Bu juda oddiy dalil. Faraz qilaylik, \(F\left( x \right)\) \(f\left( x \right)\) ga qarshi hosiladir, i.e. \(F'\left( x \right) = f\left( x \right)\). Keyin hosilalarning asosiy xususiyatlariga ko'ra, bizda ham bor:

\[{\left( {k\,F\left( x \right)} \right)^\prime } = k\,F'\left( x \right) = k\,f\left( x \right)\]

va shuning uchun \(k\,F\left( x \right)\) \(k\,f\left( x \right)\) ga qarshi hosiladir, i.e. \({\left( {k\,F\left( x \right)} \right)^\prime } = k\,f\left( x \right)\). Boshqacha aytganda,

\[\int{{k\,f\left( x \right)\,dx = k\,F\left( x \right) + c}} = k\int{{f\left( x \right)\,dx}}\]

Bu ham juda oddiy dalil Faraz qilaylik, \(F\left( x \right)\) \(f\left( x \right)\) ga qarshi hosiladir va \(G\left( x \right)\) \(g\left( x \right)\) ga qarshi hosiladir. Shunday qilib, bizda \(F'\left( x \right) = f\left( x \right)\) va \(G'\left( x \right) = g\left( x \right)\) bor. Buni hosilalarning asosiy xossalari ham aytadi

\[{\left( {F\left( x \right) \pm G\left( x \right)} \right)^\prime } = \,F'\left( x \right) \pm G'\left( x \right) = f\left( x \right) \pm g\left( x \right)\]

va shuning uchun \(\,F\left( x \right) + G\left( x \right)\) \(\,f\left( x \right) + g\left( x \right)\) ga qarshi hosiladir va \(\,F\left( x \right) - G\left( x \right)\) \(\,f\left( x \right) - g\left( x \right)\) ga qarshi hosiladir. Boshqacha aytganda,

\[\int{{f\left( x \right) \pm g\left( x \right)\,dx}} = F\left( x \right) \pm G\left( x \right) + c = \int{{f\left( x \right)\,dx}} \pm \int{{g\left( x \right)\,dx}}\]

Aniq integralning ta'rifidan bizda,

\[\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{b - a}}{n}\]

va bizda ham bor,

\[\int_{{\,b}}^{{\,a}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{a - b}}{n}\]

Shuning uchun,

\[\begin{align*}\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\frac{{b - a}}{n}} \\ & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\frac{{ - \left( {a - b} \right)}}{n}} \\ & = \mathop {\lim }\limits_{n \to \infty } \left( { - \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\frac{{a - b}}{n}} } \right)\\ & = - \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\frac{{a - b}}{n}} = - \int_{{\,b}}^{{\,a}}{{f\left( x \right)\,dx}}\end{align*}\]

Aniq integralning ta'rifidan bizda,

\[\begin{align*}\int_{{\,a}}^{{\,a}}{{f\left( x \right)\,dx}} & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{a - a}}{n} = 0\\ & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\left( 0 \right)} \\ & = \mathop {\lim }\limits_{n \to \infty } 0\\ & = 0\end{align*}\]

Aniq integralning ta'rifidan bizda,

\[\begin{align*}\int_{{\,a}}^{{\,b}}{{c\,f\left( x \right)\,dx}} & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {c\,f\left( {x_i^*} \right)\Delta x} \\ & = \mathop {\lim }\limits_{n \to \infty } c\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \\ & = c\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \\ & = c\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\end{align*}\]

Esda tutingki, biz konstantalarni yig'indilardan va chegaralardan tashqariga chiqarishimiz mumkin.

Avval "+" formulasini isbotlaymiz. Aniq integralning ta'rifidan bizda,

\[\begin{align*}\int_{{\,a}}^{{\,b}}{{f\left( x \right) + g\left( x \right)\,dx}} & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {f\left( {x_i^*} \right) + \,g\left( {x_i^*} \right)} \right)\Delta x} \\ & = \mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} + \sum\limits_{i = 1}^n {g\left( {x_i^*} \right)\Delta x} } \right)\\ & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {g\left( {x_i^*} \right)\Delta x} \\ & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} + \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}\end{align*}\]

