Question

For the function f(x)= x^2-2 defined over[1,6],

Verify L(P,f)≤U(P,f) where is the partition which divide [1,6] into five equal intervals

Answer 1

If [1,6] is divided into 5 equal partitions, the partitions will be :-

$P=\{P_i\}=\{[x_i,x_{i+1}]\}=\{[i,i+1]\}, \\(i=1,2,3,4,5)\\ \text{Now we know that,}\\ M_i= sup\{f(x) : x_i\leq x\leq x_{i+1} \}\\ =sup\{f(x) : i\leq x\leq {i+1} \}\\ \text{And, }\\ m_i= inf \{f(x) : x_i\leq x\leq x_{i+1} \}\\ =inf\{f(x) : i\leq x\leq {i+1} \}$

As f(x) is a increasing function over positive numbers so,

$M_i=f(x_{i+1})=f(i+1)=(i+1)^2-2\\ =i^2+2i-1\\ \text{and, }\\ m_i=f(x_i)=f(i)=i^2-2 \\\text{Now, }\Delta x_i=x_{i+1}-x_i=i+1-i=1$

$\therefore U(P,f)-L(,U,f)\\ =\sum_{i=1}^5 M_i\Delta x_i-\sum_{i=1}^5 m_i\Delta x_i\\ =\sum_{i=1}^5 (M_i-m_i)\Delta x_i\\ =\sum_{i=1}^5 (i^2+2i-1-i^2+2)\times 1\\ =\sum_{i=1}^5 (2i+1) \\= 2\times \dfrac{5\times 6}{2}+5=35>0 \\\Rightarrow U(P,f)>L(P,f)$

(Hence verified.)