সৃজনশীলঃ ‘g(x,y)=16×2+25y2−400, F(z)=zln⁡z.…’ চট্টগ্রাম বোর্ড ২০২২ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

এখানে, $\int f\left(x\right)dx$

$=\int x\ln xdx$

$\begin{array}{cc}& =\ln x\int xdx-\int \frac{d}{dx}(\ln x)\int xdxdx\\ & =\frac{\ln x\times x^2}{2}-\int (\frac{1}{x}\times \frac{x^2}{2})dx\\ & =\frac{x^2\ln x}{2}-\int \frac{x}{2}dx\end{array}$

$=\frac{x^2\ln x}{2}-\frac{x^2}{2\times 2}+c$[C একটি যোগত্রীকর ধ্রুবক]

$=\frac{x^2\ln x}{2}-\frac{x^2}{4}+C$

এখন ${\int }_1^{\sqrt{e}}f\left(x\right)dx$

$={[\frac{x^2\ln x}{2}-\frac{x^2}{4}]}^{\sqrt{e}}1$

$=[\frac{e\ln e}{2}-\frac{e}{4})-(\frac{\ln 1}{2}-1/4)$

$=(\frac{e}{2}-\frac{e}{4})-(0-\frac{1}{4})$

$=\frac{e}{2}-\frac{e}{4}+1/4$$=\frac{1}{4}$ Ans

দেওয়া আছে,

$g(x,y)=16x^2+25y^2-400=0$

$\Rightarrow 16x+25y^2=400$

$\frac{x^2}{25}+\frac{y^2}{16}=1$

$\Rightarrow y^2=1-\frac{x^2}{25}$

$\Rightarrow \frac{y^2}{16}=\frac{25-x^2}{25}$

$\Rightarrow y=\sqrt{16\times \frac{25-x^2}{25}}$

$\Rightarrow y=\pm 4\frac{\sqrt{25-x^2}}{5}$

$\Rightarrow y=\pm \frac{4}{5}\sqrt{25-x^2}$

'+' চিহ্ন নিয়ে, 

$y=\frac{4}{5}\sqrt{25-x^2}$

$x-$ অক্ষ্যে $y=0$ বসিয়ে 

$0=\frac{4}{5}\sqrt{25-x^2}$

$\Rightarrow x=\pm 5$

$\therefore$ ক্ষেত্রফল $=2{\int }_{-5}^5\frac{4}{5}\sqrt{25-x^2}dx$

$=2\times \frac{4}{5}{\int }_{-5}^5\sqrt{25-x^2}dx$

এখন, $\int \sqrt{25-x^2}2x$

let, $x=5\sin \theta$

$=\int \sqrt{25-25{\sin }^2\theta }dxdx=5\cos \theta d\theta$

$=\int \sqrt{25(1-{\sin }^2\theta )}dx$

$=\int 5\cos \theta dx$

$=\int 5\mathrm{cst}5\cos \theta d\theta$

$=\int 25{\cos }^2\theta d\theta$

$=\frac{25}{2}\int 2{\cos }^2\theta d\theta$

$=\frac{25}{2}\int (1+\cos 2\theta )d\theta$

$=\frac{25}{2}(\theta +\frac{\sin 2\theta }{2})+c$ [যেখানে c একটি যোগজীকরন ধ্রুবক]

$=\frac{25}{2}(\theta +\frac{\sin 2\theta }{2})+c$

এখন, ${\int }_{-5}^5\sqrt{25-x^2}$

$=\frac{25}{2}{[\theta +\frac{\sin 2\theta }{2}]}_{-\pi /2}^{\pi /2}$

$=\frac{25}{2}[(\frac{\pi }{2}+o)-(-\pi /2)+0]$

$=\frac{25}{2}(\frac{\pi }{2}+\frac{\pi }{2})$

$=\frac{25}{2}\pi$

$=\frac{25\pi }{2}$

$\therefore$ ক্ষেত্রফল $=2\times \frac{4}{5}{\int }_{-5}^5\sqrt{25-x^2}dx$

$=2\times \frac{4}{5}\times \frac{-25\pi }{2}$

$=20\pi$

$\therefore$ অর্ধক্ষেত্রফল $=10\pi$ (Ans)