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Let \(f(x)=x,\ 0\le x\le 1\) and \(p=0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\) be the partition of [0, 1]. Find U(p, f) and L(p, f).
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A) \(\Large \frac{1}{8},\frac{1}{8}\) |
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B) \(\Large \frac{3}{8},\frac{7}{8}\) |
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C) \(\Large \frac{5}{8},\frac{3}{8}\) |
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D) \(\Large \frac{7}{8},\frac{9}{8}\) |
Correct Answer:
| C) \(\Large \frac{5}{8},\frac{3}{8}\) |
Description for Correct answer:
\(\Large P=\{0,\frac{1}{4},\frac{1}{2}\frac{3}{4},1\}\)
sub intervals are
\(\Large \left[ 0,\frac{1}{4} \right],\left[ \frac{1}{4},\frac{1}{2} \right],\left[ \frac{1}{2},\frac{3}{4} \right],\left[ \frac{3}{4},1 \right]\)
since \(f(x)=x\)
\(\Large \Rightarrow m_{1}=0;\ m_{2}=\frac{1}{4},\ m_{3}=\frac{1}{2},\ m_{4}=\frac{3}{4}\)
\(\Large \Rightarrow M_{1}=\frac{1}{4};\ M_{2}=\frac{1}{2};\ M_{3}=\frac{3}{4};\ M_{4}=1\)
\(\Delta x_{1}=\Delta x_{2}=\Delta x_{3}=\Delta x_{4}=\Large\frac{1}{4}\)
\(U(P,f)=\sum\limits_{r=1}^{4}m_{r}\Delta x_{r}\)
\(\Large =\frac{1}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}+\frac{3}{4}.\frac{1}{4}+1.\frac{1}{4}\)
\(=\frac{5}{8}\)
\(L(P,f)=\sum\limits_{r=1}^{4}m_{r}\Delta x_{r}\)
\(\Large =0.\frac{1}{4}+\frac{1}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}+\frac{3}{4}.\frac{1}{4}\)
\(\Large =0+\frac{1}{16}+\frac{1}{8}+\frac{3}{16}\)
\(\Large =\frac{3}{8}\)
\(\Large P=\{0,\frac{1}{4},\frac{1}{2}\frac{3}{4},1\}\)
sub intervals are
\(\Large \left[ 0,\frac{1}{4} \right],\left[ \frac{1}{4},\frac{1}{2} \right],\left[ \frac{1}{2},\frac{3}{4} \right],\left[ \frac{3}{4},1 \right]\)
since \(f(x)=x\)
\(\Large \Rightarrow m_{1}=0;\ m_{2}=\frac{1}{4},\ m_{3}=\frac{1}{2},\ m_{4}=\frac{3}{4}\)
\(\Large \Rightarrow M_{1}=\frac{1}{4};\ M_{2}=\frac{1}{2};\ M_{3}=\frac{3}{4};\ M_{4}=1\)
\(\Delta x_{1}=\Delta x_{2}=\Delta x_{3}=\Delta x_{4}=\Large\frac{1}{4}\)
\(U(P,f)=\sum\limits_{r=1}^{4}m_{r}\Delta x_{r}\)
\(\Large =\frac{1}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}+\frac{3}{4}.\frac{1}{4}+1.\frac{1}{4}\)
\(=\frac{5}{8}\)
\(L(P,f)=\sum\limits_{r=1}^{4}m_{r}\Delta x_{r}\)
\(\Large =0.\frac{1}{4}+\frac{1}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}+\frac{3}{4}.\frac{1}{4}\)
\(\Large =0+\frac{1}{16}+\frac{1}{8}+\frac{3}{16}\)
\(\Large =\frac{3}{8}\)
Part of solved Real Analysis questions and answers : >> Elementary Mathematics >> Real Analysis
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