“-” formulasini isbotlash uchun yuqoridagi ishni ortiqcha belgisi o‘rniga minus belgisi bilan takrorlashimiz mumkin yoki biz bu haqiqatni endi bilganimizdan ortiqcha belgisi bilan va yuqorida isbotlangan xususiyatlardan quyidagi tarzda foydalanishimiz mumkin.

\[\begin{align*}\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}} & = \int_{{\,a}}^{{\,b}}{{f\left( x \right) + \left( { - g\left( x \right)} \right)\,dx}}\\ & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} + \int_{{\,a}}^{{\,b}}{{\left( { - g\left( x \right)} \right)\,dx}}\\ & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} - \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}\end{align*}\]

Agar biz \(f\left( x \right) = c\) ni aniqlasak, u holda bizda aniq integral ta'rifidan kelib chiqib,

\[\begin{align*}\int_{{\,a}}^{{\,b}}{{c\,dx}} & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\\ & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{b - a}}{n}\\ & = \mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{i = 1}^n c } \right)\frac{{b - a}}{n}\\ & = \mathop {\lim }\limits_{n \to \infty } \left( {cn} \right)\frac{{b - a}}{n}\\ & = \mathop {\lim }\limits_{n \to \infty } c\left( {b - a} \right)\\ & = c\left( {b - a} \right)\end{align*}\]

Aniq integralning ta'rifidan bizda,

\[\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{b - a}}{n}\]

Endi, taxminga ko'ra, \(f\left( x \right) \ge 0\) va bizda ham \(\Delta x > 0\) bor va shuning uchun biz buni bilamiz

\[\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \ge 0\]

Shunday qilib, biz chegaralarning asosiy xususiyatlaridan kelib chiqib,

\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \ge \mathop {\lim }\limits_{n \to \infty } 0 = 0\]

Ammo chap tomon aynan integralning ta'rifidir va shuning uchun bizda

\[\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \ge 0\]

Bizda \(f\left( x \right) \ge g\left( x \right)\) mavjud bo'lganligi sababli, biz \(f\left( x \right) - g\left( x \right) \ge 0\) on \(a \le x \le b\) va shuning uchun yuqorida 8-xususiyat tomonidan isbotlanganligini bilamiz,

\[\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}} \ge 0\]

Biz 4-mulkdan ham bilamizki,

\[\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} - \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}\]

Shunday qilib, bizda,

\[\begin{align*}\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} - \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}} & \ge 0\\ \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} & \ge \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}\end{align*}\]

\(m \le f\left( x \right) \le M\) ni hisobga olgan holda biz yozish uchun har bir tengsizlikda 9-xususiyatdan foydalanishimiz mumkin,

\[\int_{{\,a}}^{{\,b}}{{m\,dx}} \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le \int_{{\,a}}^{{\,b}}{{M\,dx}}\]

Keyin chap va o'ngdagi 7-xususiyat bo'yicha integral olish uchun,

\[m\left( {b - a} \right) \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le M\left( {b - a} \right)\]

Avval shuni ta'kidlaymizki, biz funktsiya va mutlaq qiymat haqida quyidagilarni aytishimiz mumkin,

\[ - \left| {f\left( x \right)} \right| \le f\left( x \right) \le \left| {f\left( x \right)} \right|\]

Agar biz har bir tengsizlikda 9-xususiyatdan foydalansak,

\[\int_{{\,a}}^{{\,b}}{{ - \left| {f\left( x \right)} \right|\,dx}} \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)} \right|\,dx}}\]

Biz bilamizki, chap integraldan minus belgisini ajratib ko'rsatishimiz mumkin:

\[ - \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)} \right|\,dx}} \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)} \right|\,dx}}\]

Va nihoyat, eslang, agar \(\left| p \right| \le b\) keyin \( - b \le p \le b\) va, albatta, bu teskari tartibda ishlaydi, shuning uchun bizda shunday bo'lishi kerak:

\[\left| {\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}} \right| \le \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)\,} \right|dx}}\]

Faraz qilaylik, \(x\) va \(x + h\) \(\left( {a,b} \right)\) da. Keyin bizda,

\[g\left( {x + h} \right) - g\left( x \right) = \int_{{\,a}}^{{\,x + h}}{{f\left( t \right)\,\,dt}} - \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}}\]

