5.8-bo'lim : Aniq integrallarni almashtirish qoidasi
10. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[ \int_{{ - 2}}^{1}{{{{\left( {2 - z} \right)}^3} + \sin \left( {\pi z} \right){{\left[ {3 + 2\cos \left( {\pi z} \right)} \right]}^3}\,dz}}\]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.
O'zgartirishni o'rnatishdan oldin biz integralni buzishimiz kerak, chunki har bir atama boshqa almashtirishni talab qiladi. Shunday qilib,
\[ \int_{{ - 2}}^{1}{{{{\left( {2 - z} \right)}^3} + \sin \left( {\pi z} \right){{\left[ {3 + 2\cos \left( {\pi z} \right)} \right]}^3}\,dz}} = \int_{{ - 2}}^{1}{{{{\left( {2 - z} \right)}^3}dz}} + \int_{{ - 2}}^{1}{{\sin \left( {\pi z} \right){{\left[ {3 + 2\cos \left( {\pi z} \right)} \right]}^3}\,dz}}\]Har bir integral uchun almashtirish quyidagicha bo'ladi:
\[u = 2 - z\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}v = 3 + 2\cos \left( {\pi z} \right)\]2-qadam
Mana bu birinchi integral uchun haqiqiy almashtirish ishi.
\[\begin{align*}du & = - dz\hspace{0.25in}\,\, \to \hspace{0.25in}dz = - du\\ z & = - 2:u = 4\hspace{0.75in}z = 1:u = 1\end{align*}\]Mana ikkinchi integral uchun haqiqiy almashtirish ishi.
\[\begin{align*}dv & = - 2\pi \sin \left( {\pi z} \right)\,dz\hspace{0.25in}\,\, \to \hspace{0.25in}\sin \left( {\pi z} \right)dz = - \frac{1}{{2\pi }}dv\\ z & = - 2:v = 5\hspace{0.75in}z = 1:v = 1\end{align*}\]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.
\[ \int_{{ - 2}}^{1}{{{{\left( {2 - z} \right)}^3} + \sin \left( {\pi z} \right){{\left[ {3 + 2\cos \left( {\pi z} \right)} \right]}^3}\,dz}} = - \int_{4}^{1}{{{u^3}du}} - \frac{1}{{2\pi }}\int_{5}^{1}{{{v^3}\,dv}}\]3-qadam
U holda integral,
\[\begin{align*}\int_{{ - 2}}^{1}{{{{\left( {2 - z} \right)}^3} + \sin \left( {\pi z} \right){{\left[ {3 + 2\cos \left( {\pi z} \right)} \right]}^3}\,dz}} & = \left. { - \frac{1}{4}{u^4}} \right|_4^1 - \left. {\frac{1}{{8\pi }}{v^4}} \right|_5^1\\ & = - \frac{1}{4}\left( {1 - 256} \right) - \frac{1}{{8\pi }}\left( {1 - 625} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{255}}{4} + \frac{{78}}{\pi }}}\end{align*}\]