AreaBetweenCurves | MathModel.uz

6.2-bo'lim: Egri chiziqlar orasidagi maydon

Ushbu bo'limda biz ikkita egri chiziq orasidagi maydonni topishni ko'rib chiqamiz. Aslida biz ko'rib chiqmoqchi bo'lgan ikkita holat mavjud.

Birinchi holda biz $\left[ {a,b} \right]$ oralig'ida $y = f\left( x \right)$ va $y = g\left( x \right)$ orasidagi maydonni aniqlamoqchimiz. Shuningdek, biz $f\left( x \right) \ge g\left( x \right)$ deb taxmin qilamiz. Dastlab nimani ko'rib chiqmoqchi ekanligimiz haqida tasavvurga ega bo'lish uchun quyidagi eskizga qarang.

This is the graph of two unknown functions on the domain a<x<b that are mostly in the 1st quadrant. The graph of f(x) is always over the graph of g(x) and the graph of g(x) dips briefly into the 4th quadrant just to make the point that the quadrant we are in does not matter.

The area between the two functions has been shaded in.

In Hudud va hajm formulalari Qo'shimchalar bo'limida biz bu holatda maydon uchun quyidagi formulani oldik.

\[\begin{equation}A = \int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}}\label{eq:eq1}\end{equation}\]

Ikkinchi holat birinchi holat bilan deyarli bir xil. Bu erda biz $f\left( y \right) \ge g\left( y \right)$ bilan $\left[ {c,d} \right]$ oralig'ida $x = f\left( y \right)$ va $x = g\left( y \right)$ o'rtasidagi maydonni aniqlaymiz.

This graph is similar to the first graph. The only difference is that the functions are in the form f(y) and g(y) where f(y) is always “larger”, i.e. to the right of g(y). The area between them has been shaded in.

Bu holda formula:

\[\begin{equation}A = \int_{{\,c}}^{{\,d}}{{f\left( y \right) - g\left( y \right)\,dy}}\label{eq:eq2}\end{equation}\]

Endi $\eqref{eq:eq1}$ va $\eqref{eq:eq2}$ mukammal xizmat ko'rsatadigan formulalardir, lekin ba'zida ular har doim birinchi funktsiya ikkita funktsiyadan kattaroq bo'lishini talab qilishini unutish oson. Shunday qilib, biz ushbu formulalar o'rniga quyidagi "so'z" formulalarini ishlatamiz, bu maydon har doim "kattaroq" funktsiyani minus "kichikroq" funksiya ekanligini esda tutishimizga ishonch hosil qilamiz.

Birinchi holda biz foydalanamiz,

Ikki egri chiziq orasidagi maydon, 1-holat

\[\begin{equation}A = \int_{{\,a}}^{{\,b}}{{\left( \begin{array}{c}{\mbox{upper}}\\ {\mbox{function}}\end{array} \right) - \left( \begin{array}{c}{\mbox{lower}}\\ {\mbox{function}}\end{array} \right)\,dx}},\hspace{0.5in}a \le x \le b\label{eq:eq3}\end{equation}\]

Ikkinchi holda biz foydalanamiz,

Ikki egri chiziq orasidagi maydon, 2-holat

\[\begin{equation}A = \int_{{\,c}}^{{\,d}}{{\left( \begin{array}{c}{\mbox{right}}\\ {\mbox{function}}\end{array} \right) - \left( \begin{array}{c}{\mbox{left}}\\ {\mbox{function}}\end{array} \right)\,dy}},\hspace{0.5in}c \le y \le d \label{eq:eq4}\end{equation}\]

Ushbu formulalardan foydalanish har doim bizni har bir muammo bilan nima sodir bo'layotgani haqida o'ylashga va formuladan foydalanishga kirishganimizda funktsiyalarning to'g'ri tartibiga ega ekanligimizga ishonch hosil qilishga majbur qiladi.

Keling, misol bilan ishlaylik.

1-misol $y = {x^2}$ va $y = \sqrt x $ bilan o'ralgan hudud maydonini aniqlang.

