5.7-bo'lim : Aniq integrallarni hisoblash
12. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[\int_{{\frac{\pi }{6}}}^{{\frac{\pi }{3}}}{{2{{\sec }^2}\left( w \right) - 8\csc \left( w \right)\cot \left( w \right)\,dw}}\]Yechim
Avval e'tibor bering, bu erda qandaydir "yashirin" ratsional ifoda mavjud bo'lsa ham (trig funktsiyalarining ta'riflarida) biz integrallashayotgan intervalda na kosinus, na sinus nolga teng va shuning uchun ikkala atama ham intervalda uzluksizdir.
Shuning uchun biz funktsiyani birlashtirishimiz kerak.
\[\int_{{\frac{\pi }{6}}}^{{\frac{\pi }{3}}}{{2{{\sec }^2}\left( w \right) - 8\csc \left( w \right)\cot \left( w \right)\,dw}} = \left. {\left( {2\tan \left( w \right) + 8\csc \left( w \right)} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{3}}\]Eslatib o'tamiz, biz aniq integral holatda "+c" ni qo'shishimiz shart emas, chunki u keyingi bosqichda bekor qilinadi.
2-qadam
Yakuniy bosqich - bu faqat baholashni amalga oshirishdir.
Tasdiqlash va faqat baholash natijalarini ko'rsatish uchun asosiy arifmetikani sizga qoldiramiz. Avval yuqori chegarani baholaganingizga ishonch hosil qiling va keyin pastki chegaradagi baholashni olib tashlang.
Mana bu muammoning javobi.
\[\int_{{\frac{\pi }{6}}}^{{\frac{\pi }{3}}}{{2{{\sec }^2}\left( w \right) - 8\csc \left( w \right)\cot \left( w \right)\,dw}} = \left( {\frac{{16}}{{\sqrt 3 }} + 2\sqrt 3 } \right) - \left( {16 + \frac{2}{{\sqrt 3 }}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{14}}{{\sqrt 3 }} + 2\sqrt 3 - 16}}\]