5.8-bo'lim : Aniq integrallarni almashtirish qoidasi
9. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[ \int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}}\]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,
\[ \int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}} = \int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} \,dt}} + \int_{{ - 2}}^{0}{{\frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}}\]Har bir integral uchun almashtirish quyidagicha bo'ladi:
\[u = 3 + {t^2}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}v = 6t - 1\]2-qadam
Mana bu birinchi integral uchun haqiqiy almashtirish ishi.
\[\begin{align*}du & = 2t\,dt\hspace{0.25in}\,\, \to \hspace{0.25in}t\,dt = \frac{1}{2}du\\ t & = - 2:u = 7\hspace{0.75in}t = 0:u = 3\end{align*}\]Mana ikkinchi integral uchun haqiqiy almashtirish ishi.
\[\begin{align*}dv & = 6\,dt\hspace{0.25in}\,\, \to \hspace{0.25in}dt = \frac{1}{6}dv\\ t & = - 2:v = - 13\hspace{0.75in}t = 0:v = - 1\end{align*}\]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.
\[ \int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}} = \frac{1}{2}\int_{7}^{3}{{{u^{\frac{1}{2}}}\,du}} + \frac{3}{6}\int_{{ - 13}}^{{ - 1}}{{{v^{ - 2}}\,dv}}\]3-qadam
U holda integral,
\[\int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}} = \left. {\frac{1}{3}{u^{\frac{3}{2}}}} \right|_7^3 - \left. {\frac{1}{2}{v^{ - 1}}} \right|_{ - 13}^{ - 1} = \frac{1}{3}\left( {{3^{\frac{3}{2}}} - {7^{\frac{3}{2}}}} \right) - \frac{1}{2}\left( { - 1 - \left( { - \frac{1}{{13}}} \right)} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}\left( {{3^{\frac{3}{2}}} - {7^{\frac{3}{2}}}} \right) + \frac{6}{{13}}}}\]