Yechim

Avvalo, "o'ralgan hudud" deganda nimani tushunamiz. Bu shuni anglatadiki, biz qiziqayotgan mintaqa mintaqaning har bir chegarasida ikkita egri chiziqdan biriga ega bo'lishi kerak. Shunday qilib, bu erda yopiq mintaqa soyali ikkita funktsiyaning grafigi.

$The graph of $y=\sqrt{x}$ and $y=x^{2}$ on the domain 0<x<1.2. In the domain 0<x<1 the graph of $\sqrt{x}$ is larger than the graph of $x^{2}$ and this region is shaded in. In the domain x>1 the graph of $x^{2}$ is larger than the graph of $\sqrt{x}$.$

E'tibor bering, biz ushbu ikki grafikning eng o'ngdagi kesishish nuqtasining o'ng tomonidagi mintaqaning biron bir qismini olmaymiz. Bu mintaqada o'ng tomonda chegara yo'q va shuning uchun bu hudud yopiq hududning bir qismi emas. Esda tutingki, berilgan funktsiyalardan biri yopiq hududning chegarasida bo'lishi kerak.

Bundan tashqari, ushbu grafikdan ko'rinib turibdiki, yuqori funktsiya biz foydalanadigan $x$ diapazoniga bog'liq bo'ladi. Shuning uchun siz har doim mintaqaning grafigini chizishingiz kerak. Eskizsiz ikkita funksiyadan qaysi biri kattaroq ekanligini xato qilish oson. Bunday holda, ko'pchilik $y = {x^2}$ ni yuqori funktsiya deb aytishi mumkin va ular $x$-ning aksariyat qismi uchun to'g'ri keladi. Biroq, bu holda u ikkita funktsiyaning eng pasti hisoblanadi.

Buning uchun integratsiya chegaralari ikkita egri chiziqning kesishish nuqtalari bo'ladi. Bunday holda, ular $x = 0$ va $x = 1$ da kesishishini ko'rish juda oson, shuning uchun bu integratsiya chegaralari.

Shunday qilib, maydonni topish uchun hisoblashimiz kerak bo'lgan integral,

\[\begin{align*}A & = \int_{{\,a}}^{{\,b}}{{\left( \begin{array}{c}{\mbox{upper}}\\ {\mbox{function}}\end{array} \right) - \left( \begin{array}{c}{\mbox{lower}}\\ {\mbox{function}}\end{array}\right)\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\sqrt x - {x^2}\,dx}}\\ & = \left. {\left( {\frac{2}{3}{x^{\frac{3}{2}}} - \frac{1}{3}{x^3}} \right)} \right|_0^1\\ & = \frac{1}{3}\end{align*}\]

Keyingi misolga o'tishdan oldin, bir nechta muhim narsalarni e'tiborga olish kerak.

Birinchidan, bu muammolarning deyarli barchasida grafik juda zarur. Ko'pincha integratsiya chegaralarini beradigan chegara hududini grafiksiz aniqlash qiyin.

Bundan tashqari, grafiksiz funksiyalarning qaysi biri yuqori, qaysi biri pastki funktsiya ekanligini aniqlash ko'pincha qiyin bo'lishi mumkin. Bu, ayniqsa, ushbu savolga javob aslida biz ishlatayotgan $x$ diapazoniga bog'liq bo'lgan oxirgi misol kabi holatlarda to'g'ri keladi.

Va nihoyat, oldingi bobda ko'rib chiqqan egri chiziq ostidagi maydondan farqli o'laroq, ikkita egri chiziq orasidagi maydon har doim ijobiy bo'ladi. Agar biz salbiy raqam yoki nolga ega bo'lsak, biz biror joyda xatoga yo'l qo'yganimizga amin bo'lishimiz mumkin va orqaga qaytib, uni topishimiz kerak.

Shuni ham yodda tutingki, ba'zan o'ralgan mintaqani aytish o'rniga biz chegaralangan mintaqani aytamiz. Ular bir xil narsani anglatadi.

Keling, yana bir nechta misollar bilan ishlaylik.

