5.8-bo'lim : Aniq integrallarni almashtirish qoidasi
5. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[ \int_{{ - 4}}^{{ - 1}}{{\sqrt[3]{{5 - 2y}} + \frac{7}{{5 - 2y}}\,dy}}\]Yechim
Biz qilishimiz kerak bo'lgan birinchi qadam - almashtirishni amalga oshirish.
Shu nuqtada siz almashtirishlar bilan juda qulay bo'lishingiz kerak. Agar almashtirishlar sizga mos kelmasa, siz almashtirish bo'limlariga qaytib, u erda ba'zi muammolarni hal qilishingiz kerak.
Ushbu muammoning o'rnini bosish,
\[u = 5 - 2y\]2-qadam
Mana bu muammo uchun haqiqiy almashtirish ishi.
\[\begin{align*}du & = - 2\,dy\hspace{0.25in}\,\, \to \hspace{0.25in}\,dy = - \frac{1}{2}du\\ y & = - 4:u = 13\hspace{0.75in}y = - 1:u = 7\end{align*}\]Ushbu bo'lim uchun eslatmalarda qilganimizdek, biz integralni bajargandan so'ng orqaga almashtirish bilan shug'ullanmaslik uchun chegaralarni \(u\) ga aylantiramiz.
Bu erda almashtirishdan keyingi integral.
\[ \int_{{ - 4}}^{{ - 1}}{{\sqrt[3]{{5 - 2y}} + \frac{7}{{5 - 2y}}\,dy}} = - \frac{1}{2}\int_{{13}}^{7}{{{u^{\frac{1}{3}}} + \frac{7}{u}\,du}}\]3-qadam
U holda integral,
\[\begin{align*}\int_{{ - 4}}^{{ - 1}}{{\sqrt[3]{{5 - 2y}} + \frac{7}{{5 - 2y}}\,dy}} & = \left. {\left( { - \frac{1}{2}\left[ {\frac{3}{4}{u^{\frac{4}{3}}} + 7\ln \left| u \right|} \right]} \right)} \right|_{13}^7\\ & = - \frac{3}{8}{7^{\frac{4}{3}}} - \frac{7}{2}\ln \left| 7 \right| - \left( { - \frac{3}{8}{{13}^{\frac{4}{3}}} - \frac{7}{2}\ln \left| {13} \right|} \right)\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{8}\left( {{{13}^{\frac{4}{3}}} - {7^{\frac{4}{3}}}} \right) + \frac{7}{2}\left( {\ln \left( {13} \right) - \ln \left( 7 \right)} \right)}}\end{align*}\]