15.2-bo'lim: Takroriy integrallar
Oldingi bo'limda biz ikki karrali integral ta'rifini bergan edik. Biroq, bir o'zgaruvchili integralda bo'lgani kabi, ta'rifdan to'g'ridan-to'g'ri foydalanib hisoblash juda murakkab, shuning uchun biz ikki karrali integrallarni amaliy hisoblash yo'llarini ko'rib chiqishimiz kerak. Dastlab integrallash sohasini to'g'ri to'rtburchak deb hisoblashda davom etamiz:
\[R = \left[ {a,b} \right] \times \left[ {c,d} \right]\]Keyingi bo'limda esa ixtiyoriy shakldagi sohalarni ko'rib chiqamiz.
Quyidagi teorema to'g'ri to'rtburchak soha bo'yicha ikki karrali integralni qanday hisoblash mumkinligini ko'rsatadi.
Fubini teoremasi
Agar \(f\left( {x,y} \right)\) funksiya \(R = \left[ {a,b} \right] \times \left[ {c,d} \right]\) sohada uzluksiz bo'lsa, u holda:
\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \int_{{\,a}}^{{\,b}}{{\int_{{\,c}}^{{\,d}}{{f\left( {x,y} \right)\,dy}}\,dx}} = \int_{{\,c}}^{{\,d}}{{\int_{{\,a}}^{{\,b}}{{f\left( {x,y} \right)\,dx}}\,dy}}\]Ushbu integrallar takroriy integrallar deb ataladi.
E'tibor bering, to'g'ri to'rtburchak soha bo'yicha ikki karrali integralni hisoblashning ikkita teng kuchli tartibi mavjud. Ichki differensial ichki integralning integrallash chegaralariga mos keladi, tashqi differensial esa tashqi integralning chegaralariga mos keladi. Boshqacha qilib aytganda, agar ichki differensial \(dy\) bo'lsa, ichki integralning integrallash chegaralari \(y\) ga tegishli bo'ladi; agar tashqi differensial \(dx\) bo'lsa, tashqi integral chegaralari \(x\) ga tegishli bo'ladi.
Keling, birinchi tartibdan foydalanib, ifodani biroz aniqlashtirib qavslar bilan yozaylik:
\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \int_{{\,a}}^{{\,b}}{{\left[ {\int_{{\,c}}^{{\,d}}{{f\left( {x,y} \right)\,dy}}} \right]\,dx}}\]Biz ikki karrali integralni avval ichki integralni hisoblash orqali topamiz:
\[\int_{{\,c}}^{{\,d}}{{f\left( {x,y} \right)\,dy}}\]Bunda biz \(x\) ni o'zgarmas (konstanta) deb hisoblaymiz va \(y\) ga nisbatan odatiy bir o'zgaruvchili integral kabi integrallaymiz. Bu esa natijada faqat \(x\) ga bog'liq bo'lgan funksiyani beradi va uni o'z navbatida tashqi integralda \(x\) bo'yicha integrallaymiz.
Xuddi shunday jarayonni biz xususiy hosilalarni hisoblashda ham amalga oshirgan edik. Funksiyadan \(y\) bo'yicha xususiy hosila olishda biz \(x\) ni o'zgarmas deb qarab, \(y\) ga nisbatan bir o'zgaruvchili funksiya kabi differensiallagan edik.
Ikki karrali integrallar ham aynan shu prinsip asosida hisoblanadi: biz barcha \(x\) larni o'zgarmas deb qarab, \(y\) ga nisbatan integrallaymiz yoki barcha \(y\) larni o'zgarmas deb qarab, \(x\) ga nisbatan integrallaymiz.
Keling, bir nechta misollarni ko'rib chiqaylik.
- \( \displaystyle \iint\limits_{R}{{6x{y^2}\,dA}}\), \(R = \left[ {2,4} \right] \times \left[ {1,2} \right]\)
- \( \displaystyle \iint\limits_{R}{{2x - 4{y^3}\,dA}}\), \(R = \left[ { - 5,4} \right] \times \left[ {0,3} \right]\)
- \( \displaystyle \iint\limits_{R}{{{x^2}{y^2} + \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right)\,dA}}\), \(R = \left[ { - 2, - 1} \right] \times \left[ {0,1} \right]\)
- \( \displaystyle \iint\limits_{R}{{\frac{1}{{{{\left( {2x + 3y} \right)}^2}}}\,dA}}\), \(R = \left[ {0,1} \right] \times \left[ {1,2} \right]\)
- \( \displaystyle \iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}}\), \(R = \left[ { - 1,2} \right] \times \left[ {0,1} \right]\)
a \( \displaystyle \iint\limits_{R}{{6x{y^2}\,dA}}\), \(R = \left[ {2,4} \right] \times \left[ {1,2} \right]\) Yechim
Qaysi o'zgaruvchi bo'yicha birinchi integrallashimizning ahamiyati yo'q, integrallash tartibidan qat'i nazar bir xil natija hosil bo'ladi. Bunga ishonch hosil qilish uchun ikkala tartibda ham hisoblab chiqaylik.
