5.4-bo'lim : Ko'proq almashtirish qoidasi
9. \( \displaystyle \int{{7\left( {3y + 2} \right){{\left( {4y + 3{y^2}} \right)}^3} + \sin \left( {3 + 8y} \right)\,dy}}\) ni baholang.
Yechim
Shubhasiz, har bir atama alohida almashtirishni talab qiladi. Shunday qilib, biz birinchi navbatda integralni quyidagicha ajratishimiz kerak.
\[\int{{7\left( {3y + 2} \right){{\left( {4y + 3{y^2}} \right)}^3} + \sin \left( {3 + 8y} \right)\,dy}} = \int{{7\left( {3y + 2} \right){{\left( {4y + 3{y^2}} \right)}^3}\,dy}} + \int{{\sin \left( {3 + 8y} \right)\,dy}}\]2-qadam
Har bir integral uchun zarur almashtirishlar,
\[u = 4y + 3{y^2}\hspace{0.75in}v = 3 + 8y\]Agar siz asosiy almashtirish mexanikasidan mamnun bo'lmasangiz, oldingi bo'limda ba'zi muammolar bilan shug'ullanishingiz kerak, chunki biz ushbu bo'limda asoslar haqida batafsil ma'lumot bermaymiz. Ushbu bo'limdagi muammolar almashtirishning asosiy mexanikasi bilan ancha qulay bo'lganlar uchun mo'ljallangan va bir nechta "ilg'or" almashtirishlarni o'z ichiga oladi.
3-qadam
Bu erda har bir almashtirish uchun differentsial ish.
\[\begin{align*}du & = \left( {4 + 6y} \right)dy = 2\left( {3y + 2} \right)dy\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\,\left( {3y + 2} \right)dy = \frac{1}{2}du\\ dv & = 8\,dy\hspace{0.5in} \to \hspace{0.25in}\,dy = \frac{1}{8}dv\end{align*}\]Endi almashtirishlarni bajarish va integrallarni baholash,
\[\begin{align*}\int{{7\left( {3y + 2} \right){{\left( {4y + 3{y^2}} \right)}^3} + \sin \left( {3 + 8y} \right)\,dy}} & = \frac{7}{2}\int{{{u^3}du}} + \frac{1}{8}\int{{\sin \left( v \right)\,dv}} = \frac{7}{8}{u^4} - \frac{1}{8}\cos \left( v \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{7}{8}{{\left( {4y + 3{y^2}} \right)}^4} - \frac{1}{8}\cos \left( {3 + 8y} \right) + c}}\end{align*}\]Integralni baholagandan so'ng asl o'zgaruvchiga qaytishni unutmang!