Endi, foydalanish Mulk 5 Integral xossalaridan biz birinchi integralni qayta yozishimiz va keyin quyidagi tarzda biroz soddalashtirishimiz mumkin.

\[\begin{align*}g\left( {x + h} \right) - g\left( x \right) = \left( {\int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}} + \int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,\,dt}}} \right) - \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}}\\ & = \int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,\,dt}}\end{align*}\]

Nihoyat, \(h \ne 0\) deb faraz qilamiz va biz quyidagilarni olamiz:

\[\begin{equation}\frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} = \frac{1}{h}\int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,\,dt}} \label{eq:eq1} \end{equation}\]

Keling, \(h > 0\) deb faraz qilaylik va biz hali ham \(x + h\) \(\left( {a,b} \right)\) da deb faraz qilganimiz uchun biz \(f\left( x \right)\) \(\left[ {x,x + h} \right]\) da uzluksiz ekanligini bilamiz va shuning uchun Ekstremal qiymat teoremasi biz \(\left[ {x,x + h} \right]\) da \(c\) va \(d\) raqamlari mavjudligini bilamiz, shuning uchun \(f\left( c \right) = m\) \(\left[ {x,x + h} \right]\) da \(f\left( x \right)\) ning mutlaq minimumi va \(f\left( d \right) = M\) \(f\left( x \right)\)2 da \(f\left( x \right)\) ning mutlaq maksimali.

So, by Mulk 10 Integral xususiyatlardan keyin biz bilamiz,

\[mh \le \int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,dt}} \le Mh\]

Or,

\[f\left( c \right)h \le \int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,dt}} \le f\left( d \right)h\]

Endi buning ikkala tomonini h ga bo'ling,

\[f\left( c \right) \le \frac{1}{h}\int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,dt}} \le f\left( d \right)\]

va keyin olish uchun \(\eqref{eq:eq1}\) dan foydalaning,

\[\begin{equation}f\left( c \right) \le \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} \le f\left( d \right) \label{eq:eq2} \end{equation}\]

Keyinchalik, agar \(h < 0\) bo'lsa, biz yuqoridagi bir xil argumentdan o'tishimiz mumkin, bundan tashqari biz yuqoridagi tengsizlikka erishish uchun \(\left[ {x + h,x} \right]\) ustida ishlaymiz. Boshqacha qilib aytganda, \(\eqref{eq:eq2}\), \(h \ne 0\) sharti bilan rost.

Endi, agar biz \(h \to 0\) ni olsak, bizda \(c \to x\) va \(d \to x\) ham bor, chunki \(c\) va \(d\) ham \(x\) va \(x + h\) orasida. Bu bizda quyidagi ikkita chegara borligini anglatadi.

\[\mathop {\lim }\limits_{h \to 0} f\left( c \right) = \mathop {\lim }\limits_{c \to x} f\left( c \right) = f\left( x \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{h \to 0} f\left( d \right) = \mathop {\lim }\limits_{d \to x} f\left( d \right) = f\left( x \right)\]

The Siqish teoremasi keyin bizga aytadi,

\[\begin{equation}\mathop {\lim }\limits_{h \to 0} \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} = f\left( x \right) \label{eq:eq3} \end{equation}\]

lekin buning chap tomoni aynan \(g\left( x \right)\) hosilasining ta'rifi va shuning uchun biz buni olamiz,

\[g'\left( x \right) = f\left( x \right)\]

Shunday qilib, biz \(g\left( x \right)\) \(\left( {a,b} \right)\) da farqlanishini ko'rsatdik.

Endi, Teorema Hosila ta'rifi bo'limining oxirida \(g\left( x \right)\) \(\left( {a,b} \right)\) da ham uzluksiz ekanligini aytadi. Nihoyat, agar biz \(x = a\) yoki \(x = b\) ni olsak, xuddi shunday natijani olish uchun bir tomonlama chegaralar yordamida \(\eqref{eq:eq3}\) ni olish uchun ishlatgan shunga o'xshash argumentdan o'tishimiz mumkin va shuning uchun Hosilaning ta'rifi bo'limining oxiridagi teorema ham \(g\left( x \right)\) ning \(g\left( x \right)\) va \(g\left( x \right)\) uzluksiz ekanligini ma'lum qiladi. shuning uchun aslida \(g\left( x \right)\) \(\left[ {a,b} \right]\) da uzluksizdir.

