Using only the definition of Riemann sum and your knowledge of limits, compute t

lwfrgin 2021-10-22 Answered
Using only the definition of Riemann sum and your knowledge of limits, compute the exact area under the curve \(\displaystyle{2}{x}^{{2}}\) between x=-2 and x=1.

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Expert Answer

krolaniaN
Answered 2021-10-23 Author has 12799 answers
Divide the are in n strips of rectangles with breath:
\(\displaystyle\triangle{x}={\frac{{{1}-{\left(-{2}\right)}}}{{{n}}}}\)
\(\displaystyle={\frac{{{3}}}{{{n}}}}\)
Now using right hand Riemann sums the area under the curve \(\displaystyle{2}{x}^{{2}}\) is:
\(\displaystyle{S}_{{n}}=\triangle{x}{\sum_{{{i}={1}}}^{{n}}}{2}{\left(-{2}+{i}\triangle{x}\right)}^{{2}}\)
\(\displaystyle={\frac{{{3}}}{{{n}}}}{\sum_{{{i}={1}}}^{{n}}}{2}{\left(-{2}+{i}{\frac{{{3}}}{{{n}}}}\right)}^{{2}}\)
\(\displaystyle={\frac{{{3}}}{{{n}}}}{\sum_{{{i}={1}}}^{{n}}}{2}{\left({4}-{\frac{{{12}}}{{{n}}}}{i}+{\frac{{{9}}}{{{n}^{{2}}}}}{i}^{{2}}\right)}\)
\(\displaystyle={\frac{{{24}}}{{{n}}}}{\sum_{{{i}={1}}}^{{n}}}{1}-{\frac{{{72}}}{{{n}^{{2}}}}}{\sum_{{{i}={1}}}^{{n}}}{i}+{\frac{{{54}}}{{{n}^{{3}}}}}{\sum_{{{i}={1}}}^{{n}}}{i}^{{2}}\)
\(\displaystyle={\frac{{{24}}}{{{n}}}}\times{n}-{\frac{{{72}}}{{{n}^{{2}}}}}\times{\frac{{{n}{\left({n}+{1}\right)}}}{{{2}}}}+{\frac{{{54}}}{{{n}^{{3}}}}}\times{\frac{{{n}{\left({n}+{1}\right)}{\left({2}{n}+{1}\right)}}}{{{6}}}}\)
\(\displaystyle={24}-{36}{\left({1}+{\frac{{{1}}}{{{n}}}}\right)}+{9}{\left({1}+{\frac{{{1}}}{{{n}}}}\right)}{\left({2}+{\frac{{{1}}}{{{n}}}}\right)}\)
Now to calculate the area decrease the width of the rectangular strips to infidecimal value thereby increasing the number of rectangles to infinity.
Finally the required area of the rectangle will be:
\(\displaystyle\lim_{{{n}\to\infty}}{S}_{{n}}=\lim_{{{n}\to\infty}}{24}-{36}{\left({1}+{\frac{{{1}}}{{{n}}}}\right)}_{{{1}+{\frac{{{1}}}{{{n}}}}}}{\left({2}+{\frac{{{1}}}{{{n}}}}\right)}\)
\(\displaystyle={24}-{36}{\left({1}+{0}\right)}+{9}{\left({1}+{0}\right)}{\left({2}+{0}\right)}\)
\(\displaystyle={24}-{36}+{18}\)
\(\displaystyle={6}\)
Thus, area under the curve \(\displaystyle{2}{x}^{{2}}\) is 6
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