2-misol $y = x{{\bf{e}}^{ - {x^2}}}$, $y = x + 1$, $x = 2$ va $y$ o'qi bilan chegaralangan hudud maydonini aniqlang.

Yechim

Bunday holda, oxirgi ikki ma'lumot, $x = 2$ va $y$ o'qi bizga mintaqaning o'ng va chap chegaralarini bildiradi. Shuningdek, $y$ o'qi $x = 0$ chizig'i bilan berilganligini eslang. Mana, yopiq hudud soyalangan grafik.

$The graph of $y=x+1$ and $y=x{{\mathbf{e}}^{-{{x}^{2}}}}$ on the domain 0<x<2. The two graphs never intersect on this domain and the graph of $y=x+1$ is always larger than $y=x{{\mathbf{e}}^{-{{x}^{2}}}}$. The area between the two graphs has been shaded in.$

Bu erda, birinchi misoldan farqli o'laroq, ikkita egri chiziq uchrashmaydi. Buning o'rniga biz yuqorida aytib o'tganimizdek, mintaqaning chap va o'ng tomonlarini bog'lash uchun ikkita vertikal chiziqqa tayanamiz.

Mana, maydonni beradigan integral.

\[\begin{align*}A & = \int_{{\,a}}^{{\,b}}{{\left( \begin{array}{c}{\mbox{upper}}\\ {\mbox{function}}\end{array} \right) - \left( \begin{array}{c}{\mbox{lower}}\\ {\mbox{function}}\end{array} \right)\,dx}}\\ & = \int_{{\,0}}^{{\,2}}{{x + 1 - x{{\bf{e}}^{ - {x^2}}}\,dx}}\\ & = \left. {\left( {\frac{1}{2}{x^2} + x + \frac{1}{2}{{\bf{e}}^{ - {x^2}}}} \right)} \right|_0^2\\ & = \frac{7}{2} + \frac{{{{\bf{e}}^{ - 4}}}}{2} = 3.5092\end{align*}\]

3-misol $y = 2{x^2} + 10$ va $y = 4x + 16$ bilan chegaralangan hudud maydonini aniqlang.

Yechim

Bunday holda, kesishish nuqtalarini (oxir-oqibat bizga kerak bo'ladi) grafikdan osongina aniqlab bo'lmaydi, shuning uchun ularni hoziroq davom ettiramiz. E'tibor bering, ushbu muammolarning aksariyati uchun siz kesishish nuqtalarini grafikdan aniq aniqlay olmaysiz va shuning uchun ularni qo'lda aniqlashingiz kerak bo'ladi. Bu holda biz ikkita tenglamani teng o'rnatish orqali kesishish nuqtalarini olishimiz mumkin.

\[\begin{align*}2{x^2} + 10 & = 4x + 16\\ 2{x^2} - 4x - 6 & = 0\\ 2\left( {x + 1} \right)\left( {x - 3} \right) & = 0\end{align*}\]

Shunday qilib, ikkita egri chiziq $x = - 1$ va $x = 3$ da kesishadi. Agar bizga kerak bo'lsa, biz qiymatlarni tenglamalardan biriga qayta ulash orqali ularning har biriga mos keladigan $y$ qiymatlarini olishimiz mumkin. Grafikdagi ikkita kesishish nuqtasining koordinatalari $\left( { - 1,12} \right)$ va $\left( {3,28} \right)$ ekanligini tekshirishni sizga qoldiramiz.

Shuni ham yodda tutingki, agar siz grafik chizishda yaxshi bo'lmasangiz, kesishish nuqtalarini bilish hech bo'lmaganda grafikni boshlashga yordam beradi. Mana mintaqaning grafigi.

