Asosiy tarkibga o'tish

7.10-bo'lim : Aniq integrallarni taxmin qilish

2. \(n = 6\) dan foydalanib, \(\displaystyle \int_{{ - 1}}^{2}{{\sqrt {{{\bf{e}}^{ - \,{x^{\,2}}}} + 1} \,dx}}\) qiymatini taxminiy hisoblang

  1. O'rta nuqta qoidasi,
  2. Trapezoid qoidasi va
  3. Simpson qoidasi.

Ishingiz uchun kamida 6 kasrdan iborat aniqlikdan foydalaning.

a O'rta nuqta qoidasi Yechim

Muammoni hal qilish uchun haqiqatan ham kerak bo'lmasa-da, bu erda grafikning eskizi.

The graph of \(\sqrt {{{\bf{e}}^{ - {x^2}}} + 1} \) on the interval \(\left[ { - 1,2} \right]\). The graph rises gently from about 1.15 on the left to a peak of about 1.41 at the origin and then decreases, flattening out towards 1 at the right end. The seven points that divide the interval into 6 subintervals are marked along the x-axis.

Biz \(\left[ { - 1,2} \right]\) oralig'ini har birining kengligi bo'lgan 6 ta subintervalga bo'lish kerakligini bilamiz,

\[\Delta x = \frac{{2 - \left( { - 1} \right)}}{6} = \frac{1}{2}\]

Ushbu kichik intervallarning har birining so'nggi nuqtalari yuqoridagi grafikdagi \(x\) o'qidagi nuqtalar bilan ifodalanadi.

Har bir nuqta orasidagi belgi har bir subintervalning o'rta nuqtasini ifodalaydi. Har bir kichik oraliq uchun o'rta nuqtalarning \(x\) qiymatlari quyidagicha:

\[ - \frac{3}{4},\hspace{0.25in} - \frac{1}{4},\hspace{0.25in}\frac{1}{4},\hspace{0.25in}\frac{3}{4},\hspace{0.25in}\frac{5}{4},\hspace{0.25in}\frac{7}{4}\]

Shunday qilib, integralning qiymatini taxmin qilish uchun O'rta nuqta qoidasidan foydalanish uchun formulaga ulanishimiz kerak. Shunday qilib,

\[\begin{align*}\int_{{ - 1}}^{2}{{\sqrt {{{\bf{e}}^{ - \,{x^{\,2}}}} + 1} \,dx}} & \approx \left( {\frac{1}{2}} \right)\left[ {f\left( { - \frac{3}{4}} \right) + f\left( { - \frac{1}{4}} \right) + f\left( {\frac{1}{4}} \right) + f\left( {\frac{3}{4}} \right) + f\left( {\frac{5}{4}} \right) + f\left( {\frac{7}{4}} \right)} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{3.70700857}}\end{align*}\]

b Trapetsiya qoidasi Yechim

Midpoint Rule ishidan bilamizki, har bir kichik oraliqning kengligi \(\Delta x = \frac{1}{2}\) va mos yozuvlar uchun grafikning eskizi va nuqtalar bilan belgilangan har bir kichik intervalning oxirgi nuqtalari quyida ko'rsatilgan.

The graph of \(\sqrt {{{\bf{e}}^{ - {x^2}}} + 1} \) on the interval \(\left[ { - 1,2} \right]\). The graph rises gently from about 1.15 on the left to a peak of about 1.41 at the origin and then decreases, flattening out towards 1 at the right end. The seven points that divide the interval into 6 subintervals are marked along the x-axis.

Shunday qilib, integralning qiymatini taxmin qilish uchun Trapezoid qoidasidan foydalanish uchun formulaga ulanishimiz kerak. Shunday qilib,

\[\begin{align*}\int_{{ - 1}}^{2}{{\sqrt {{{\bf{e}}^{ - \,{x^{\,2}}}} + 1} \,dx}} & \approx \left( {\frac{{{\scriptstyle{}^{1}/{}_{2}}}}{2}} \right)\left[ {f\left( { - 1} \right) + 2f\left( { - \frac{1}{2}} \right) + 2f\left( 0 \right) + 2f\left( {\frac{1}{2}} \right) + 2f\left( 1 \right) + 2f\left( {\frac{3}{2}} \right) + f\left( 2 \right)} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{3.69596543}}\end{align*}\]

c Simpson qoidasi Yechim

Midpoint Rule ishidan biz bilamizki, har bir kichik oraliqning kengligi \(\Delta x = \frac{1}{2}\) va mos yozuvlar uchun grafikning eskizi va nuqtalar bilan belgilangan har bir subintervalning so'nggi nuqtalari quyida ko'rsatilgan.

The graph of \(\sqrt {{{\bf{e}}^{ - {x^2}}} + 1} \) on the interval \(\left[ { - 1,2} \right]\). The graph rises gently from about 1.15 on the left to a peak of about 1.41 at the origin and then decreases, flattening out towards 1 at the right end. The seven points that divide the interval into 6 subintervals are marked along the x-axis.

Birinchi ikki qismda bo'lgani kabi, integralning taxminiy qiymatini aniqlash uchun Simpson qoidasidan foydalanish uchun formulaga ulanishimiz kerak. Shunday qilib,

\[\begin{align*}\int_{{ - 1}}^{2}{{\sqrt {{{\bf{e}}^{ - \,{x^{\,2}}}} + 1} \,dx}} & \approx \left( {\frac{{{\scriptstyle{}^{1}/{}_{2}}}}{3}} \right)\left[ {f\left( { - 1} \right) + 4f\left( { - \frac{1}{2}} \right) + 2f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + 2f\left( 1 \right) + 4f\left( {\frac{3}{2}} \right) + f\left( 2 \right)} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{3.70358145}}\end{align*}\]