Avval \(g\left( x \right) = \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}}\) ga ruxsat bering va keyin biz \(g'\left( x \right) = f\left( x \right)\) va shuning uchun \(g\left( x \right)\) \(\left[ {a,b} \right]\) da \(f\left( x \right)\) ning anti-hosilasi ekanligini Hisoblashning asosiy teoremasining I qismidan bilamiz. Bundan tashqari, \(F\left( x \right)\) biz tanlamoqchi bo'lgan \(\left[ {a,b} \right]\) da \(f\left( x \right)\) ning har qanday anti-hosilasi deb faraz qilaylik. Demak, bu bizda bo'lishi kerak degan ma'noni anglatadi,

\[g'\left( x \right) = F'\left( x \right)\]

Keyin, tomonidan 2-fakt O'rtacha qiymat teoremasi bo'limida biz \(g\left( x \right)\) va \(F\left( x \right)\) \(\left( {a,b} \right)\) da qo'shimcha doimiydan ko'p bo'lmagan farq qilishi mumkinligini bilamiz. Boshqacha qilib aytganda, \(a < x < b\) uchun bizda,

\[F\left( x \right) = g\left( x \right) + c\]

Endi \(g\left( x \right)\) va \(F\left( x \right)\) \(\left[ {a,b} \right]\) da uzluksiz bo'lganligi sababli, agar biz bu chegarani \(x \to {a^ + }\) va \(x \to {b^ - }\) deb olsak, bu \(x = a\) va \(x = b\) bo'lsa ham amal qilishini ko'rishimiz mumkin.

Shunday qilib, \(a \le x \le b\) uchun biz \(F\left( x \right) = g\left( x \right) + c\) ekanligini bilamiz. Keling, buni va \(g\left( x \right)\) ta'rifini quyidagi amallarni bajarish uchun ishlatamiz.

\[\begin{align*}F\left( b \right) - F\left( a \right) & = \left( {g\left( b \right) + c} \right) - \left( {g\left( a \right) + c} \right)\\ & = g\left( b \right) - g\left( a \right)\\ & = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,\,dt}} + \int_{{\,a}}^{{\,a}}{{f\left( t \right)\,\,dt}}\\ & = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,\,dt}} + 0\\ & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,\,dx}}\end{align*}\]

E'tibor bering, oxirgi bosqichda biz integralda ishlatiladigan o'zgaruvchi muhim emasligidan foydalandik va shuning uchun biz \(t\)-ni \(x\)-ga o'zgartirishimiz mumkin edi.

Biz bilamizki, \(n\) raqamlarining o'rtacha qiymati shunchaki \(n\) ga bo'lingan barcha raqamlar yig'indisidir, shuning uchun keling, bundan boshlaylik. Keling, \(\left[ {a,b} \right]\) oralig'ini olaylik va uni har bir uzunlikdagi \(n\) subintervallariga ajratamiz,

\[\Delta x = \frac{{b - a}}{n}\]

Endi bu oraliqlarning har biridan \(x_1^*,x_2^*, \ldots ,x_n^*\) nuqtalarini tanlang va shuni yodda tutingki, bu raqamlarning har birini qanday tanlashimiz muhim emas, agar ular tegishli intervaldan kelsa. Keyin \(f\left( {x_1^*} \right),f\left( {x_2^*} \right), \ldots ,f\left( {x_n^*} \right)\) funktsiya qiymatlarining o'rtacha qiymatini hisoblash orqali hisoblashimiz mumkin,

\[\begin{equation}\frac{{f\left( {x_1^*} \right) + f\left( {x_2^*} \right) + \cdots + f\left( {x_n^*} \right)}}{n} \label{eq:eq4} \end{equation}\]

Endi \(\Delta x\) ta'rifimizdan \(n\) uchun quyidagi formulani olishimiz mumkin.

