সৃজনশীলঃ ‘দৃশ্যকল্প-১: g(x)=cos⁡x,h(x)=x4দৃশ্যকল্প-২: x216+y29=1…’ রাজশাহী বোর্ড ২০২২ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছ, g(x)=cos x

এবং, $h\left(x\right)=x^4$

এখন, ${\int }_0^{\pi /2}\left\{g\right(x){\}}^{2}dx$

$\begin{array}{cc}& =\frac{1}{2}{\int }_0^{\pi /2}2\left\{g\right(x){\}}^{2}dx\\ & =\frac{1}{2}{\int }_0^{\pi /2}2{\cos }^{2}xdx\end{array}$

$\begin{array}{cc}& =\frac{1}{2}{\int }_0^{\pi /2}(1+\cos 2x)dx\\ & =\frac{1}{2}{[x+\frac{\sin 2x}{2}]}_0^{\pi /2}\\ & =\frac{1}{2}[(\frac{\pi }{2}+0]-(0+0\left)\right]\\ & =\frac{\pi }{4}\end{array}$

এখন,

ধরিহলেহলে$\begin{array}{cc}& {\int }_0^1\frac{x}{1+h\left(x\right)}dx={\int }_0^3\frac{x}{1+x^4}dx\\ & ={\int }_0^1\frac{x}{1+{\left(x^2\right)}^2}dx\\ & =\frac{1}{2}{\int }_0^1\frac{2x}{1+{\left(x^2\right)}^2}dx\\ & =\frac{1}{2}{\int }_0^1\frac{dz}{1+z^2}\\ & \text{ ধরি, }{x}^2=z\\ & 2xdx=dz\\ & x=1\text{ হলে }\\ & =\frac{1}{2}[{\tan }^{-1}z]\delta \\ & z=1\\ & x=0\text{ হলে }\\ & z=0\\ & =\frac{1}{2}=1({\tan }^{-1}1-{\tan }^{-1}0)\\ & =\frac{1}{2}{\tan }^{-1}1=\frac{1}{2}\cdot \frac{\pi }{4}=\frac{\pi }{8}\\ & \therefore {\int }_0^{\pi /2}{\cos }^{2}xdx+{\int }_0^1\frac{x}{1+h\left(x\right)}dx\\ & =\frac{\pi }{4}+\frac{\pi }{8}=\frac{3\pi }{8}\text{ (Ans) }\\ & \end{array}$

$\frac{x^2}{16}+\frac{y^2}{9}=1\Rightarrow \frac{x^2}{4^2}+\frac{y^2}{3^2}=1\dots \dots \dots \dots \left(i\right)$

এবং, $x^2+y^2=9\Rightarrow x^2+y^2=3^2\dots$ (ii)

(ii) নং থেকে,

$\begin{array}{cc}& \frac{x^2+y^2}{9}=1\\ & \frac{x^2}{9}+\frac{y^2}{9}=1\end{array}$

(i) - (ii)

নংএবসিয়ে$\begin{array}{cc}& \frac{x^2}{4}-\frac{x^2}{9}=0\\ & x^2\frac{5}{36}=0\\ & \therefore x=0\\ & x=0\text{ (i) নং এ বসিয়ে, }\\ & y^2=9\\ & \Rightarrow y=\pm 3\end{array}$

$\therefore$ (0,3) ও (0,-3) তে দ্বারা ছেদ করবে।

(i) নং এ x = 0 ও y = 0 বসিয়ে

$y=\pm 3\bullet x=\pm 4$

$\therefore$ উপবৃত্তটি $(0,3),(0,3)$ ও $(4,0),(-4,0)$বিন্দু দিয়ে যাবে

(ii) নং থেকে

$y_1=y=\sqrt{9-x^2}$

এবং (i) নং থেকে

$\begin{array}{cc}& y_2=\sqrt[9]{1-\frac{x^2}{16}}\\ & =\frac{3}{4}\sqrt{16-x^2}\end{array}$

$\therefore$আবদ্ধ অংশের ক্ষেত্রফল, $={\int }_0^4{y}_2^{dx}{\int }_0^3{y}_{1}dx$

একক$\begin{array}{cc}& =\frac{3}{4}{\int }_0^4\sqrt{16-x^2}\mathrm{dx}-{\int }_0^3\sqrt{3^2-x^2}\mathrm{dx}\\ & =\frac{3}{4}{[\frac{x\sqrt{16-x^2}}{2}+\frac{4^2}{2}{\sin }^2\frac{x}{4}]}_0^4-[\frac{x\sqrt{3^2-x^2}}{2}+\frac{3^2}{2}{\sin }^{-1}\frac{x}{3}]\\ & =\frac{3}{4}[(0+8\cdot \frac{\pi }{2})-(0+0\left)\right]-[(0+\frac{9}{2}\times \frac{\pi }{2})-(0-0\left)\right]\\ & =\frac{3}{4}\times \frac{8\pi }{2}-\frac{9}{2}\times \frac{\pi }{2}\\ & =3\pi -\frac{9\pi }{4}\\ & =\frac{12\pi -9\pi }{4}\\ & =\frac{3\pi }{4}\text{ একক (Ans) }\end{array}$