$The graph of $y=4x+16$ and $y=2{{x}^{2}}+10$ on the domain -1<x<3. The two graphs only intersect at the endpoints of the domain and in this domain the line is always larger than the parabola. The area between the two graphs has been shaded in.$

Grafik yordamida biz endi yuqori va pastki funktsiyani aniqlashimiz mumkin va shuning uchun endi yopiq maydonni topishimiz mumkin.

\[\begin{align*}A &= \int_{{\,a}}^{{\,b}}{{\left( \begin{array}{c}{\mbox{upper}}\\ {\mbox{function}}\end{array} \right) - \left( \begin{array}{c}{\mbox{lower}}\\ {\mbox{function}}\end{array} \right)\,dx}}\\ & = \int_{{\, - 1}}^{{\,3}}{{4x + 16 - \left( {2{x^2} + 10} \right)\,dx}}\\ & = \int_{{\, - 1}}^{{\,3}}{{ - 2{x^2} + 4x + 6\,dx}}\\ & = \left. {\left( { - \frac{2}{3}{x^3} + 2{x^2} + 6x} \right)} \right|_{ - 1}^3\\ & = \frac{{64}}{3}\end{align*}\]

Ushbu muammolarda qavslar bilan ehtiyot bo'ling. Talabalarning bu masalalarda eng ko'p uchraydigan xatolaridan biri bu ikkinchi davrdagi qavslarni e'tiborsiz qoldirishdir.

4-misol $y = 2{x^2} + 10$, $y = 4x + 16$, $x = - 2$ va $x = 5$ bilan chegaralangan hudud maydonini aniqlang.

Yechim

Demak, bu masalada ishlatiladigan funksiyalar birinchi masaladagi funksiyalar bilan bir xil. Farqi shundaki, biz chegaralangan hududni kesishgan nuqtalardan tashqariga kengaytirdik. Bu avvalgi misolda biz ishlatgan funksiyalar bo'lgani uchun biz kesishish nuqtalarini qayta topish bilan bezovtalanmaymiz.

Mana bu hududning grafigi.

$The graph of $y=4x+16$ and $y=2{{x}^{2}}+10$ on the domain -2<x<5. The two graphs intersect at x=-1 and x=3. In the domain -1<x<3 the line is larger than the parabola while in the domains -2<x<-1 and 3<x<5 the parabola is larger than the line. The area between the two graphs has been shaded in.$

Mayli, bizda kichik muammo bor. Bizning formulamiz bitta funktsiya har doim yuqori funktsiya bo'lishini va boshqa funktsiya har doim pastki funktsiya bo'lishini talab qiladi va bu erda bizda aniq emas. Biroq, bu aslida birinchi qarashda ko'rinadigan muammo emas. Bir funksiya har doim yuqori funktsiya, ikkinchisi esa har doim pastki funktsiya bo'lgan uchta hudud mavjud. Shunday qilib, biz qilishimiz kerak bo'lgan yagona narsa, biz qila oladigan uchta mintaqaning har birining maydonini topib, keyin hammasini qo'shishdir.

Mana hudud.

\[\begin{align*}A & = \int_{{\, - 2}}^{{\, - 1}}{{2{x^2} + 10 - \left( {4x + 16} \right)\,dx}} + \int_{{\, - 1}}^{{\,3}}{{4x + 16 - \left( {2{x^2} + 10} \right)\,dx}} + \int_{{\,3}}^{{\,5}}{{2{x^2} + 10 - \left( {4x + 16} \right)\,dx}}\\ & = \int_{{\, - 2}}^{{\, - 1}}{{2{x^2} - 4x - 6\,dx}} + \int_{{\, - 1}}^{{\,3}}{{ - 2{x^2} + 4x + 6\,dx}} + \int_{{\,3}}^{{\,5}}{{2{x^2} - 4x - 6\,dx}}\\ & = \left. {\left( {\frac{2}{3}{x^3} - 2{x^2} - 6x} \right)} \right|_{ - 2}^{ - 1} + \left. {\left( { - \frac{2}{3}{x^3} + 2{x^2} + 6x} \right)} \right|_{ - 1}^3 + \left. {\left( {\frac{2}{3}{x^3} - 2{x^2} - 6x} \right)} \right|_3^5\\ & = \frac{{14}}{3} + \frac{{64}}{3} + \frac{{64}}{3}\\ & = \frac{{142}}{3}\end{align*}\]

5-misol $y = \sin x$, $y = \cos x$, $x = \frac{\pi }{2}$ va $y$ o'qi bilan o'ralgan hudud maydonini aniqlang.

