \frac{5}{4} = \frac{400}{4} = \boxed{100} \] --- ### **Final Answers:** 1. \( f(x) = 2 \) on \([3, 6]\): **6** 2. \( f(x) = 2x + 9 \) on \([-1, 4]\): **60** 3. \( f(x) = 5x^3 \) on \([1, 3]\): **100**

The Fundamental Theorem of Calculus connects differentiation and integration, allowing us to evaluate definite integrals by finding antiderivatives of functions. It states that if \(F(x)\) is an antiderivative of \(f(x)\), then the definite integral of \(f(x)\) over \([a, b]\) is \(F(b) - F(a)\). This theorem simplifies the process of calculating areas under curves by reducing it to basic algebraic evaluations of antiderivatives at interval endpoints.