1-usul:
Bu holda biz avval \(y\) ga nisbatan integrallaymiz. Hisoblanishi kerak bo'lgan takroriy integral quyidagicha bo'ladi:
Integrallash chegaralarini qo'yishda differensiallarga to'liq mos kelishiga e'tibor bering. \(dy\) ichki differensial bo'lgani uchun, ichki integral \(y\) ning chegaralariga (\(1\) dan \(2\) gacha) ega bo'ladi.
Buni hisoblash uchun avval ichki integralni hisoblaymiz, tashqi integralni esa o'zgarishsiz qoldiramiz:
\[\begin{align*}\iint\limits_{R}{{6x{y^2}\,dA}} & = \int_{{\,2}}^{{\,4}}{{\left. {\left( {2x{y^3}} \right)} \right|_1^2\,dx}}\\ & = \int_{{\,2}}^{{\,4}}{{16x - 2x\,dx}}\\ & = \int_{{\,2}}^{{\,4}}{{14x\,dx}}\end{alignScroll*}\]Yodingizda bo'lsin, birinchi integralni hisoblayotganda \(x\) o'zgarmas deb qaraladi. Endi hosil bo'lgan oddiy bir o'zgaruvchili integralni hisoblaymiz:
\[\iint\limits_{R}{{6x{y^2}\,dA}} = \left. {7{x^2}} \right|_2^4 = 84\]2-usul:
Bu holda biz avval \(x\) bo'yicha, so'ngra \(y\) bo'yicha integrallaymiz:
Albatta, birinchi usul bilan bir xil natija hosil bo'ldi.
Demak, to'g'ri to'rtburchak sohada integrallashni istalgan tartibda bajarish mumkin.
b \( \displaystyle \iint\limits_{R}{{2x - 4{y^3}\,dA}}\), \(R = \left[ { - 5,4} \right] \times \left[ {0,3} \right]\) Yechim
Ushbu integralni avval \(y\) ga nisbatan integrallaymiz:
\[\begin{align*}\iint\limits_{R}{{2x - 4{y^3}\,dA}} & = \int_{{\, - 5}}^{{\,4}}{{\int_{{\,0}}^{{\,3}}{{2x - 4{y^3}\,dy}}\,dx}}\\ & = \int_{{\, - 5}}^{{\,4}}{{\left. {\left( {2xy - {y^4}} \right)} \right|_0^3\,dx}}\\ & = \int_{{\, - 5}}^{{\,4}}{{6x - 81\,dx}}\\ & = \left. {\left( {3{x^2} - 81x} \right)} \right|_{ - 5}^4\\ & = - 756\end{align*}\]Yodingizda bo'lsin, \(y\) ga nisbatan integrallashda barcha \(x\) lar o'zgarmas (konstanta) deb qaraladi. Ichki integral nuqtayi nazaridan \(2x\) ifodasi ham o'zgarmasdir va biz bilamizki, o'zgarmasni \(y\) ga nisbatan integrallaganda shunchaki \(y\) qo'shiladi, natijada birinchi haddan \(2xy\) hosil bo'ladi.
c \( \displaystyle \iint\limits_{R}{{{x^2}{y^2} + \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right)\,dA}}\), \(R = \left[ { - 2, - 1} \right] \times \left[ {0,1} \right]\) Yechim
Bu holda avval \(x\) bo'yicha integrallaymiz:
\[\begin{align*}\iint\limits_{R}{{{x^2}{y^2} + \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right)\,dA}} & = \int_{{\,0}}^{{\,1}}{{\int_{{\, - 2}}^{{\, - 1}}{{{x^2}{y^2} + \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right)\,dx}}\,dy}}\\ & = \int_{{\,0}}^{{\,1}}{{\left. {\left( {\frac{1}{3}{x^3}{y^2} + \frac{1}{\pi }\sin \left( {\pi x} \right) + x\sin \left( {\pi y} \right)} \right)} \right|_{ - 2}^{ - 1}\,dy}}\\ & = \int_{{\,0}}^{{\,1}}{{\frac{7}{3}{y^2} + \sin \left( {\pi y} \right)\,dy}}\\ & = \left. {\frac{7}{9}{y^3} - \frac{1}{\pi }\cos \left( {\pi y} \right)} \right|_0^1\\ & = \frac{7}{9} + \frac{2}{\pi }\end{align*}\]Calculus I kursidagi asosiy o'zgaruvchini almashtirish qoidasini unutmang!
