Answer to Question #285295 in Real Analysis for Sarita bartwal

Question

For the function f(x)= x^2-2 defined over[1,5],

Verify L(P,f)≤U(P,f) where is the partition which divide [1,5] into four equal intervals

Answer 1

$\text{Given} \, f(x) = x^2 - 2 \text{over} [1,5] \\ \text{into four equal intervals i.e n=4}\\ \text{Recall that standard partition is of the form}\\ P_n = {\{a+ \dfrac{i(b-a)}{n} \, i = 0, 1, 2,...n}\} \\ \text{where a and b are ends of the interval}\\ \text{Hence,}\\ P_4 = {\{1 + i\dfrac{5-1}{4}, i = 0,1,2,...n}\}\\ P_4 ={\{1 + i, i = 0,1,2,...,n}\} \\ P_4 = {\{1, 2, 3, 4, 5}\}\\ P_4 ={\{(1,2), (2,3), (3,4), (4,5)}\} \text{with} \\ \delta x = x_5 - x_4 = x_4 - x_3 \\ = x_3 - x_2 = x_2 - x_1 = 1 \\ \text{It has been given that} \, f(x) = x^2 \\ \text{Now, we have that,}\\ f(x) = {\{(-1,2), (2,7), (7,14), (14, 23)}\} \\ \text{where} \, m_1 = -1, M_1 = 2, m_2 = 2, M_2 = 7 \\ m_3 = 7, M_3 = 14, m_4 = 14, M_4 = 23 \\ \text{To find the} \, L(P,f) \\ \text{We know that} \, L(P,f) = \sum _{n = 1}^{4} m_i \delta x_i\\ = -1(1) + 2(1) + 7(1) + 14(1)\\ \text{Hence,}\\ L(P,f) = 22\\ \text{To find the} \, U(P,f) \\ \text{We know that} \, U(P,f) = \sum _{n = 1}^{4} M_i \delta x_i\\ = 2(1) + 7(1) + 14(1) + 23(1)\\ \text{Hence,}\\ U(P,f) = 46 \\ \text{Thus, from the above results. We see that,}\\ L(P,f) < U(P,f)$