7.1-bo'lim : Qismlar bo'yicha integratsiya
1. \( \displaystyle \int{{4x\cos \left( {2 - 3x} \right)\,dx}}\) ni baholang.
Yechim
Bu erda birinchi qadam \(u\) va \(dv\) ni tanlashdir. Biz \(u\) va \(dv\) ni tanlamoqchimiz, shunda biz \(du\) va \(v\) ni hisoblab, hamma narsani Qismlar bo‘yicha integratsiya formulasiga ulaganimizda, biz qo‘lga kiritadigan yangi integral qila olamiz.
Buni hisobga olsak, \(u\) va \(dv\) uchun quyidagi tanlovlar biz uchun ishlashi kerak.
\[u = 4x\hspace{0.5in}dv = \cos \left( {2 - 3x} \right)\,dx\]2-qadam
Keyinchalik, \(du\) (\(u\)ni farqlash orqali) va \(v\) (\(dv\) ni birlashtirish orqali) hisoblashimiz kerak.
\[\begin{align*}u & = 4x & \hspace{0.5in} & \to & \hspace{0.25in}du & = 4dx\\ dv & = \cos \left( {2 - 3x} \right)\,dx & \hspace{0.25in} & \to & \hspace{0.25in}v & = - \frac{1}{3}\sin \left( {2 - 3x} \right)\end{align*}\]3-qadam
\(u\), \(du\), \(v\) va \(dv\) ni qismlar bo‘yicha integratsiyaga ulash formulasini beradi:
\[\begin{align*}\int{{4x\cos \left( {2 - 3x} \right)\,dx}} & = \left( {4x} \right)\left( { - \frac{1}{3}\sin \left( {2 - 3x} \right)} \right) - \int{{ - \frac{4}{3}\sin \left( {2 - 3x} \right)\,dx}}\\ & = - \frac{4}{3}x\sin \left( {2 - 3x} \right) + \frac{4}{3}\int{{\sin \left( {2 - 3x} \right)\,dx}}\end{align*}\]4-qadam
Yaxshi, biz olgan yangi integral osonlik bilan bajariladi va shuning uchun bu muammoni hal qilish uchun biz qilishimiz kerak bo'lgan yagona narsa integralni bajarishdir.
\[ \int{{4x\cos \left( {2 - 3x} \right)\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{4}{3}x\sin \left( {2 - 3x} \right) + \frac{4}{9}\cos \left( {2 - 3x} \right) + c}}\]