5.8-bo'lim : Aniq integrallarni almashtirish qoidasi
6. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[ \int_{{ - 1}}^{2}{{{x^3} + {{\bf{e}}^{\frac{1}{4}x}}\,dx}}\]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 birinchi a'zo almashtirishga muhtoj emas. Shunday qilib,
\[ \int_{{ - 1}}^{2}{{{x^3} + {{\bf{e}}^{\frac{1}{4}x}}\,dx}} = \int_{{ - 1}}^{2}{{{x^3}\,dx}} + \int_{{ - 1}}^{2}{{{{\bf{e}}^{\frac{1}{4}x}}\,dx}}\]Ikkinchi integralning o'rnini bosish shunday bo'ladi:
\[u = \frac{1}{4}x\]2-qadam
Mana bu ikkinchi integral uchun haqiqiy almashtirish ishi.
\[\begin{align*}du & = \frac{1}{4}dx\hspace{0.25in}\,\, \to \hspace{0.25in}dx = 4du\\ x & = - 1:u = - \frac{1}{4}\hspace{0.75in}x = 2:u = \frac{1}{2}\end{align*}\]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.
\[ \int_{{ - 1}}^{2}{{{x^3} + {{\bf{e}}^{\frac{1}{4}x}}\,dx}} = \int_{{ - 1}}^{2}{{{x^3}\,dx}} + 4\int_{{ - \,\frac{1}{4}}}^{{\frac{1}{2}}}{{{{\bf{e}}^u}\,du}}\]3-qadam
U holda integral,
\[\int_{{ - 1}}^{2}{{{x^3} + {{\bf{e}}^{\frac{1}{4}x}}\,dx}} = \left. {\frac{1}{4}{x^4}} \right|_{ - 1}^2 + \left. {4{{\bf{e}}^u}} \right|_{ - \,\frac{1}{4}}^{\frac{1}{2}} = \left( {4 - \frac{1}{4}} \right) + \left( {4{{\bf{e}}^{\frac{1}{2}}} - 4{{\bf{e}}^{ - \,\,\frac{1}{4}}}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{15}}{4} + 4{{\bf{e}}^{\frac{1}{2}}} - 4{{\bf{e}}^{ - \,\,\frac{1}{4}}}}}\]