5.6-bo'lim : Aniq integralning ta'rifi
2. Integralni baholash uchun aniq integralning ta'rifidan foydalaning. \(x_{\,i}^*\) uchun har bir oraliqning o'ng oxirgi nuqtasidan foydalaning.
\[\int_{0}^{1}{{6x\left( {x - 1} \right)\,dx}}\]Yechim
Har bir subintervalning kengligi,
\[\Delta x = \frac{{1 - 0}}{n} = \frac{1}{n}\]\(\left[ {0,1} \right]\) oralig'i uchun subintervallar shunday bo'ladi:
\[\left[ {0,\frac{1}{n}} \right],\left[ {\frac{1}{n},\frac{2}{n}} \right],\left[ {\frac{2}{n},\frac{3}{n}} \right], \ldots ,,\left[ {\frac{{i - 1}}{n},\frac{i}{n}} \right], \ldots ,,\left[ {\frac{{n - 1}}{n},1} \right]\]Bundan u umumiy subintervalning to'g'ri yakuniy nuqtasiga o'xshaydi va shuning uchun \(x_i^*\),
\[x_i^* = \frac{i}{n}\]2-qadam
Aniq integralning ta'rifidagi yig'indi shunday bo'ladi:
\[\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \sum\limits_{i = 1}^n {\left[ {\left( {\frac{{6i}}{n}} \right)\left( {\frac{i}{n} - 1} \right)} \right]\left[ {\frac{1}{n}} \right]} = \sum\limits_{i = 1}^n {\left[ {\frac{{6{i^2}}}{{{n^3}}} - \frac{{6i}}{{{n^2}}}} \right]} \]3-qadam
Endi biz formulalardan foydalanishimiz kerak Xulosa belgisi yig'indini "baholash" uchun Qo'shimchalar bo'limidagi bo'lim.
\[\begin{align*}\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} & = \sum\limits_{i = 1}^n {\left[ {\frac{{6{i^2}}}{{{n^3}}}} \right]} - \sum\limits_{i = 1}^n {\left[ {\frac{{6i}}{{{n^2}}}} \right]} = \frac{6}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} - \frac{6}{{{n^2}}}\sum\limits_{i = 1}^n i \\ & = \frac{6}{{{n^3}}}\left( {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) - \frac{6}{{{n^2}}}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right) = \frac{{2{n^2} + 3n + 1}}{{{n^2}}} - \frac{{3n + 3}}{n}\end{align*}\]4-qadam
Nihoyat, biz integralning qiymatini aniqlash uchun aniq integralning ta'rifidan foydalanishimiz mumkin.
\[\int_{0}^{1}{{6x\left( {x - 1} \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{2{n^2} + 3n + 1}}{{{n^2}}} - \frac{{3n + 3}}{n}} \right] = 2 - 3 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 1}}\]