7.8-bo'lim : Noto'g'ri integrallar
10. Quyidagi integral yaqinlashish yoki ayrilish kerakligini aniqlang. Agar integral yaqinlashsa, uning qiymatini aniqlang.
\[\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\]Yechim
Endi bu erda cheksiz chegara bor, lekin \(x = 0\)-da biz hal qilishimiz kerak bo'lgan uzilishlar mavjudligiga e'tibor bering.
Izohlardagi materiallarga asoslanib, ikkala integral birlashishi sharti bilan, biz istalgan nuqtada integralni bo'lishimiz kerakligini tushunishimiz kerak. Bunday holda, keling, integralni \(x = - 1\) da ajratamiz. Shunday qilib,
\[\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = \int_{{ - \infty }}^{{ - 1}}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} + \int_{{ - 1}}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\]Shuni yodda tutingki, integralni shunday bo'lish faqat ikkala integral yaqinlashganda amalga oshirilishi mumkin! Agar ulardan bittasi ham divergent ekanligi aniqlansa, biz buni qila olmaganmiz va asl integral divergent bo'ladi.
Shunday qilib, buni amalga oshirish haqiqatan ham mumkinmi yoki yo'qmi, deb tashvishlanmang, keling, muammoni hal qilaylik.
Endi biz muammolarni quyidagicha bartaraf etishimiz mumkin,
\[\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = \mathop {\lim }\limits_{t \to \,\, - \infty } \int_{t}^{{ - 1}}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} + \mathop {\lim }\limits_{s \to {0^ - }} \int_{{ - 1}}^{s}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\]2-qadam
Keyinchalik, integralni bajaramiz. Biz integratsiya jarayoniga ko'p tushuntirishlar/tafsilotlar kiritmaymiz. Shu nuqtaga kelib, sizning integratsiya qobiliyatingiz juda yaxshi bo'ladi deb taxmin qilinadi. Agar integratsiya qobiliyatingiz biroz zanglagan deb topsangiz, orqaga qaytib, oldingi tegishli bo'limlardan ba'zi mashqlarni bajarishingiz kerak.
Bunday holda biz oddiy Calc I almashtirishni amalga oshirishimiz mumkin. Mana integratsiya ishi.
\[\int{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = - {{\bf{e}}^{\frac{1}{x}}} + c\]E'tibor bering, biz bu erda aniq integral qilmadik. Limitlar bizning integralni qanday bajarishimizga ta'sir qilmaydi va har biri uchun integral bir xil edi, faqat chegaralar boshqacha edi, shuning uchun integralni ikki marta bajarish uchun hech qanday sabab yo'q.
3-qadam
OK, endi integralning chegaralari haqida qayg'uraylik.
\[\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = \mathop {\lim }\limits_{t \to \,\, - \infty } \left. {\left( { - {{\bf{e}}^{\frac{1}{x}}}} \right)} \right|_t^{ - 1} + \mathop {\lim }\limits_{s \to {0^ - }} \left. {\left( { - {{\bf{e}}^{\frac{1}{x}}}} \right)} \right|_{ - 1}^s = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - {{\bf{e}}^{ - 1}} + {{\bf{e}}^{\frac{1}{t}}}} \right) + \mathop {\lim }\limits_{s \to {0^ - }} \left( { - {{\bf{e}}^{\frac{1}{s}}} + {{\bf{e}}^{ - 1}}} \right)\]4-qadam
Endi biz oldingi bosqichdagi javobimizdagi chegaralarni baholashimiz kerak. Mana ish chegarasi
\[\begin{align*}\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - {{\bf{e}}^{ - 1}} + {{\bf{e}}^{\frac{1}{t}}}} \right) + \mathop {\lim }\limits_{s \to {0^ - }} \left( { - {{\bf{e}}^{\frac{1}{s}}} + {{\bf{e}}^{ - 1}}} \right)\\ & = \,\,\,\,\,\,\,\,\,\left[ { - {{\bf{e}}^{ - 1}} + {{\bf{e}}^0}} \right]\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {0 + {{\bf{e}}^{ - 1}}} \right]\end{align*}\]E'tibor bering,
\[\mathop {\lim }\limits_{s \to {0^ - }} \frac{1}{s} = - \infty \]chunki biz chap qo'l chegarasini qilamiz va shuning uchun \(s\) salbiy bo'ladi. Bu o'z navbatida shuni anglatadiki,
\[\mathop {\lim }\limits_{s \to {0^ - }} \left( { - {{\bf{e}}^{\frac{1}{s}}}} \right) = 0\]5-qadam
Yakuniy qadam - javobni yozish!
Endi oldingi bosqichdagi ish chegarasidan shuni ko'ramizki,
\[\int_{{ - \infty }}^{{ - 1}}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = - {{\bf{e}}^{ - 1}} + 1\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\int_{{ - 1}}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = {{\bf{e}}^{ - 1}}\]Shuning uchun integrallarning har biri yaqinlashuvchi va yuqorida ko'rsatilgan qiymatlarga ega. Bu shuni anglatadiki, biz 1-bosqichda bo'lgani kabi integralni ham parchalashimiz mumkin. Bundan tashqari, asl integral endi,
\[\begin{align*}\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} & = \int_{{ - \infty }}^{{ - 1}}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} + \int_{{ - 1}}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\\ & = \,\,\, - {{\bf{e}}^{ - 1}} + 1\,\,\, + \,\,\,\,\,\,\,{{\bf{e}}^{ - 1}}\\ & = 1\end{align*}\]Shuning uchun integral birlashadi va uning qiymati 1.