সৃজনশীলঃ ‘Q(x,y)=9×2+16y2−144f(x)=x−2,g(x)=x(x−1)(x2+1)…’ কুমিল্লা বোর্ড ২০২৫ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $g\left(x\right)=\frac{x}{(x-1)(x^2+1)}$

$\therefore {\int }_g\left(x\right)=\int \frac{x}{(x-1)(x^2+1)}dx=\int \frac{(x-1)+1}{(x-1)(x^2+1)}dx$

$=\int (\frac{1}{x^2+1}+\frac{1}{(x-1)(x^2+1)})dx$

$=\int \frac{1}{x^2+1}dx+\int \frac{1}{(x-1)(x^2+1)}dx$

$={\tan }^{-1}x+\int \frac{1}{(x-1)(x^2+1)}dx$

$={\tan }^{-1}x+\frac{1}{2}\int \{\frac{1}{x-1}-\frac{x+1}{x^2+1}\}dx$

$={\tan }^{-1}x+\frac{1}{2}\int \frac{dx}{x-1}-\frac{1}{2}\int \frac{x+1}{x^2+1}dx$

$={\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{2}\int \frac{x+1}{x^2+1}dx$

$={\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{2}\int \frac{x}{x^2+1}dx-\frac{1}{2}\int \frac{1}{x^2+1}dx$

$={\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{4}\int \frac{2x}{x^2+1}dx-\frac{1}{2}{\tan }^{-1}x$

$=\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln |x^2+1|+c$

নির্ণেয় যোজিত ফল : $\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln |x^2+1|+c$.

দেওয়া আছে, $Q(x,y)=9x^2+16y^2-144$

$f\left(x\right)=x-2$

এখানে, $Q(x,y)=0$

বা, $9x^2+16y^2-144=0$

বা, $16y^2=144-9x^2$

বা, $y^2=\frac{9(16-x^2)}{16}$

$\therefore y=\frac{3}{4}\sqrt{16-x^2}$

$f\left(x\right)=0$

বা, $x-2=0\therefore x=2$

$\therefore Q(x,y)=0$ এবং $f\left(x\right)=0$ দ্বারা আবদ্ধ ক্ষুদ্রতর অংশের ক্ষেত্রফল

ধরি, $x=4\sin \theta$

$\therefore dx=4\cos \theta d\theta$

$x=2$ হলে, $\theta =\frac{\pi }{6}$

$x=4$ হলে, $\theta =\frac{\pi }{2}$

$=2{\int }_2^{4}ydx=2\sqrt{\frac{3}{4}}\sqrt{16-x^2}dx$

$=\frac{3}{2}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\sqrt{16-(4\sin \theta {)}^2}\cdot 4\cos \theta d\theta$

$=6{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\sqrt{16(1-{\sin }^2\theta )}\cdot \cos \theta d\theta$

$=6{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\sqrt{16{\cos }^2\theta }\cdot \cos \theta =6{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}4\cos \theta \cdot \cos \theta d\theta$

$=12{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}2{\cos }^2\theta d\theta =12{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}(1+\cos 2\theta )d\theta$

$=12{[\theta +\frac{\sin 2\theta }{2}]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}=12(\frac{\pi }{2}+0-\frac{\pi }{6}-\frac{\sqrt{3}}{4})$

$=12(\frac{\pi }{3}-\frac{\sqrt{3}}{4})=4\pi -3\sqrt{3}$

নির্ণেয় ক্ষেত্রফল $(4\pi -3\sqrt{3})$ বর্গ একক।