Har bir subintervalning kengligi,

\[\Delta x = \frac{{4 - 1}}{n} = \frac{3}{n}\]

\(\left[ {1,4} \right]\) oralig'i uchun subintervallar shunday bo'ladi:

\[\left[ {1,1 + \frac{3}{n}} \right],\left[ {1 + \frac{3}{n},1 + \frac{6}{n}} \right],\left[ {1 + \frac{6}{n},1 + \frac{9}{n}} \right], \ldots ,,\left[ {1 + \frac{{3\left( {i - 1} \right)}}{n},1 + \frac{{3i}}{n}} \right], \ldots ,,\left[ {1 + \frac{{3\left( {n - 1} \right)}}{n},4} \right]\]

Bundan u umumiy subintervalning to'g'ri yakuniy nuqtasiga o'xshaydi va shuning uchun \(x_i^*\),

\[x_i^* = 1 + \frac{{3i}}{n}\]

2-qadam

Aniq integralning ta'rifidagi yig'indi shunday bo'ladi:

\[\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \sum\limits_{i = 1}^n {\left[ {2\left( {1 + \frac{{3i}}{n}} \right) + 3} \right]\left[ {\frac{3}{n}} \right]} = \sum\limits_{i = 1}^n {\left[ {\frac{{15}}{n} + \frac{{18i}}{{{n^2}}}} \right]} \]

3-qadam

Endi biz formulalardan foydalanishimiz kerak Xulosa belgisi yig'indini "baholash" uchun Qo'shimchalar bo'limidagi bo'lim.

\[\begin{align*}\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} & = \sum\limits_{i = 1}^n {\frac{{15}}{n}} + \sum\limits_{i = 1}^n {\frac{{18i}}{{{n^2}}}} = \frac{1}{n}\sum\limits_{i = 1}^n {15} + \frac{{18}}{{{n^2}}}\sum\limits_{i = 1}^n i \\ & = \frac{1}{n}\left( {15n} \right) + \frac{{18}}{{{n^2}}}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right) = 15 + \frac{{9n + 9}}{n}\end{align*}\]

4-qadam

Nihoyat, biz integralning qiymatini aniqlash uchun aniq integralning ta'rifidan foydalanishimiz mumkin.

\[\int_{1}^{4}{{2x + 3\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \mathop {\lim }\limits_{n \to \infty } \left[ {15 + \frac{{9n + 9}}{n}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {24 + \frac{9}{n}} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{24}}\]