5.2-bo'lim : Noaniq integrallarni hisoblash
23. \(g'\left( z \right) = 3{z^3} + \frac{7}{{2\sqrt z }} - {{\bf{e}}^z}\) va \(g\left( 1 \right) = 15 - {\bf{e}}\) ekanligini hisobga olib, \(g\left( z \right)\) ni aniqlang.
Maslahat: Biz bilamizki, integratsiya oddiygina integralni olish uchun qanday funktsiyani farqlaganimizni so'raydi va shuning uchun biz funktsiyaning umumiy formulasiga kelish uchun bu fikrdan foydalanishimiz kerak.
Yechim
Ushbu bo'limdagi eslatmalardan biz ko'rganimizni eslang,
\[g\left( z \right) = \int{{g'\left( z \right)\,\,dz}}\]Shunday qilib, \(g\left( z \right)\) ning umumiy formulasiga erishish uchun biz qilishimiz kerak bo'lgan narsa muammo bayonotida berilgan hosilani birlashtirishdir.
\[g\left( z \right) = \int{{3{z^3} + \frac{7}{2}{z^{ - \,\frac{1}{2}}} - {{\bf{e}}^z}\,dz}} = \frac{3}{4}{z^4} + 7{z^{\,\frac{1}{2}}} - {{\bf{e}}^z} + c\]"+c" ni unutmang!
Maslahat: \(c\) integrasiya konstantasining qiymatini aniqlash uchun biz \(z = 1\) funksiyasining qiymatiga egamiz.
2-qadam
Bizda \(g\left( 1 \right) = 15 - {\bf{e}}\) sharti borligi uchun biz avvalgi qadamdagi javobimizga \(z = 1\) ni ulab, natijani \(15 – \bf{e}\) ga tenglashtirib, \(c\) uchun hosil bo‘lgan tenglamani yechishimiz mumkin.
Shunday qilib,
\[15 - {\bf{e}} = g\left( 1 \right) = \frac{{31}}{4} - {\bf{e}} + c\hspace{0.25in} \Rightarrow \hspace{0.5in}c = \frac{{29}}{4}\]Keyin funktsiya,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{g\left( z \right) = \frac{3}{4}{z^4} + 7{z^{\,\frac{1}{2}}} - {{\bf{e}}^z} + \frac{{29}}{4}}}\]