Kassandra Mccall

2022-10-02

Using the definition of the integral, find ${\int }_{0}^{a}{x}^{2}dx$

Erika Gomez

As you have that ${x}^{2}$ is continuous all partitions will give the same integral. Hence
${\int }_{0}^{a}{x}^{2}=\underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^{n}\frac{a}{n}\cdot \frac{\left(ai{\right)}^{2}}{{n}^{2}}=\underset{n\to \mathrm{\infty }}{lim}\frac{a}{{n}^{3}}\sum _{i=0}^{n}\left(ai{\right)}^{2}=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}^{3}}{{n}^{3}}\cdot \frac{n\cdot \left(n+1\right)\cdot \left(2n+1\right)}{6}$
The limit of this is$\frac{{a}^{3}}{3}$
which we expected.
We used that
$\sum _{i=1}^{n}{i}^{2}=\frac{n\cdot \left(n+1\right)\cdot \left(2n+1\right)}{6}$
To see that this partiation is allowed we can use that ${x}^{2}$ is strict monoton increasing, So
$\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}\frac{a}{n}{\left(\frac{i}{n}\right)}^{2}\le {\int }_{0}^{a}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\le \underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^{n}{\left(\frac{i}{n}\right)}^{2}$

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