\[n = \frac{{b - a}}{{\Delta x}}\]

va biz buni olish uchun \(\eqref{eq:eq4}\) ga ulashimiz mumkin,

\[\begin{align*}\frac{{f\left( {x_1^*} \right) + f\left( {x_2^*} \right) + \cdots + f\left( {x_n^*} \right)}}{{\frac{{b - a}}{{\Delta x}}}} & = \frac{{\left[ {f\left( {x_1^*} \right) + f\left( {x_2^*} \right) + \cdots + f\left( {x_n^*} \right)} \right]\Delta x}}{{b - a}}\\ & = \frac{1}{{b - a}}\left[ {f\left( {x_1^*} \right)\Delta x + f\left( {x_2^*} \right)\Delta x + \cdots + f\left( {x_n^*} \right)\Delta x} \right]\\ & = \frac{1}{{b - a}}\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \end{align*}\]

Endi \(n\) ni oshiramiz. Bu shuni anglatadiki, biz oraliqdagi ko'proq va ko'proq funktsiya qiymatlarining o'rtacha qiymatini olamiz va shuning uchun biz \(n\) ni qanchalik katta tanlasak, bu funktsiyaning o'rtacha qiymatiga yaqinroq bo'ladi.

Agar biz \(n\) cheksizlikka o'tishi bilan chegarani olsak, biz o'rtacha funktsiya qiymatini olamiz. Yoki,

\[{f_{avg}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{b - a}}\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \frac{1}{{b - a}}\,\,\,\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \]

Biz \(\frac{1}{{b - a}}\) ni chegaradan chiqarib tashlashimiz mumkin va endi yig'indi chegarasi tanish ko'rinishi kerak, chunki bu aniq integralning ta'rifi. Shunday qilib, aniq integralni qo'yib, biz kutgan formulani olamiz.

\[{f_{avg}} = \frac{1}{{b - a}}\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\]

Keling, aniqlashdan boshlaylik,

\[F\left( x \right) = \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,dt}}\]

\(f\left( x \right)\) uzluksiz bo'lgani uchun biz dan bilamiz Hisoblashning asosiy teoremasi, I qism \(F\left( x \right)\) \(\left[ {a,b} \right]\) da uzluksiz, \(\left( {a,b} \right)\) da differentsiallanadi va \(F'\left( x \right) = f\left( x \right)\).

Endi, dan O'rtacha qiymat teoremasi biz \(c\) raqami borligini bilamiz, shunday qilib \(a < c < b\) va bu,

\[F\left( b \right) - F\left( a \right) = F'\left( c \right)\left( {b - a} \right)\]

Biroq, biz \(F'\left( c \right) = f\left( c \right)\) va,

\[F\left( b \right) = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,dt}} = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\hspace{0.25in}\hspace{0.25in}F\left( a \right) = \int_{{\,a}}^{{\,a}}{{f\left( t \right)\,dt}} = 0\]

Shunday qilib, bizda,

\[\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = f\left( c \right)\left( {b - a} \right)\]

Keling, \(a \le x \le b\) diapazonini \(\Delta x\) kenglikdagi \(n\) subintervallariga bo'lishdan boshlaylik va bu intervallarning har biridan \(x_1^*,x_2^*, \ldots ,x_n^*\) nuqtalarini tanlang.

Endi, agar \(n\) katta bo'lsa va \(F\left( x \right)\) uzluksiz bo'lsa, \(F\left( x \right)\) har bir intervalda va \(i\) da ko'p o'zgarmaydi deb taxmin qilishimiz mumkin.^th oraliqda biz kuch \(F\left( x \right) \approx F\left( {x_i^*} \right)\) qiymati bilan taxminan doimiy deb taxmin qilishimiz mumkin. Har bir intervaldagi ish taxminan,

\[{W_i} \approx F\left( {x_i^*} \right)\Delta x\]

\(a \le x \le b\) bo'yicha umumiy ish taxminan,

\[W \approx \sum\limits_{i = 1}^n {{W_i}} = \sum\limits_{i = 0}^n {F\left( {x_i^*} \right)\Delta x} \]

Nihoyat, agar biz \(n\) cheksizlikka o'tgani uchun bu chegarani olsak, biz aniq ishni bajaramiz. Shunday qilib,

\[W = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 0}^n {F\left( {x_i^*} \right)\Delta x} \]

Biroq, bu aniq integralning ta'rifidan boshqa narsa emas va shuning uchun \(F\left( x \right)\) kuchining \(a \le x \le b\) ustidan qilgan ishi,

\[W = \int_{{\,a}}^{{\,b}}{{F\left( x \right)\,dx}}\]