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

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.

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

krolaniaN
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