5.8-bo'lim : Aniq integrallarni almashtirish qoidasi
4. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[ \int_{1}^{4}{{\sqrt w \,{{\bf{e}}^{1 - \sqrt {{w^{\,3}}} }}\,dw}}\]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,
\[u = 1 - {w^{\,\frac{3}{2}}}\]2-qadam
Mana bu muammo uchun haqiqiy almashtirish ishi.
\[\begin{align*}du & = - \frac{3}{2}{w^{\,\frac{1}{2}}}dw\hspace{0.25in}\,\, \to \hspace{0.25in}\sqrt w \,dw = - \frac{2}{3}du\\ w & = 1:u = 0\hspace{0.75in}w = 4:u = - 7\end{align*}\]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.
\[ \int_{1}^{4}{{\sqrt w \,{{\bf{e}}^{1 - \sqrt {{w^{\,3}}} }}\,dw}} = - \frac{2}{3}\int_{0}^{{ - 7}}{{{{\bf{e}}^u}\,du}}\]3-qadam
U holda integral,
\[ \int_{1}^{4}{{\sqrt w \,{{\bf{e}}^{1 - \sqrt {{w^{\,3}}} }}\,dw}} = \left. { - \frac{2}{3}{{\bf{e}}^u}} \right|_0^{ - 7} = - \frac{2}{3}{{\bf{e}}^{ - 7}} - \left( { - \frac{2}{3}{{\bf{e}}^0}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{2}{3}\left( {1 - {{\bf{e}}^{ - 7}}} \right)}}\]