We are asked to show that (f(x) = 2x^2 + 3x + 5), (x \in [0,1]), is Riemann integrable using both definitions and verify that both give the same result. Let’s solve step by step. --- Step 1: Using the Definition of Riemann Integrability via Upper and Lower Sums 1. Partition ([0,1]) into (n) subintervals of equal width: [ \Delta x = \frac{1-0}{n} = \frac{1}{n}, \quad x_i = \frac{i}{n}, \ i=0,1,2,\dots,n ] 2. Function values on subintervals ([x_{i-1}, x_i]): Since (f(x) = 2x^2 + 3x + 5) is increasing on ([0,1]) (because (f'(x) = 4x+3 > 0)), we have: [ m_i = f(x_{i-1}) = 2\left(\frac{i-1}{n}\right)^2 + 3\frac{i-1}{n} + 5 ] [ M_i = f(x_i) = 2\left(\frac{i}{n}\right)^2 + 3\frac{i}{n} + 5 ] 3. Lower sum (L_n): [ L_n = \sum_{i=1}^n m_i \Delta x = \frac{1}{n} \sum_{i=1}^n \left[ 2\left(\frac{i-1}{n}\right)^2 + 3\frac{i-1}{n} + 5 \right] ] [ = \frac{1}{n} \sum_{i=1}^n \left( \frac{2(i-1)^2}{n^2} + \frac{3(i-1)}{n} + 5 \right) ] [ L_n = \sum_{i=1}^n \left( \frac{2(i-1)^2}{n^3} + \frac{3(i-1)}{n^2} + \frac{5}{n} \right) ] Compute sums using formulas: (\sum_{i=1}^{n} (i-1) = \frac{n(n-1)}{2}), (\sum_{i=1}^{n} (i-1)^2 = \frac{(n-1)n(2n-1)}{6}) [ L_n = \frac{2}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} + \frac{3}{n^2} \cdot \frac{n(n-1)}{2} + 5 ] [ L_n = \frac{(n-1)(2n-1)}{3n^2} + \frac{3(n-1)}{2n} + 5 ] [