7.8-bo'lim : Noto'g'ri integrallar
9. Quyidagi integral yaqinlashadimi yoki ayiriladimi, aniqlang. Agar integral yaqinlashsa, uning qiymatini aniqlang.
\[\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}}\]Yechim
Birinchidan, \(x = 2\) da integranda nolga bo'linish (va shuning uchun uzilish) mavjudligiga e'tibor bering va bu integral chegaralari orasida ekanligini unutmang. Biz bilamizki, bu uzilish mavjud ekan, biz integralni bajara olmaymiz.
Biroq, ushbu bo'limdagi eslatmalardan biz faqat uzilishlar bilan shug'ullanishimiz mumkinligini eslaylik, agar ular integral chegaralaridan birida sodir bo'lsa. Shunday qilib, biz \(x = 2\) da integralni buzishimiz kerak.
\[\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}} = \int_{1}^{2}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} + \int_{2}^{4}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}}\]Shuni ham yodda tutingki, har ikkala yangi integral konvergent bo'lsa, biz faqat shunday integralni buzishimiz mumkin! Agar ulardan bittasi ham divergent ekanligi aniqlansa, biz buni qila olmaganmiz va asl integral divergent bo'ladi.
Shunday qilib, buni amalga oshirish mumkinmi yoki yo'qmi, deb tashvishlanmang, keling, muammoni hal qilaylik.
Har biridagi uzilishlarni quyidagicha bartaraf etishimiz mumkin,
\[\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}} = \mathop {\lim }\limits_{t \to \,{2^ - }} \int_{1}^{t}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} + \mathop {\lim }\limits_{s \to \,{2^ + }} \int_{s}^{4}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}}\]Shuni unutmangki, bu turdagi integrallarning chegaralari bir tomonlama chegaralar bo'lishi kerak.
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 integral bo'lish uchun ba'zi qisman kasrlarni bajarishimiz kerak bo'ladi. Mana qisman kasr ishi.
\[\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = \frac{A}{{x + 3}} + \frac{B}{{x - 2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,1 = A\left( {x - 2} \right) + B\left( {x + 3} \right)\] \[\begin{array}{ll}{x = 2:} & {1 = 5B}\\ & \\ {x = - 3:} & {1 = - 5A}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\begin{array}{l}{\displaystyle A = - \frac{1}{5}}\\{\displaystyle B = \frac{1}{5}}\end{array}\]Keyin integratsiya ishi,
\[\int{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} = \int{{\frac{{\frac{1}{5}}}{{x - 2}} - \frac{{\frac{1}{5}}}{{x + 3}}\,dx}} = \frac{1}{5}\ln \left| {x - 2} \right| - \frac{1}{5}\ln \left| {x + 3} \right| + c\]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.
\[\begin{align*}\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}} & = \mathop {\lim }\limits_{t \to \,{2^ - }} \left. {\left( {\frac{1}{5}\ln \left| {x - 2} \right| - \frac{1}{5}\ln \left| {x + 3} \right|} \right)} \right|_1^t + \mathop {\lim }\limits_{s \to \,{2^ + }} \left. {\left( {\frac{1}{5}\ln \left| {x - 2} \right| - \frac{1}{5}\ln \left| {x + 3} \right|} \right)} \right|_s^4\\ & = \mathop {\lim }\limits_{t \to \,{2^ - }} \left( {\frac{1}{5}\ln \left| {t - 2} \right| - \frac{1}{5}\ln \left| {t + 3} \right| - \left( {\frac{1}{5}\ln \left( 1 \right) - \frac{1}{5}\ln \left( 4 \right)} \right)} \right)\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \mathop {\lim }\limits_{s \to \,{2^ + }} \left( {\frac{1}{5}\ln \left( 2 \right) - \frac{1}{5}\ln \left( 7 \right) - \left( {\frac{1}{5}\ln \left| {s - 2} \right| - \frac{1}{5}\ln \left| {s + 3} \right|} \right)} \right)\end{align*}\]4-qadam
Endi biz oldingi bosqichdagi javobimizdagi chegaralarni baholashimiz kerak. Mana ish chegarasi.
\[\begin{align*}\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}} & = \mathop {\lim }\limits_{t \to \,{2^ - }} \left( {\frac{1}{5}\ln \left| {t - 2} \right| - \frac{1}{5}\ln \left| {t + 3} \right| - \left( {\frac{1}{5}\ln \left( 1 \right) - \frac{1}{5}\ln \left( 4 \right)} \right)} \right)\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \mathop {\lim }\limits_{s \to \,{2^ + }} \left( {\frac{1}{5}\ln \left( 2 \right) - \frac{1}{5}\ln \left( 7 \right) - \left( {\frac{1}{5}\ln \left| {s - 2} \right| - \frac{1}{5}\ln \left| {s + 3} \right|} \right)} \right)\\ & = \,\,\,\,\,\,\,\,\,\left[ { - \infty - \frac{1}{5}\ln \left( 5 \right) + \frac{1}{5}\ln \left( 4 \right)} \right]\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\left[ {\frac{1}{5}\ln \left( 2 \right) - \frac{1}{5}\ln \left( 7 \right) + \frac{1}{5}\ln \left( 5 \right) + \infty } \right]\end{align*}\]E'tibor bering, biz har bir chegara nima ekanligini tushunish uchun qavs ichiga javoblarni joylashtiramiz. Bu keyingi qadam uchun muhim bo'ladi.
5-qadam
Yakuniy qadam - javobni yozish!
Endi oldingi bosqichdagi ish chegarasidan shuni ko'ramizki,
\[\begin{align*}\int_{1}^{2}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} & = \mathop {\lim }\limits_{t \to \,{2^ - }} \left( {\frac{1}{5}\ln \left| {t - 2} \right| - \frac{1}{5}\ln \left| {t + 3} \right| - \left( {\frac{1}{5}\ln \left( 1 \right) - \frac{1}{5}\ln \left( 4 \right)} \right)} \right) = \, - \infty \\ & \\ \int_{2}^{4}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} & = \mathop {\lim }\limits_{s \to \,{2^ + }} \left( {\frac{1}{5}\ln \left( 2 \right) - \frac{1}{5}\ln \left( 7 \right) - \left( {\frac{1}{5}\ln \left| {s - 2} \right| - \frac{1}{5}\ln \left| {s + 3} \right|} \right)} \right) = \infty \end{align*}\]Shuning uchun bu integrallarning har biri bir-biridan farq qiladi. Bu shuni anglatadiki, biz 1-bosqichda bo'lgani kabi, integralni sindira olmadik.
Bu o'z navbatida integral degan ma'noni anglatadi farqlanadi.