Yechim

Birinchidan, mintaqaning grafigini olamiz.

$The graph of $y=\cos(x)$ and $y=\sin(x)$ on the domain 0<x<$\frac{\pi}{2}$. The two graphs intersect at $x=\frac{\pi}{4}$. In the domain 0<x<$\frac{\pi}{4}$ the cosine is larger than sine and in the domain $\frac{\pi}{4}$<x<\frac{\pi}{2}$ sine is larger than cosine. The area between the two graphs has been shaded in.$

Shunday qilib, bizda maydonni olish uchun ikkita integral qilish kerak bo'lgan yana bir holat mavjud. Kesishish nuqtasi qaerda bo'ladi

\[\sin x = \cos x\]

oraliqda. Bu $x = \frac{\pi }{4}$ ekanligini tekshirishni sizga qoldiramiz. Keyin hudud,

\[\begin{align*}A & = \int_{{\,0}}^{{\,\frac{\pi }{4}}}{{\cos x - \sin x\,dx}} + \int_{{\,{\pi }/{4}\;}}^{{\,{\pi }/{2}\;}}{{\sin x - \cos x\,dx}}\\ & = \left. {\left( {\sin x + \cos x} \right)} \right|_0^{\frac{\pi }{4}} + \left. {\left( { - \cos x - \sin x} \right)} \right|_{{\pi }/{4}\;}^{{\pi }/{2}\;}\\ & = \sqrt 2 - 1 + \left( {\sqrt 2 - 1} \right)\\ & = 2\sqrt 2 - 2 = 0.828427\end{align*}\]

Ushbu keyingi misol bilan ehtiyot bo'lishimiz kerak.

6-misol $x = \frac{1}{2}{y^2} - 3$ va $y = x - 1$ bilan o'ralgan hududning maydonini aniqlang.

Yechim

Birinchi tenglama sizni xafa qilishiga yo'l qo'ymang. Biz vaqti-vaqti bilan bunday tenglamalar bilan shug'ullanishimiz kerak, shuning uchun biz ular bilan shug'ullanishga ko'nikishimiz kerak.

Har doimgidek, bu ikkita egri chiziqning kesishish nuqtalariga ega bo'lsak yordam beradi. Bu holda biz $x$ uchun ikkinchi tenglamani yechish va keyin ularni tenglashtirish orqali kesishish nuqtalarini olamiz. Mana bu ish,

\[\begin{align*}y + 1 & = \frac{1}{2}{y^2} - 3\\ 2y + 2 & = {y^2} - 6\\ 0 & = {y^2} - 2y - 8\\ 0 & = \left( {y - 4} \right)\left( {y + 2} \right)\end{align*}\]

Shunday qilib, ikkita egri chiziq $y = - 2$ va $y = 4$ da kesishadi yoki agar bizga to'liq koordinatalar kerak bo'lsa, ular quyidagicha bo'ladi: $\left( { - 1, - 2} \right)$ va $\left( {5,4} \right)$.

Mana, ikkita egri chiziqning eskizi.

$The graph of $x=y+1$ and $x=\frac{1}{2}{{y}^{2}}-3$ on the domain -4<y<4.5. The two graphs intersect at y=-2 and y=4. In the domain -2<y<4 the graph of the line is larger (i.e. to the right) than the graph of the parabola. This is the region we are interested in for this problem.$

Endi, agar ehtiyot bo'lmasak, bizda jiddiy muammo bo'ladi. Shu paytgacha biz yuqori va pastki funksiyadan foydalandik. Buni amalga oshirish uchun bu erda, aslida, mintaqaning turli xil pastki funktsiyalarga ega bo'lgan ikkita qismi mavjudligiga e'tibor bering. $\left[ { - 3, - 1} \right]$ diapazonida parabola aslida ham yuqori, ham pastki funktsiyadir.

