সৃজনশীলঃ ‘P = (x–4)2…’ সিলেট বোর্ড ২০২২ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

প্রশ্নমতে,

$\int \frac{x}{p}dx$

$=\int \frac{x}{(x-4{)}^2(x-3)}dx$

$=\int \frac{1}{(x-4)}\left[\frac{x}{(x-4)(x-3)}\right]dx$

$=\int \frac{1}{(x-4)}[\frac{4}{(x-4)(4-3)}+\frac{3}{(x-3)(3-4)}]dx$

$=\int \frac{4}{(x-4{)}^2}dx-\int \frac{3}{(x-3)(x-4)}dx$

$=4\cdot \frac{(x-4)}{-2+1}-3\int \{\frac{1}{(x-3)(3-4)}+\frac{1}{(x-4)(4-3)}\}dx$

$=\frac{-4}{x-4}-3\int (\frac{-1}{x-3}+\frac{1}{x-4})dx$

$=\frac{-4}{x-4}-3\int (\frac{1}{x-4}-\frac{1}{x-3})dx$

$=\frac{-4}{x-4}-3\ln |x-4|-3\ln |x-3|+c$

দেওয়া আছে, $g(x,y)=x^2+y^2$ এবং $g(x,y)=100,x=5$

$\therefore x^2+y^2=100$

$◻y=\pm \sqrt{100-x^2}$

$◻y=\pm \sqrt{{10}^2-x^2}$.

$x=5$ ......... (ii)

এখন, (i) ও (ii) নং দ্বারা আবদ্ধ অংশের ক্ষেত্রফল,

$2{\int }_5^{10}ydx$

$=2{\int }_5^{10}\sqrt{{10}^2-x^2}dx$

$=2{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\sqrt{{10}^2-{10}^2{\sin }^2\theta }\cdot 10\cos \theta d\theta$

$=2{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}10\sqrt{1-{\sin }^2\theta }\cdot 10\cos \theta d\theta$

$=20{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}10{\cos }^2\theta d\theta$

$=200{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta$

$=200{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{1}{2}\cdot 2{\cos }^2\theta d\theta$

$=100{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{1}{2}\cdot (1+{\cos }^2\theta )d\theta$

$=100{[\theta +\frac{\sin 2\theta }{2}]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}$

$=100[(\frac{\pi }{2}+\frac{sin\frac{\pi }{2}}{2})-(\frac{\pi }{6}+\frac{sin\frac{\pi }{3}}{2})]$

$=100[\frac{\pi }{2}+0-\frac{\pi }{6}-\frac{\sqrt{3}}{4}]$

$=100[\frac{\pi }{3}-\frac{\sqrt{3}}{4}]$

$=\frac{100\pi }{3}-25\sqrt{3}$