15.3-bo'lim : Umumiy hududlar bo'yicha ikki karrali integrallar
11. Avval integratsiya tartibini o'zgartirish orqali \(\displaystyle \int_{0}^{1}{{\int_{{ - \sqrt y }}^{{{y^{\,2}}}}{{6x - y\,dx}}\,dy}}\) ni baholang.
Yechim
E'tibor bering, ushbu muammo bilan oldingi muammodan farqli o'laroq, ushbu integratsiya tartibi bilan bog'liq muammolar yo'q. Bu integralni berilgan tartibda bemalol bajarishimiz mumkin. Biz faqat muammodagi tartibni o'zgartirmoqdamiz, chunki bizga muammo bayonotida shunday qilish kerakligi aytilgan.
Mana biz integraldan oladigan chegaralar.
\[\begin{array}{c}0 \le y \le 1\\ - \sqrt y \le x \le {y^2}\end{array}\]Mana bu chegaralar tasvirlangan hududning qisqacha eskizi.
Integratsiya tartibini o'zgartirganda, teskari tartib uchun to'g'ri chegaralarni olishimizga ishonch hosil qilish uchun mintaqaning eskizini olish juda foydali bo'ladi. Bu, ayniqsa, bu muammo bilan bog'liq.
2-qadam
Yaxshi, agar biz integratsiya tartibini o'zgartirsak, avval \(y\) ga integratsiya qilishimiz kerak. Biroq, bu kichik muammoga olib keladi. Pastki funktsiya o'zgaradi va shuning uchun biz buni ikkita mintaqaga bo'lishimiz kerak, bu esa o'z navbatida tartib integratsiyasini o'zgartirganda ikkita integralni bildiradi.
Har bir hudud uchun teskari tartibda chegaralar quyidagicha bo'ladi:
\[\begin{array}{*{20}{c}}\begin{array}{c} - 1 \le x \le 0\\ \,{x^2} \le y \le 1\end{array}&{\hspace{0.5in}}&\begin{array}{c}0 \le x \le 1\\ \sqrt x \le y \le 1\end{array}\end{array}\]Teskari tartibli integrallar,
\[\int_{0}^{1}{{\int_{{ - \sqrt y }}^{{{y^{\,2}}}}{{6x - y\,dx}}\,dy}} = \int_{{ - 1}}^{0}{{\int_{{{x^2}}}^{1}{{6x - y\,dy}}\,dx}} + \int_{0}^{1}{{\int_{{\sqrt x }}^{1}{{6x - y\,dy}}\,dx}}\]3-qadam
Endi biz faqat integrallarni baholashimiz kerak. Mana har biri uchun \(y\) integratsiyasi.
\[\begin{align*}\int_{0}^{1}{{\int_{{ - \sqrt y }}^{{{y^{\,2}}}}{{6x - y\,dx}}\,dy}} & = \int_{{ - 1}}^{0}{{\left. {\left( {6xy - \frac{1}{2}{y^2}} \right)} \right|_{{x^2}}^1\,dx}} + \int_{0}^{1}{{\left. {\left( {6xy - \frac{1}{2}{y^2}} \right)} \right|_{\sqrt x }^1\,dx}}\\ & = \int_{{ - 1}}^{0}{{\frac{1}{2}{x^4} - 6{x^3} + 6x - \frac{1}{2}\,dx}} + \int_{0}^{1}{{ - 6{x^{\frac{3}{2}}} + \frac{{13}}{2}x - \frac{1}{2}\,dx}}\end{align*}\]4-qadam
Va nihoyat, \(x\) integratsiyasi.
\[\begin{align*}\int_{0}^{1}{{\int_{{ - \sqrt y }}^{{{y^{\,2}}}}{{6x - y\,dx}}\,dy}} & = \left. {\left( {\frac{1}{{10}}{x^5} - \frac{3}{2}{x^4} + 3{x^2} - \frac{1}{2}x} \right)} \right|_{ - 1}^0 + \left. {\left( { - \frac{{12}}{5}{x^{\frac{5}{2}}} + \frac{{13}}{4}{x^2} - \frac{1}{2}x} \right)} \right|_0^1\\ & = \hspace{1.0in} - \frac{{19}}{{10}}\hspace{1.15in} + \hspace{1.0in}\frac{7}{{20}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{31}}{{20}}}}\end{align*}\]