d \( \displaystyle \iint\limits_{R}{{\frac{1}{{{{\left( {2x + 3y} \right)}^2}}}\,dA}}\), \(R = \left[ {0,1} \right] \times \left[ {1,2} \right]\) Yechim
\(x\) bo'yicha chegaralar juda qulay (nol va bir) bo'lganligi sababli, avval \(x\) bo'yicha integrallaymiz. Integrallashni osonlashtirish uchun integral ostidagi ifodani daraja ko'rinishida yozib olamiz:
\[\begin{align*}\iint\limits_{R}{{{{\left( {2x + 3y} \right)}^{ - 2}}\,dA}} & = \int_{{\,1}}^{{\,2}}{{\int_{{\,0}}^{{\,1}}{{{{\left( {2x + 3y} \right)}^{ - 2}}\,dx}}\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{\left. {\left( { - \frac{1}{2}{{\left( {2x + 3y} \right)}^{ - 1}}} \right)} \right|_0^1\,dy}}\\ & = - \frac{1}{2}\int_{{\,1}}^{{\,2}}{{\frac{1}{{2 + 3y}} - \frac{1}{{3y}}\,dy}}\\ & = - \left. {\frac{1}{2}\left( {\frac{1}{3}\ln \left| {2 + 3y} \right| - \frac{1}{3}\ln \left| y \right|} \right)} \right|_1^2\\ & = - \frac{1}{6}\left( {\ln 8 - \ln 2 - \ln 5} \right)\end{align*}\]e \( \displaystyle \iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}}\), \(R = \left[ { - 1,2} \right] \times \left[ {0,1} \right]\) Yechim
Ikkala o'zgaruvchi bo'yicha ham integrallash mumkin bo'lsa-da, ba'zan bir tartib ikkinchisiga qaraganda ancha oson kechadi. Ushbu misolda avval \(y\) bo'yicha integrallash ancha yengil:
\[\iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}} = \int_{{ - 1}}^{2}{{\int_{0}^{1}{{x{{\bf{e}}^{xy}}\,dy}}\,dx}}\]\(y\) bo'yicha integralni oddiy almashtirish orqali hisoblash mumkin:
\[u = xy\hspace{0.5in}du = x\,dy\]Bu quyidagini beradi:
\[\begin{align*}\iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}} & = \int_{{ - 1}}^{2}{{\left. {{{\bf{e}}^{xy}}} \right|_0^1\,dx}}\\ & = \int_{{ - 1}}^{2}{{{{\bf{e}}^x} - 1\,dx}}\\ & = \left. {\left( {{{\bf{e}}^x} - x} \right)} \right|_{ - 1}^2\\ & = {{\bf{e}}^2} - 2 - \left( {{{\bf{e}}^{ - 1}} + 1} \right)\\ & = {{\bf{e}}^2} - {{\bf{e}}^{ - 1}} - 3\end{align*}\]Ko'rib turganingizdek, to'g'ri almashtirish tanlansa, integral juda oddiy yechiladi. Endi avval \(x\) bo'yicha integrallagan taqdirimizda nima bo'lishini ko'rib chiqaylik:
\[\iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}} = \int_{0}^{1}{{\int_{{ - 1}}^{2}{{x{{\bf{e}}^{xy}}\,dx}}\,dy}}\]Buni hisoblash uchun biz bo'laklab integrallash formulasidan foydalanishimiz kerak bo'ladi:
\[\begin{align*}u & = x & \hspace{0.25in} dv & = {{\bf{e}}^{xy}}\,dx\\ du & = dx & \hspace{0.25in} v &= \frac{1}{y}{{\bf{e}}^{xy}}\end{align*}\]U holda integral quyidagi ko'rinishga keladi:
\[\begin{align*}\iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}} & = \int_{0}^{1}{{\left. {\left( {\frac{x}{y}{{\bf{e}}^{xy}} - \int_{{}}^{{}}{{\frac{1}{y}{{\bf{e}}^{xy}}\,dx}}} \right)} \right|_{ - 1}^2\,dy}}\\ & = \int_{0}^{1}{{\left. {\left( {\frac{x}{y}{{\bf{e}}^{xy}} - \frac{1}{{{y^2}}}{{\bf{e}}^{xy}}} \right)} \right|_{ - 1}^2\,dy}}\\ & = \int_{0}^{1}{{\left( {\frac{2}{y}{{\bf{e}}^{2y}} - \frac{1}{{{y^2}}}{{\bf{e}}^{2y}}} \right) - \left( { - \frac{1}{y}{{\bf{e}}^{ - y}} - \frac{1}{{{y^2}}}{{\bf{e}}^{ - y}}} \right)\,dy}}\end{align*}\]Biz buni oxirigacha yetkazishni ham xohlamaymiz, chunki hosil bo'lgan integralni hisoblash nihoyatda og'ir (agar imkoni bo'lsa ham). Demak, integrallash tartibini to'g'ri tanlash juda muhim!
