Asosiy tarkibga o'tish

15.3-bo'lim : Umumiy hududlar bo'yicha ikki karrali integrallar

10. Avval integratsiya tartibini teskari o'zgartirish orqali \(\displaystyle \int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}}\) ni baholang.

Yechim

Keling, shuni ta'kidlab boshlaylikki, agar biz \(y\) ga nisbatan integratsiyalashgan bo'lsak, birinchi navbatda trigni almashtirishni amalga oshirishimiz kerak (buni qilish kerak bo'lmasa, barchamiz minnatdor bo'lamiz!), shuning uchun bu holda tartibni almashtirish bizni chalkash integraldan qutqarishi mumkin.

Shunday qilib, biz integraldan oladigan chegaralar.

\[\begin{array}{c}0 \le x \le 3\\ 2x \le y \le 6\end{array}\]

Mana bu chegaralar tasvirlangan hududning qisqacha eskizi.

A sketch of the region described by the limits. It is a triangle bounded above by the horizontal line \(y = 6\), below by the line \(y = 2x\), which can also be written \(x = \frac{1}{2}y\), and on the left by the y-axis, with vertices at the origin, \(\left( {0,6} \right)\) and \(\left( {3,6} \right)\). The inside is shaded green.

Integratsiya tartibini o'zgartirganda, teskari tartib uchun to'g'ri chegaralarni olishimizga ishonch hosil qilish uchun mintaqaning eskizini olish juda foydali bo'ladi.

2-qadam

Yaxshi, agar integratsiya tartibini o'zgartirish uchun biz avval \(x\) ga integratsiya qilishimiz kerak. Teskari tartib uchun chegaralar, keyin,

\[\begin{array}{c}0 \le y \le 6\\ 0 \le x \displaystyle \le \frac{1}{2}y\end{array}\]

Teskari tartibli integral,

\[\int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}} = \int_{0}^{6}{{\int_{0}^{{\frac{1}{2}y}}{{\sqrt {{y^2} + 2} \,dx}}\,dy}}\]
3-qadam

Endi biz faqat integrallarni baholashimiz kerak. Mana \(x\) integratsiyasi.

\[\int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}} = \int_{0}^{6}{{\left. {\left( {x\sqrt {{y^2} + 2} } \right)} \right|_0^{\frac{1}{2}y}\,dy}} = \int_{0}^{6}{{\frac{1}{2}y\sqrt {{y^2} + 2} \,dy}}\]
4-qadam

E'tibor bering, \(x\) integratsiyasi tufayli biz triglarni almashtirishni amalga oshirishimiz shart emas. Bizga kerak bo'lgan yagona narsa - oddiy Hisob I integrali. Mana \(y\) integratsiyasi (almashtirish tafsilotlarini tekshirishni sizga qoldiramiz).

\[\int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}} = \left. {\left( {\frac{1}{6}{{\left( {{y^2} + 2} \right)}^{\frac{3}{2}}}} \right)} \right|_0^6 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{6}\left( {{{38}^{\frac{3}{2}}} - {2^{\frac{3}{2}}}} \right) = 38.5699}}\]