Let \(f(x)=\frac{1}{x}\ 0\le x\le 2\ P=\{ 1,1.2,1.4,1.6,1.8,2 \}\) Find \(w(P,f)\)
Correct Answer: Description for Correct answer:
\(P=\left[1,1.2,1.4,1.6,1.8,2\right]\)
Subintervals of \(P\) are
\([1,1.2][1.2,1.4][1.4,1.6][1.6,1.8][1.6,1.8][1.8,2]\)
\(\Large m_{1}=\frac{1}{1.2};\ m_{2}=\frac{1}{1.4};\ m_{3}=\frac{1}{1.8};\ m_{4}=\frac{1}{1.8};\ m_{5}=\frac{1}{2}\)
\(\Large M_{1}=1;\ M_{2}=\frac{1}{1.2};\ M_{3}=\frac{1}{1.4};\ M_{4}=\frac{1}{1.6};\ M_{5}=\frac{1}{1.8}\)
\(\Delta x_{1}=\Delta x_{2}=\Delta x_{3}=\Delta x_{4}=\Delta x_{5}=0.2\)
\(w(P,f)=\sum\limits_{r=1}^{5}(M_{r}-m_{r})\Delta x_{r}\)
\(\Large = \left(1-\frac{1}{2}\right) \times 0.2+ \left(\frac{1}{1.2}-\frac{1}{1.4}\right)0.2+ \left(\frac{1}{1.4}-\frac{1}{1.6}\right) \times 0.2\)
\(\Large + \left(\frac{1}{1.6}-\frac{1}{1.8}\right) \times 0.2+ \left(\frac{1}{1.8}-\frac{1}{2}\right) \times 0.2 \)
\(=0.2\left[ 1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.4}+\frac{1}{1.4}-\frac{1}{1.6}+\frac{1}{1.6}-\frac{1}{1.8}+\frac{1}{1.8}-\frac{1}{2}\right]\)
\(\Large =0.2 \left(1-\frac{1}{2}\right) \)
\(\Large =0.2 \times \frac{1}{2}=0.1\)
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