Oldingi misollar to'plamida ko'rganimizdek, integralni ikkala yo'nalishda ham hisoblash mumkin. Biroq ba'zida integrallashning bir tartibi ikkinchisidan ancha oson bo'ladi, shuning uchun hisoblashga kirishishdan oldin qaysi o'zgaruvchi bo'yicha birinchi integrallash qulayroq ekanligi haqida o'ylab ko'rish zarur.
Ushbu bo'limning keyingi mavzusi — ayrim takroriy integrallarni ancha oson hisoblash imkonini beruvchi qulay faktdir.
Bunday integrallarning ajoyib xususiy holati mavjud. Dastlab \(f\left( {x,y} \right) = g\left( x \right)h\left( y \right)\) deb faraz qilaylik va integrallash sohasi \(R = \left[ {a,b} \right] \times \left[ {c,d} \right]\) to'g'ri to'rtburchak bo'lsin. U holda integral quyidagicha yoziladi:
\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{R}{{g\left( x \right)h\left( y \right)\,dA}} = \int_{c}^{d}{{\int_{a}^{b}{{g\left( x \right)h\left( y \right)\,dx}}\,dy}}\]E'tibor bering, bu holda qaysi o'zgaruvchi bo'yicha birinchi integrallashimiz muhim emas, ikkala tartib ham bir xil mehnat bilan bir xil natijaga olib keladi.
Endi e'tibor bering: ichki integral \(x\) ga nisbatan bo'lib, \(h\left( y \right)\) faqat \(y\) ning funksiyasi bo'lganligi sababli, \(x\) bo'yicha integrallash jarayonida uni "o'zgarmas" deb hisoblash mumkin (\(x\) ning o'zgarishi \(y\) ning qiymatiga mutlaqo ta'sir qilmaydi!). Shuningdek, u \(g(x)\) ga ko'paytirilganligi sababli, \(h\left( y \right)\) ni ichki integraldan tashqariga chiqarish mumkin:
\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{R}{{g\left( x \right)h\left( y \right)\,dA}} = \int_{c}^{d}{{h\left( y \right)\left[ \int_{a}^{b}{{g\left( x \right)\,dx}} \right]\,dy}}\]Endi \(\int_{a}^{b}{{g\left( x \right)\,dx}}\) ifodasi Calculus I kursidan ma'lum bo'lgan standart aniq integraldir va uning qiymati oddiy sondan (o'zgarmasdan) iborat. Shuning uchun uni tashqi \(y\) integraldan ham tashqariga chiqarish mumkin:
\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{R}{{g\left( x \right)h\left( y \right)\,dA}} = \left( \int_{a}^{b}{{g\left( x \right)\,dx}} \right) \left( \int_{c}^{d}{{h\left( y \right)\,dy}} \right)\]Boshqacha qilib aytganda, agar biz ikki o'zgaruvchili funksiyani faqat \(x\) ga bog'liq funksiya bilan faqat \(y\) ga bog'liq funksiyaning ko'paytmasiga ajrata olsak, ikkita bir o'zgaruvchili integralni alohida hisoblab, ularni bir-biriga ko'paytirish kifoyadir.
Mana ushbu muhim qoidaning qisqacha bayoni:
Fakt
Agar \(f\left( {x,y} \right) = g\left( x \right)h\left( y \right)\) bo'lsa va biz \(R = \left[ {a,b} \right] \times \left[ {c,d} \right]\) to'g'ri to'rtburchak soha bo'yicha integrallayotgan bo'lsak, u holda:
\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{R}{{g\left( x \right)h\left( y \right)\,dA}} = \left( {\int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}} \right)\left( {\int_{{\,c}}^{{\,d}}{{h\left( y \right)\,dy}}} \right)\]Keling, ushbu qoidani qo'llashga doir misol ko'raylik.
