5.7-bo'lim : Aniq integrallarni hisoblash
15. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\(\displaystyle \int_{0}^{4}{{f\left( t \right)\,dt}}\), bu erda \(f\left( t \right) = \left\{ {\begin{array}{*{20}{c}}{2t}&{t > 1}\\{1 - 3{t^2}}&{t \le 1}\end{array}} \right.\)“\(c\)” 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 \(t = 1\) funksiyasining aniq o‘zgarishini beradi. Biroq, esda tutingki, biz har doim istalgan nuqtada integralni buzishimiz mumkin va \(t = 1\) buni amalga oshirish uchun yaxshi nuqta bo'lib tuyuladi.
\(t = 1\) da integralni ajratish,
\[\int_{0}^{4}{{f\left( t \right)\,dt}} = \int_{0}^{1}{{f\left( t \right)\,dt}} + \int_{1}^{4}{{f\left( t \right)\,dt}}\]Shunday qilib, birinchi integralda bizda \(0 \le t \le 1\) bor va shuning uchun biz birinchi integralda \(f\left( t \right) = 1 - 3{t^2}\) dan foydalanishimiz mumkin. Xuddi shunday, ikkinchi integralda bizda \(1 \le t \le 4\) bor va shuning uchun biz ikkinchi integralda \(f\left( t \right) = 2t\) dan foydalanishimiz mumkin.
Ushbu funktsiyani almashtirishni amalga oshirish,
\[\int_{0}^{4}{{f\left( t \right)\,dt}} = \int_{0}^{1}{{1 - 3{t^2}\,dt}} + \int_{1}^{4}{{2t\,dt}}\]2-qadam
Bu nuqtada qilishimiz kerak bo'lgan narsa - har bir integralni baholash. Mana o'sha ish.
\[\int_{0}^{4}{{f\left( t \right)\,dt}} = \int_{0}^{1}{{1 - 3{t^2}\,dt}} + \int_{1}^{4}{{2t\,dt}} = \left. {\left( {t - {t^3}} \right)} \right|_0^1 + \left. {{t^2}} \right|_1^4 = \left[ {0 - 0} \right] + \left[ {16 - 1} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{15}}\]