Kassandra Mccall

2022-10-02

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

Erika Gomez

Beginner2022-10-03Added 4 answers

As you have that ${x}^{2}$ is continuous all partitions will give the same integral. Hence

$${\int}_{0}^{a}{x}^{2}=\underset{n\to \mathrm{\infty}}{lim}\sum _{i=0}^{n}\frac{a}{n}\cdot \frac{(ai{)}^{2}}{{n}^{2}}=\underset{n\to \mathrm{\infty}}{lim}\frac{a}{{n}^{3}}\sum _{i=0}^{n}(ai{)}^{2}=\underset{n\to \mathrm{\infty}}{lim}\frac{{a}^{3}}{{n}^{3}}\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}$$

The limit of this is$$\frac{{a}^{3}}{3}$$

which we expected.

We used that

$$\sum _{i=1}^{n}{i}^{2}=\frac{n\cdot (n+1)\cdot (2n+1)}{6}$$

To see that this partiation is allowed we can use that ${x}^{2}$ is strict monoton increasing, So

$$\underset{n\to \mathrm{\infty}}{lim}\sum _{i=1}^{n}\frac{a}{n}{\left(\frac{i}{n}\right)}^{2}\le {\int}_{0}^{a}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\le \underset{n\to \mathrm{\infty}}{lim}\sum _{i=0}^{n}{\left(\frac{i}{n}\right)}^{2}$$

$${\int}_{0}^{a}{x}^{2}=\underset{n\to \mathrm{\infty}}{lim}\sum _{i=0}^{n}\frac{a}{n}\cdot \frac{(ai{)}^{2}}{{n}^{2}}=\underset{n\to \mathrm{\infty}}{lim}\frac{a}{{n}^{3}}\sum _{i=0}^{n}(ai{)}^{2}=\underset{n\to \mathrm{\infty}}{lim}\frac{{a}^{3}}{{n}^{3}}\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}$$

The limit of this is$$\frac{{a}^{3}}{3}$$

which we expected.

We used that

$$\sum _{i=1}^{n}{i}^{2}=\frac{n\cdot (n+1)\cdot (2n+1)}{6}$$

To see that this partiation is allowed we can use that ${x}^{2}$ is strict monoton increasing, So

$$\underset{n\to \mathrm{\infty}}{lim}\sum _{i=1}^{n}\frac{a}{n}{\left(\frac{i}{n}\right)}^{2}\le {\int}_{0}^{a}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\le \underset{n\to \mathrm{\infty}}{lim}\sum _{i=0}^{n}{\left(\frac{i}{n}\right)}^{2}$$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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