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

\(\displaystyle\triangle{x}={\frac{{{1}-{\left(-{2}\right)}}}{{{n}}}}\)

\(\displaystyle={\frac{{{3}}}{{{n}}}}\)

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