Let's solve the problem step by step using the **definition of the definite integral** (Riemann sum). --- ### Step 1: Set Up the Riemann Sum The definite integral from 0 to 1 of \( 1 + x \) dx is defined as: \[ \int_0^1 (1 + x)\, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x \] Where: - \( \Delta x = \frac{1-0}{n} = \frac{1}{n} \) - \( x_i^* \) is a sample point in the i-th subinterval. For simplicity, use the right endpoint: \( x_i^* = \frac{i}{n} \) Plug into the function: \[ f(x_i^*) = 1 + x_i^* = 1 + \frac{i}{n} \] So the sum becomes: \[ \sum_{i=1}^n \left(1 + \frac{i}{n}\right) \Delta x = \sum_{i=1}^n \left(1 + \frac{i}{n}\right) \cdot \frac{1}{n} \] **Explanation:** We broke the interval [0,1] into \( n \) equal parts and wrote the Riemann sum using the right endpoints. --- ### Step 2: Simplify the Sum Expand the sum: \[ \sum_{i=1}^n \left(1 + \frac{i}{n}\right) \cdot \frac{1}{n} = \sum_{i=1}^n \frac{1}{n} + \sum_{i=1}^n \frac{i}{n^2} \] \[ = \frac{1}{n} \sum_{i=1}^n 1 + \frac{1}{n^2} \sum_{i=1}^n i \] Recall the sum formulas: - \( \sum_{i=1}^n 1 = n \) - \( \sum_{i=1}^n i = \frac{n(n+1)}{2} \) Substitute: \[ = \frac{1}{n} \cdot n + \frac{1}{n^2} \cdot \frac{n(n+1)}{2} = 1 + \frac{n+1}{2n} \] Now, take the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \left[1 + \frac{n+1}{2n}\right] = 1 + \frac{1}{2} = \frac{3}{2} \] **Explanation:** We simplified the sum using standard formulas, then took the limit as \( n \) becomes large. --- ### **Final Answer:** \[ \boxed{\int_0^1 (1 + x)\, dx = \frac{3}{2}} \] This matches the required proof using the definition!