Да се реши интеграла.

[aifC]

[tex]\int\limits_{0}^{n} \{x^{2}\}dx + \frac{2}{3}n^{3} = ?[/tex]

[Добромир Глухаров]

Аз получавам $\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}+\cdots+\sqrt{n^2-1}+\sqrt{n^2}$, но не мога да го опростя повече.

[jordan.dimitrov]

[tex]\int\limits_{0}^{n}x^{2}dx = [\frac{x^3}{3} ]_{0}^{n} = \frac{n^3}{3} - \frac{0^3}{3} = \frac{n^3}{3}[/tex]

[tex]\int\limits_{0}^{n}x^{2}dx + \frac{2}{3}n^3 = n^3[/tex]

[aifC]

Важно е да се отбележи че : [tex]\{x^{2}\} = x^{2} - \lfloor x^{2}\rfloor[/tex]

След преработка на интеграла достигаме до:[tex]\sum_{j=1}^{n^{2} } j^{\frac{1}{2}} - \frac{2}{3}n^{3}[/tex]
[tex]\implies \sum_{j=1}^{n^{2}}(j^{\frac{3}{2}} - (j - 1)^{\frac{3}{2}}) = \sum_{j=1}^{n^{2}}(j^{\frac{3}{2}}) - \sum_{j=0}^{n^{2} - 1}(j^{\frac{3}{2}}) \\
= (n^{2})^{\frac{3}{2}} + \sum_{j=1}^{n^{2} - 1}(j^{\frac{3}{2}}) - \sum_{j=1}^{n^{2} - 1}(j^{\frac{3}{2}}) - (0)^{\frac{3}{2}} = n^{3};[/tex]

[matst]

Нека [tex]n=2[/tex].
[tex]\int\limits_{0}^{2} \{x^2\} dx+\frac{2}{3}.2^3= \int\limits_{0}^{1} \{x^2\} dx+\int\limits_{1}^{\sqrt{2}} \{x^2\} dx+\int\limits_{\sqrt{2}}^{\sqrt{3}} \{x^2\} dx+\int\limits_{\sqrt{3}}^{2} \{x^2\} dx+\frac{2}{3}.2^3=[/tex]
[tex]=\int\limits_{0}^{1} x^2 dx+\int\limits_{1}^{\sqrt{2}} (x^2-1) dx+\int\limits_{\sqrt{2}}^{\sqrt{3}} (x^2-2) dx+\int\limits_{\sqrt{3}}^{2} (x^2-3) dx+\frac{2}{3}.2^3=[/tex]
[tex]=\int\limits_{0}^{2} x^2 dx-\int\limits_{1}^{\sqrt{2}} dx-2\int\limits_{\sqrt{2}}^{\sqrt{3}} dx-3\int\limits_{\sqrt{3}}^{2} dx+\frac{2}{3}.2^3=[/tex]
[tex]=\left. \frac{1}{3} x^3 \right|_{x=0}^{x=2} -(\sqrt{2}-1 )-2(\sqrt{3}-\sqrt{2}) -3(2-\sqrt{3})+\frac{2}{3}.2^3=[/tex]
[tex]=2^3-\sqrt{2}+1-2\sqrt{3}+2\sqrt{2}-3.2+3.\sqrt{3}=[/tex]
[tex]=3+\sqrt{2}+\sqrt{3}=[/tex]
[tex]=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}.[/tex]

В общият случай се получава отговора на Добромир Глухаров от Съб Мар 31, 2018 7:33 pm.

[aifC]

[tex]\int\limits_{0}^{n} \{x^{2}\} dx = \sum_{j=1}^{n^{2} }\int\limits_{\sqrt{j-1}}^{\sqrt{j}} \{x^{2}\} dx = \sum_{j=1}^{n^{2}} \int\limits_{j-1}^{j} \frac{\{u\}}{2u^{\frac{1}{2}}}du[/tex]
[tex]= \sum_{j=1}^{n^{2}} \int\limits_{0}^{1} \frac{y}{2(j-1+y)^{\frac{1}{2}}} dy = \frac{1}{2} \sum_{j=1}^{n^{2}} \left[\frac{2}{3}(j-1+y)^{\frac{3}{2}} - 2(j-1)(j-1+y)^{\frac{1}{2}}\right]_0^{1} = \frac{1}{2} \sum_{j=1}^{n^{2}} \left[\frac{2}{3}(j^{\frac{3}{2}} - (j-1)^{\frac{3}{2}}) - 2(j-1)(j^{\frac{1}{2}} - (j-1)^{\frac{1}{2}}) \right][/tex]
[tex]= \frac{1}{2} \sum_{j=1}^{n^{2}} \left[\frac{2}{3}(j^{\frac{3}{2}} - (j-1)^{\frac{3}{2}}) + 2j^{\frac{1}{2}} - 2(j^{\frac{3}{2}} - (j-1)^{\frac{3}{2}})\right] = \frac{1}{2} \sum_{j=1}^{n^{2}}j^{\frac{1}{2}} - \frac{2}{3}\sum_{j=1}^{n^{2}}(j^{\frac{3}{2}} - (j-1)^{\frac{3}{2}})[/tex]
[tex]= \sum_{j=1}^{n^{2}}j^{\frac{1}{2}} - \frac{2}{3}n^{3};[/tex]