Biz hozirgacha ishlatgan formuladan foydalanish uchun $y$ uchun parabolani yechishimiz kerak. Bu beradi,

\[y = \pm \sqrt {2x + 6} \]

Bu erda "+" parabolaning yuqori qismini va "-" pastki qismini beradi.

Mana, agar biz birinchi formuladan foydalanmoqchi bo'lsak, kerak bo'ladigan har bir mintaqa soyali to'liq maydonning eskizi.

$The graph of $x=y+1$ and $x=\frac{1}{2}{{y}^{2}}-3$ on the domain -4<y<4.5. The two graphs intersect at y=-2 and y=4. Unlike the first graph in this example the graphs are done assuming they are functions of x. So, the line is graphed as $y=x-1$ and the top of the parabola is graphed as $y=\sqrt{2x+6}$ and the bottom of the parabola is graphed as $y=-\sqrt{2x+6}$. The area between the curves is shaded but is shaded based on the functions being functions of x instead of functions of y. So, in the range -3<x<-1 we get one shading because the top/bottom of the graph are the top/bottom “branches” of the parabola. In the range -1<x<5 we get a different shading because the top graph is the top “branch” of the parabola and the bottom graph is the line.$

U holda maydon uchun integrallar quyidagicha bo'ladi:

\[\begin{align*}A & = \int_{{\, - 3}}^{{\, - 1}}{{\sqrt {2x + 6} - \left( { - \sqrt {2x + 6} } \right)\,dx}} + \int_{{\, - 1}}^{{\,5}}{{\sqrt {2x + 6} - \left( {x - 1} \right)\,dx}}\\ & = \int_{{\, - 3}}^{{\, - 1}}{{2\sqrt {2x + 6} \,dx}} + \int_{{\, - 1}}^{{\,5}}{{\sqrt {2x + 6} - x + 1\,dx}}\\ & = \int_{{\, - 3}}^{{\, - 1}}{{2\sqrt {2x + 6} \,dx}} + \int_{{\, - 1}}^{{\,5}}{{\sqrt {2x + 6} \,dx}} + \int_{{\, - 1}}^{{\,5}}{{ - x + 1\,dx}}\\ & = \left. {\frac{2}{3}{u^{\frac{3}{2}}}} \right|_0^4 + \left. {\frac{1}{3}{u^{\frac{3}{2}}}} \right|_4^{16} + \left. {\left( { - \frac{1}{2}{x^2} + x} \right)} \right|_{ - 1}^5\\ & = 18\end{align*}\]

Ushbu integrallar unchalik qiyin bo'lmasa-da, ular kerak bo'lgandan ko'ra qiyinroq.

Eslatib o'tamiz, maydonni aniqlash uchun yana bir formula mavjud. Bu,

\[A = \int_{{\,c}}^{{\,d}}{{\left( \begin{array}{c}{\mbox{right}}\\ {\mbox{function}}\end{array} \right) - \left( \begin{array}{c}{\mbox{left}}\\ {\mbox{function}}\end{array} \right)\,dy}},\hspace{0.5in}c \le y \le d\]

va bizning holatlarimizda har doim chapda, ikkinchisi esa o'ngda bo'lgan bitta funktsiyaga egamiz. Shunday qilib, bu holda, bu, albatta, borish kerak bo'lgan yo'l. E'tibor bering, biz chiziq tenglamasini qayta yozishimiz kerak, chunki u $x = f\left( y \right)$ ko'rinishida bo'lishi kerak, ammo buni qilish juda oson. Mana bu formuladan foydalanish uchun grafik.

