5.5-bo'lim : Hudud muammosi
1. \(n = 6\) yordamida \(\left[ {1,4} \right]\) da \(f\left( x \right) = {x^3} - 2{x^2} + 4\) \(x\) o'qi o'rtasidagi mintaqa maydonini hisoblang,
- to'rtburchaklar balandligi uchun subintervallarning o'ng oxirgi nuqtalari,
- to'rtburchaklar balandligi uchun pastki intervallarning chap oxirgi nuqtalari va,
- to'rtburchaklar balandligi uchun subintervallarning o'rta nuqtalari.
a To'rtburchaklar balandligi uchun subintervallarning o'ng oxirgi nuqtalari. Yechim
Ushbu muammo uchun har bir kichik intervalning kengligi:
\[\Delta x = \frac{{4 - 1}}{6} = \frac{1}{2}\]Ushbu muammoni hal qilish uchun funktsiyaning grafikasini chizishimiz shart emas. Biroq, subintervallarni ko'rsatadigan raqam chizig'iga ega bo'lish yordam berishi mumkin. Mana bu raqam qatori.
![A number line for the interval \(\left[ {1,4} \right]\) divided into 6 equal subintervals, with tick marks labeled 1, \(\frac{3}{2}\), 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\) and 4.](/Solutions/CalcI/AreaProblem//Solutions/CalcI/AreaProblem/AreaProblem_Files/image001.webp)
Bunday holda, biz har bir to'rtburchakning balandligini aniqlash uchun ushbu subintervallarning har birining o'ng oxirgi nuqtalaridan foydalanamiz.
Funktsiya va \(x\) o'qi orasidagi maydon taxminan,
\[\begin{align*}{\mbox{Area}} & \approx \frac{1}{2}f\left( {\frac{3}{2}} \right) + \frac{1}{2}f\left( 2 \right) + \frac{1}{2}f\left( {\frac{5}{2}} \right) + \frac{1}{2}f\left( 3 \right) + \frac{1}{2}f\left( {\frac{7}{2}} \right) + \frac{1}{2}f\left( 4 \right)\\ & = \frac{1}{2}\left( {\frac{{23}}{8}} \right) + \frac{1}{2}\left( 4 \right) + \frac{1}{2}\left( {\frac{{57}}{8}} \right) + \frac{1}{2}\left( {13} \right) + \frac{1}{2}\left( {\frac{{179}}{8}} \right) + \frac{1}{2}\left( {36} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{683}}{{16}} = 42.6875}}\end{align*}\]b To'rtburchaklar balandligi uchun subintervallarning chap oxirgi nuqtalari. Yechim
Oldingi qismda topganimizdek, har bir subintervalning kengligi \(\Delta x = \frac{1}{2}\).
Bu erda muammoni hal qilishda yordam berish uchun pastki intervallarni ko'rsatadigan raqamlar qatorining nusxasi.
Bunday holda, biz har bir to'rtburchakning balandligini aniqlash uchun ushbu subintervallarning har birining chap uchi nuqtalaridan foydalanamiz.
Funktsiya va \(x\) o'qi orasidagi maydon taxminan,
\[\begin{align*}{\mbox{Area}} & \approx \frac{1}{2}f\left( 1 \right) + \frac{1}{2}f\left( {\frac{3}{2}} \right) + \frac{1}{2}f\left( 2 \right) + \frac{1}{2}f\left( {\frac{5}{2}} \right) + \frac{1}{2}f\left( 3 \right) + \frac{1}{2}f\left( {\frac{7}{2}} \right)\\ & = \frac{1}{2}\left( 3 \right) + \frac{1}{2}\left( {\frac{{23}}{8}} \right) + \frac{1}{2}\left( 4 \right) + \frac{1}{2}\left( {\frac{{57}}{8}} \right) + \frac{1}{2}\left( {13} \right) + \frac{1}{2}\left( {\frac{{179}}{8}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{419}}{{16}} = 26.1875}}\end{align*}\]c To'rtburchaklar balandligi uchun subintervallarning o'rta nuqtalari. Yechim
Birinchi qismda topganimizdek, har bir subintervalning kengligi \(\Delta x = \frac{1}{2}\).
Bu erda muammoni hal qilishda yordam berish uchun pastki intervallarni ko'rsatadigan raqamlar qatorining nusxasi.
Bunday holda, biz har bir to'rtburchakning balandligini aniqlash uchun ushbu subintervallarning har birining o'rta nuqtalaridan foydalanamiz.
Funktsiya va \(x\) o'qi orasidagi maydon taxminan,
\[\begin{align*}{\mbox{Area}} & \approx \frac{1}{2}f\left( {\frac{5}{4}} \right) + \frac{1}{2}f\left( {\frac{7}{4}} \right) + \frac{1}{2}f\left( {\frac{9}{4}} \right) + \frac{1}{2}f\left( {\frac{{11}}{4}} \right) + \frac{1}{2}f\left( {\frac{{13}}{4}} \right) + \frac{1}{2}f\left( {\frac{{15}}{4}} \right)\\ & = \frac{1}{2}\left( {\frac{{181}}{{64}}} \right) + \frac{1}{2}\left( {\frac{{207}}{{64}}} \right) + \frac{1}{2}\left( {\frac{{337}}{{64}}} \right) + \frac{1}{2}\left( {\frac{{619}}{{64}}} \right) + \frac{1}{2}\left( {\frac{{1101}}{{64}}} \right) + \frac{1}{2}\left( {\frac{{1831}}{{64}}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1069}}{{32}} = 33.40625}}\end{align*}\]