সৃজনশীলঃ ‘…’ বরিশাল বোর্ড ২০১৭ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

(i) $\begin{array}{cc}& {\int }_0^{\pi /2}\sqrt{1+g\left(\theta \right)}d\theta \\ & ={\int }_0^{\pi /2}\sqrt{1+\sin \theta }d\theta \\ & ={\int }_0^{\pi /2}\sqrt{{\sin }^2\frac{\theta }{2}+{\cos }^2\frac{\theta }{2}+2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}d\theta \\ & ={\int }_0^{\pi /2}\sqrt{{(\sin \frac{\theta }{2}+\cos \frac{\theta }{2})}^2}d\theta \\ & ={\int }_0^{\pi /2}(\sin \frac{\theta }{2}+\cos \frac{\theta }{2})d\theta \\ & ={-\frac{\cos \frac{\theta }{2}}{\frac{1}{2}}+\frac{\sin \frac{\theta }{2}}{\frac{1}{2}}]}_0^{\frac{\pi }{2}}\\ & =-2\cos \frac{\pi }{4}+2\sin \frac{\pi }{4}+2\cos 0-2\sin 0\\ & =-2\cdot \frac{1}{\sqrt{2}}+2\cdot \frac{1}{\sqrt{2}}+2-0\\ & =2\left(\text{ Ans. }\right)\end{array}$

ধরি, $z=\sin \theta$

$\therefore \mathrm{dz}=\cos \theta d\theta$

$\begin{array}{cc}& \text{ (ii) }{\int }_0^{\pi /2}f\left(\theta \right)\sqrt[3]{g\left(\theta \right)}d\theta \\ & ={\int }_0^{\pi /2}{\cos }^3\theta \sqrt[3]{\sin \theta }d\theta \\ & ={\int }_0^{\pi /2}(1-{\sin }^2\theta )\sqrt[3]{\sin \theta }\cdot \cos \theta d\theta \\ & ={\int }_0^1(1-z^2)z^{\frac{1}{3}}dz\\ & ={\int }_0^1(z^{\frac{1}{3}}-z^{\frac{7}{3}})dz\end{array}$

$\begin{array}{cc}& ={[\frac{z^{\frac{1}{3}+1}}{\frac{1}{3}+1}-\frac{z^{\frac{7}{3}}+1}{\frac{7}{3}+1}]}_0^1\\ & ={[\frac{3}{4}\cdot z^{\frac{4}{3}}-\frac{3}{10}{z}^{\frac{10}{3}}]}_0^1\\ & =\frac{3}{4}-\frac{3}{10}-0+0\\ & =\frac{15-6}{20}=\frac{9}{20}\text{ (Ans.) }\end{array}$

নির্ণেয় ক্ষেত্রফল $=4\times (y=\sqrt{36-x^2},x$ অক্ষ এবং কোটি $x=0$ ও $x=6$ দ্বারা আবদ্ধ ক্ষুত্রের ক্ষেত্রফল)

ধরি, $x=6\sin \theta$

$\therefore dx=6\cos \theta d\theta$

এবং $\theta ={\sin }^{-1}\frac{x}{6}$

$\begin{array}{cc}& =4{\int }_0^{6}ydx\\ & =4{\int }_0^6\sqrt{36-x^2}dx\\ & =4{\int }_0^{\pi /2}\sqrt{36-36{\sin }^2\theta }\cdot 6\cos \theta d\theta \\ & =144{\int }_0^{\pi /2}\sqrt{{\cos }^2\theta }\cos \theta d\theta \\ & =144{\int }_0^{\pi /2}{\cos }^2\theta d\theta \\ & =72{\int }_0^{\pi /2}2{\cos }^2\theta d\theta \\ & =72{\int }_0^{\pi /2}(1+\cos 2\theta )d\theta \\ & =72{[\theta +\frac{1}{2}\sin 2\theta ]}_0^{\frac{\pi }{2}}\\ & =72[\frac{\pi }{2}+0-0]\end{array}$

$=36\pi$ বগ একক। (Ans.)