$The graph of $x=y+1$ and $x=\frac{1}{2}{{y}^{2}}-3$ on the domain -4<y<4.5. The two graphs intersect at y=-2 and y=4. In this graph the area between the two graphs are shaded given that the functions are functions of y. In this case the line is always on the right and the parabola is always on the left.$

Hudud,

\[\begin{align*}A & = \int_{{\,c}}^{{\,d}}{{\left( \begin{array}{c}{\mbox{right}}\\ {\mbox{function}}\end{array} \right) - \left( \begin{array}{c}{\mbox{left}}\\ {\mbox{function}}\end{array} \right)\,dy}}\\ & = \int_{{\, - 2}}^{{\,4}}{{\left( {y + 1} \right) - \left( {\frac{1}{2}{y^2} - 3} \right)\,dy}}\\ & = \int_{{\, - 2}}^{{\,4}}{{ - \frac{1}{2}{y^2} + y + 4\,dy}}\\ & = \left. {\left( { - \frac{1}{6}{y^3} + \frac{1}{2}{y^2} + 4y} \right)} \right|_{ - 2}^4\\ & = 18\end{align*}\]

Bu biz birinchi formuladan foydalanganimiz bilan bir xil va bu birinchi usulga qaraganda osonroq edi.

Shunday qilib, ushbu oxirgi misolda biz maydonni topish uchun ikkala formuladan foydalanishimiz mumkin bo'lgan holatni ko'rdik. Biroq, ikkinchisi, albatta, osonroq edi.

Talabalar ko'pincha hisob-kitob darsiga funksiyalar bilan ishlashning yagona oson yo'li ularni $y = f\left( x \right)$ shaklida ishlatishdir, degan fikr bilan kelishadi. Biroq, biz oldingi misolda ko'rganimizdek, $x = f\left( y \right)$ ko'rinishidagi funktsiyalar bilan ishlash osonroq bo'ladigan vaqtlar bor. Darhaqiqat, muammoni hal qilishning yagona yo'li bo'ladigan holatlar bo'ladi, shuning uchun siz ushbu shakldagi funktsiyalar bilan shug'ullanishingiz mumkinligiga ishonch hosil qiling.

Keling, ushbu shakldagi funktsiyalar bilan shug'ullanishimiz mumkinligiga ishonch hosil qilish uchun yana bir misolni ko'rib chiqaylik.

7-misol $x = - {y^2} + 10$ va $x = {\left( {y - 2} \right)^2}$ bilan chegaralangan hudud maydonini aniqlang.

Yechim

Birinchidan, bizga kesishish nuqtalari kerak bo'ladi.

\[\begin{align*} - {y^2} + 10 & = {\left( {y - 2} \right)^2}\\ - {y^2} + 10 & = {y^2} - 4y + 4\\ 0 & = 2{y^2} - 4y - 6\\ 0 & = 2\left( {y + 1} \right)\left( {y - 3} \right)\end{align*}\]

Kesishish nuqtalari $y = - 1$ va $y = 3$. Bu erda mintaqaning eskizi.

$The graph of $x=-{{y}^{2}}+10$ and $x={{\left( y-2 \right)}^{2}}$ on the domain -2<y<4. The two graphs intersect at y=-1 and y=3 and the region we are interested in is in the domain -1<y<3. The parabola given by $x=-{{y}^{2}}+10$ is always on the right of the region and the parabola given by $x={{\left( y-2 \right)}^{2}}$ is always on the left of the region.$

Bu, albatta, ikkinchi maydon formulasi osonroq bo'ladigan mintaqadir. Agar biz birinchi formuladan foydalansak, biz ko'rib chiqishimiz kerak bo'lgan uchta turli mintaqa bo'lar edi.

Bu holda hudud,

\[\begin{align*}A & = \int_{{\,c}}^{{\,d}}{{\left( \begin{array}{c}{\mbox{right}}\\ {\mbox{function}}\end{array} \right) - \left( \begin{array}{c}{\mbox{left}}\\ {\mbox{function}}\end{array} \right)\,dy}}\\ & = \int_{{\, - 1}}^{{\,3}}{{ - {y^2} + 10 - {{\left( {y - 2} \right)}^2}\,dy}}\\ & = \int_{{\, - 1}}^{{\,3}}{{ - 2{y^2} + 4y + 6\,dy}}\\ & = \left. {\left( { - \frac{2}{3}{y^3} + 2{y^2} + 6y} \right)} \right|_{ - 1}^3 = \frac{{64}}{3}\end{align*}\]