Yechim
Integral ostidagi ifoda \(x\) ga bog'liq funksiya bilan \(y\) ga bog'liq funksiyaning ko'paytmasidan iborat bo'lganligi sababli, biz yuqoridagi faktdan foydalanishimiz mumkin:
\[\begin{align*}\iint\limits_{R}{{x{{\cos }^2}\left( y \right)\,dA}} & = \left( {\int_{{\, - 2}}^{{\,3}}{{x\,dx}}} \right)\left( {\int_{{\,0}}^{{\,\frac{\pi }{2}}}{{{{\cos }^2}\left( y \right)\,dy}}} \right)\\ & = \left. {\left( {\frac{1}{2}{x^2}} \right)} \right|_{ - 2}^3\left( {\frac{1}{2}\int_{{\,0}}^{{\,\frac{\pi }{2}}}{{1 + \cos \left( {2y} \right)\,dy}}} \right)\\ & = \left( {\frac{5}{2}} \right)\left( {\frac{1}{2}\left. {\left( {y + \frac{1}{2}\sin \left( {2y} \right)} \right)} \right|_0^{\frac{\pi }{2}}} \right)\\ & = \frac{{5\pi }}{8}\end{align*}\]Ushbu bo'limda muhokama qiladigan yana bir muhim mavzumiz bor. Bu mavzu takroriy integrallar bilan bevosita bog'liq bo'lmasa-da, hozirgi bosqichda buni tushuntirish juda o'rinlidir.
Biz ikki o'zgaruvchili funksiyadan bir o'zgaruvchi bo'yicha olinadigan aniqmas integrallarni ko'rib chiqmoqchimiz. Boshqacha aytganda, biz quyidagi ko'rinishdagi integrallarni o'rganamiz:
\[\begin{align*}& \int{{x{{\sec }^2}\left( {2y} \right) + 4xy\,dy}}\\ & \int{{{x^3} - {{\bf{e}}^{ - \frac{x}{y}}}\,dx}}\end{align*}\]Matematik analizdan bilamizki, aniqmas integrallar hosilasi integral ostidagi ifodaga teng bo'lgan boshlang'ich funksiyani topishni talab qiladi. Biroq, bu holda biz integraldagi differensialga (\(dy\) yoki \(dx\)) alohida e'tibor berishimiz kerak, chunki integrallash aynan qaysi o'zgaruvchi bo'yicha borayotganini u belgilaydi.
Birinchi integralda biz hosilasi integral ostidagi ifodaga teng bo'ladigan funksiyani \(y\) ga nisbatan qidiramiz, ikkinchi integralda esa \(x\) ga nisbatan qidiramiz. Asosiy e'tiborni integrallash o'zgarmasiga (doimiysiga) qanday yondashish kerakligiga qaratish lozim.
Integrallarning natijalari quyidagicha bo'ladi:
\[\begin{align*}\int{{x{{\sec }^2}\left( {2y} \right) + 4xy\,dy}} & = \frac{x}{2}\tan \left( {2y} \right) + 2x{y^2} + g\left( x \right)\\ \int{{{x^3} - {{\bf{e}}^{ - \frac{x}{y}}}\,dx}} & = \frac{1}{4}{x^4} + y\,{{\bf{e}}^{ - \frac{x}{y}}} + h\left( y \right)\end{align*}\]E'tibor bering, integrallash "doimiylari" endi qarama-qarshi o'zgaruvchining funksiyalaridan iboratdir. Birinchi integralda biz \(y\) ga nisbatan integrallayapmiz va bilamizki, faqat \(x\) ga bog'liq bo'lgan har qanday funksiyadan \(y\) bo'yicha olingan hosila nolga teng bo'ladi. Shu sababli, \(y\) ga nisbatan integrallashda funksiya tarkibida faqat \(x\) ga bog'liq had mavjud bo'lgan bo'lishi mumkinligini hisobga olishimiz kerak va shuning uchun integrallash "doimiysi" \(g(x)\) ko'rinishidagi \(x\) ning funksiyasi bo'ladi.
Xuddi shunday, ikkinchi integralda integrallash "doimiysi" \(h(y)\) ko'rinishidagi \(y\) ning funksiyasi bo'lishi kerak, chunki biz \(x\) ga nisbatan integrallayapmiz. Agar hosil bo'lgan natijadan \(x\) bo'yicha hosila olsak, faqat \(y\) ga bog'liq bo'lgan har qanday funksiyaning hosilasi nolga aylanadi.