5.7-bo'lim : Aniq integrallarni hisoblash
16. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\(\displaystyle \int_{{ - 6}}^{1}{{g\left( z \right)\,dz}}\), bu erda \(g\left( z \right) = \left\{ {\begin{array}{*{20}{c}}{2 - z}&{z > - 2}\\{4{{\bf{e}}^z}}&{z \le - 2}\end{array}} \right.\)“\(c\)” uchun ushbu integralni amalga oshirish mumkin bo'lgan yaxshi tanlovni topa olasizmi yoki yo'qligini ko'ring.
Yechim
Bu integralni bitta integral sifatida bajarish mumkin emas, biz integrallashayotgan oraliqda joylashgan \(z = - 2\) funksiyasining aniq o‘zgarishini beradi. Biroq, esda tutingki, biz har doim istalgan nuqtada integralni buzishimiz mumkin va \(z = - 2\) buni amalga oshirish uchun yaxshi nuqta bo'lib tuyuladi.
\(z = - 2\) da integralni ajratish,
\[\int_{{ - 6}}^{1}{{g\left( z \right)\,dz}} = \int_{{ - 6}}^{{ - 2}}{{g\left( z \right)\,dz}} + \int_{{ - 2}}^{1}{{g\left( z \right)\,dz}}\]Shunday qilib, birinchi integralda bizda \( - 6 \le z \le - 2\) bor va shuning uchun biz birinchi integralda \(g\left( z \right) = 4{{\bf{e}}^z}\) dan foydalanishimiz mumkin. Xuddi shunday, ikkinchi integralda bizda \( - 2 \le z \le 1\) bor va shuning uchun biz ikkinchi integralda \(g\left( z \right) = 2 - z\) dan foydalanishimiz mumkin.
Ushbu funktsiyani almashtirishni amalga oshirish,
\[\int_{{ - 6}}^{1}{{g\left( z \right)\,dz}} = \int_{{ - 6}}^{{ - 2}}{{4{{\bf{e}}^z}\,dz}} + \int_{{ - 2}}^{1}{{2 - z\,dz}}\]2-qadam
Bu nuqtada qilishimiz kerak bo'lgan narsa - har bir integralni baholash. Mana o'sha ish.
\[\begin{align*}\int_{{ - 6}}^{1}{{g\left( z \right)\,dz}} & = \int_{{ - 6}}^{{ - 2}}{{4{{\bf{e}}^z}\,dz}} + \int_{{ - 2}}^{1}{{2 - z\,dz}} = \left. {\left( {4{{\bf{e}}^z}} \right)} \right|_{ - 6}^{ - 2} + \left. {\left( {2z - \frac{1}{2}{z^2}} \right)} \right|_{ - 2}^1\\ & = \left[ {4{{\bf{e}}^{ - 2}} - 4{{\bf{e}}^{ - 6}}} \right] + \left[ {\frac{3}{2} - \left( { - 6} \right)} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{4{{\bf{e}}^{ - 2}} - 4{{\bf{e}}^{ - 6}} + \frac{{15}}{2}}}\end{align*}\]