If \(\displaystyle f(x) = 3x^{2} − 2x\), \(\displaystyle [0,3]\) given \(\displaystyle [a,b]\)
evaluate the Riemann sum with \(\displaystyle n = 6\), taking the sample points to be right endpoints.
\(\displaystyle \sum\limits_{i=6}^n \Delta x [f(a + i \Delta x)]\)
\(\displaystyle \Delta x = \dfrac{b - a}{n}\)
\(\displaystyle \Delta x = \dfrac{3 - 0}{6} = \dfrac{1}{3}\)
\(\displaystyle n = 6\)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[f(0 + (1)(\dfrac{1}{3})] + [f(0 + (2)(\dfrac{1}{3})] + [f(0 + (3)(\dfrac{1}{3})] + [f(0 + (4)(\dfrac{1}{3})] + [f(0 + (5)(\dfrac{1}{3})] + [f(0 + ( 6)(\dfrac{1}{3})]]\)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[f((1)(\dfrac{1}{3}))] + [f((2)(\dfrac{1}{3}))] + [f((3)(\dfrac{1}{3}))] + [f((4)(\dfrac{1}{3}))] + [f( (5)(\dfrac{1}{3}))] + [f(( 6)(\dfrac{1}{3}))]]\)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[f(\dfrac{1}{3})] + [[f(\dfrac{2}{3})] + [[f(\dfrac{3}{3})] + [[f(\dfrac{4}{3})] + [[f(\dfrac{5}{3})] + [[f(\dfrac{6}{3})]] \)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[f(\dfrac{1}{3})] + [[f(\dfrac{2}{3})] + [f(1)] + [[f(\dfrac{4}{3})] + [[f(\dfrac{5}{3})] + [f(2)]] \)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[3(\dfrac{1}{3})^{2} - 2(\dfrac{1}{3})] + [[3(\dfrac{2}{3})^{2} - 2(\dfrac{2}{3})] + [3(1)^{2} - 2(1)] + [[3((\dfrac{4}{3})^{2} - 2(\dfrac{4}{3})] + [[3(\dfrac{5}{3})^{2} - 2(\dfrac{5}{3})] + [3(2)^{2} - 2(2))]] \)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[3(\dfrac{1}{9}) - 2(\dfrac{1}{3})] + [[3(\dfrac{4}{9}) - 2(\dfrac{2}{3})] + [3(1) - 2(1)] + [[3((\dfrac{16}{9}) - 2(\dfrac{4}{3})] + [[3(\dfrac{25}{9}) - 2(\dfrac{5}{3})] + [3(4) - 2(2))]] \)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[(\dfrac{3}{9}) - (\dfrac{2}{3})] + [(\dfrac{12}{9}) - (\dfrac{4}{3})] + [1] + [(\dfrac{48}{9}) - (\dfrac{8}{3})] + [(\dfrac{75}{9}) - (\dfrac{10}{3})] + [8]] \)]
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[( - \dfrac{1}{3})] + [0] + [1] + [(\dfrac{24}{9})] + [(\dfrac{45}{9})] + [8]] \)]
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[( - \dfrac{1}{3})] + [0] + [1] + [(\dfrac{8}{3})] + [5] + [8]] \)]
\(\displaystyle \sum\limits_{i=6}^n [-\dfrac{1 }{ 9}] + [0] + [\dfrac{1}{ 3}] + [\dfrac{8}{9}] + [\dfrac{5}{3 }] + [\dfrac{8}{3}\)]
Right track?
evaluate the Riemann sum with \(\displaystyle n = 6\), taking the sample points to be right endpoints.
\(\displaystyle \sum\limits_{i=6}^n \Delta x [f(a + i \Delta x)]\)
\(\displaystyle \Delta x = \dfrac{b - a}{n}\)
\(\displaystyle \Delta x = \dfrac{3 - 0}{6} = \dfrac{1}{3}\)
\(\displaystyle n = 6\)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[f(0 + (1)(\dfrac{1}{3})] + [f(0 + (2)(\dfrac{1}{3})] + [f(0 + (3)(\dfrac{1}{3})] + [f(0 + (4)(\dfrac{1}{3})] + [f(0 + (5)(\dfrac{1}{3})] + [f(0 + ( 6)(\dfrac{1}{3})]]\)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[f((1)(\dfrac{1}{3}))] + [f((2)(\dfrac{1}{3}))] + [f((3)(\dfrac{1}{3}))] + [f((4)(\dfrac{1}{3}))] + [f( (5)(\dfrac{1}{3}))] + [f(( 6)(\dfrac{1}{3}))]]\)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[f(\dfrac{1}{3})] + [[f(\dfrac{2}{3})] + [[f(\dfrac{3}{3})] + [[f(\dfrac{4}{3})] + [[f(\dfrac{5}{3})] + [[f(\dfrac{6}{3})]] \)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[f(\dfrac{1}{3})] + [[f(\dfrac{2}{3})] + [f(1)] + [[f(\dfrac{4}{3})] + [[f(\dfrac{5}{3})] + [f(2)]] \)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[3(\dfrac{1}{3})^{2} - 2(\dfrac{1}{3})] + [[3(\dfrac{2}{3})^{2} - 2(\dfrac{2}{3})] + [3(1)^{2} - 2(1)] + [[3((\dfrac{4}{3})^{2} - 2(\dfrac{4}{3})] + [[3(\dfrac{5}{3})^{2} - 2(\dfrac{5}{3})] + [3(2)^{2} - 2(2))]] \)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[3(\dfrac{1}{9}) - 2(\dfrac{1}{3})] + [[3(\dfrac{4}{9}) - 2(\dfrac{2}{3})] + [3(1) - 2(1)] + [[3((\dfrac{16}{9}) - 2(\dfrac{4}{3})] + [[3(\dfrac{25}{9}) - 2(\dfrac{5}{3})] + [3(4) - 2(2))]] \)
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[(\dfrac{3}{9}) - (\dfrac{2}{3})] + [(\dfrac{12}{9}) - (\dfrac{4}{3})] + [1] + [(\dfrac{48}{9}) - (\dfrac{8}{3})] + [(\dfrac{75}{9}) - (\dfrac{10}{3})] + [8]] \)]
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[( - \dfrac{1}{3})] + [0] + [1] + [(\dfrac{24}{9})] + [(\dfrac{45}{9})] + [8]] \)]
\(\displaystyle \sum\limits_{i=6}^n (\dfrac{1}{3}) [[( - \dfrac{1}{3})] + [0] + [1] + [(\dfrac{8}{3})] + [5] + [8]] \)]
\(\displaystyle \sum\limits_{i=6}^n [-\dfrac{1 }{ 9}] + [0] + [\dfrac{1}{ 3}] + [\dfrac{8}{9}] + [\dfrac{5}{3 }] + [\dfrac{8}{3}\)]
Right track?
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