Question

Suppose f(x)=x^2 -1 for x belongs to [-1,1] find the upper and lower darboux sum of

(i) four equally space interval

(ii) Ten equally space interval

(iii) n equally space interval

Answer 1

$iii)\\ f(x)=x²-1\text{ and the partition } P_n \text{ of [-1, 1] is}:\\ P_n=\{x_i=\frac{2i}{n}-1:i=0,1,2,\cdots, n\}\\ \text{Thus, we have that:}\\ \quad\\ U(f, P_n)=\sum^{n-1}_{i=0}f(x_{i+1})(x_{i+1}-x_i)=\sum^{n-1}_{i=0}(x_{i+1}^2-1)(x_{i+1}-x_i)\\ \qquad\quad\quad=\sum_{i=0}^{n-1}[\frac{4(i+1)^2}{n^2}-\frac{4(i+1)}{n}]\frac{2}{n}=\frac{8}{n^3}\sum_{i=0}^{n-1}(i+1)^2-\frac{8}{n^2}\sum_{i=0}^{n-1}(i+1)\\ \qquad\quad\quad=\frac{8}{n^3}\sum_{i=1}^{n}i^2-\frac{8}{n^2}\sum_{i=1}^{n}i=\frac{8}{n^3}[\frac{n(n+1)(2n+1)}{6}]-\frac{8}{n^2}[\frac{n(n+1)}{2}]\\ \qquad\qquad=\frac{8(n+1)(2n+1)}{6n^2}-\frac{8n(n+1)}{2n^2}=\frac{4}{3n^2}-\frac{4}{3}\\ \quad\\ L(f, P_n)=\sum^{n-1}_{i=0}f(x_{i})(x_{i+1}-x_i)=\sum^{n-1}_{i=0}(x_{i}^2-1)(x_{i+1}-x_i)\\ \qquad\quad\quad=\sum_{i=0}^{n-1}[\frac{4i^2}{n^2}-\frac{4i}{n}]\frac{2}{n}=\frac{8}{n^3}\sum_{i=0}^{n-1}i^2-\frac{8}{n^2}\sum_{i=0}^{n-1}i\\ \qquad\quad\quad=\frac{8}{n^3}\sum_{i=0}^{n-1}i^2-\frac{8}{n^2}\sum_{i=0}^{n-1}i=\frac{8}{n^3}[\frac{n(n-1)(2n-1)}{6}]-\frac{8}{n^2}[\frac{n(n-1)}{2}]\\ \qquad\qquad=\frac{8(n-1)(2n-1)}{6n^2}-\frac{8n(n-1)}{2n^2}=\frac{4}{3n^2}-\frac{4}{3}\\ \quad\\ \quad\\ ii)\\ \text{Put n=10 in question number (iii) above, we have:}\\ U(f, P_{10})=U(f, P_{10})=-\frac{33}{25}\\ \quad\\ \quad\\ i)\\ \text{Put n=4 in question number (iii) above, we have:}\\ U(f, P_{4})=U(f, P_{4})=-\frac{5}{4}$