Question #249064

Show that š‘ˆ(āˆ’š‘“, š‘) = āˆ’šæ(š‘“, š‘) and šæ(āˆ’š‘“, š‘) = āˆ’š‘ˆ(š‘“, š‘). 


Expert's answer

š‘ˆ(š‘“, š‘) and šæ(š‘“, š‘) are upper and lower Riemann sums for the partition p.

š‘ˆ(š‘“,š‘)=āˆ‘MiĪ”xi, L(š‘“,š‘)=āˆ‘miĪ”xiš‘ˆ(š‘“, š‘)=\sum M_i\Delta x_i,\ L(š‘“, š‘)=\sum m_i\Delta x_i

Mi=sup{f(x):xiāˆ’1≤x≤xi}, mi=inf{f(x):xiāˆ’1≤x≤xi}M_i=sup\{f(x):x_{i-1}\le x \le x_i\},\ m_i=inf\{f(x):x_{i-1}\le x \le x_i\}


sup{āˆ’f(x):xiāˆ’1≤x≤xi}=āˆ’inf{f(x):xiāˆ’1≤x≤xi}=āˆ’misup\{-f(x):x_{i-1}\le x \le x_i\}=-inf\{f(x):x_{i-1}\le x \le x_i\}=-m_i

U(āˆ’f,p)=āˆ’āˆ‘miĪ”xi=āˆ’L(f,p)U(-f,p)=-\sum m_i\Delta x_i=-L(f,p)


inf{āˆ’f(x):xiāˆ’1≤x≤xi}=āˆ’sup{f(x):xiāˆ’1≤x≤xi}=āˆ’Miinf\{-f(x):x_{i-1}\le x \le x_i\}=-sup\{f(x):x_{i-1}\le x \le x_i\}=-M_i

L(āˆ’f,p)=āˆ’āˆ‘MiĪ”xi=āˆ’U(f,p)L(-f,p)=-\sum M_i\Delta x_i=-U(f,p)


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