সৃজনশীলঃ ‘f(x,y) = 4×2…’ কুমিল্লা বোর্ড ২০২২ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$\begin{array}{cc}& h\left(x\right)=\tan x\\ & h^{\prime }\left(x\right)={\sec }^{2}xdx\end{array}$

প্রদত্ত রাশি,

$\begin{array}{cc}& {\int }_0^{\frac{\pi }{4}}\frac{30h^{\prime }\left(x\right)\mathrm{dx}}{\left[\right\{h\left(x\right){\}}^2-9]\left\{h\right(x)-2\}}\\ & ={\int }_0^{\frac{\pi }{4}}\frac{3{\sec }^{2}xdx}{({\tan }^{2}x-9)(\tan x-2)}\end{array}$

$\begin{array}{cc}& ={\int }_0^1\frac{30dy}{(y^2-9)(y-2)}\\ & =30{\int }_0^1\frac{1dy}{(y+3)(y-3)(y-2)}\\ & =30{\int }_0^1[\frac{1}{(y+3)(-3-3)(-3-2)}+\frac{1}{(y-3)(3+3)(3-2)}+\frac{1}{(y-2)(2+3)(2-3)}]dy\\ & =30{\int }_0^1[\frac{1}{30(y+3)}+\frac{1}{6(y-3)}-\frac{1}{5(y-2)}]dy\\ & {=[\ln (y+z){]}_0^1+5[\ln (y-3)\left]\right]}_0-6[\ln (y-2){]}_0^1\\ & =\ln 4-\ln 3+5\left[\right(n|-2|-\ln |-3|]-16[|n|+||-\ln |-2|]\\ & =\ln \frac{4}{3}+5\ln \frac{2}{3}-6\ln \frac{1}{2}\\ & =\ln \frac{4}{3}+\ln {\left(\frac{2}{3}\right)}^5-\ln {\left(\frac{1}{2}\right)}^6\\ & =\ln (\frac{4}{3}\times {\left(\frac{2}{3}\right)}^5÷\frac{1}{2^6})\\ & =\ln (\frac{4}{3}\times {\left(\frac{2}{5}\right)}^5\times 2^6)\\ & =\ln \frac{2^{13}}{3^6}\\ & =\ln 2^{13}-\ln 3^6\\ & =13\ln 2-6\ln 3\\ & \end{array}$

দেওয়া আছে.

$\begin{array}{cc}& f(x,y)=4x^2+9y^2\\ & \therefore 36=4x^2+9y^2\\ & \Rightarrow \frac{4}{36}{x}^2+\frac{9}{36}{y}^2=\frac{36}{36}\\ & \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1\end{array}$

এখানে, $a=3,b=2$

আবদ্ধ ক্ষেত্রের ক্ষেত্রফল, $=4x$ এক চতর্থাংশের ক্ষেত্রফল,

এখন$\text{ এখন, }\begin{array}{cc}\frac{y^2}{4}& =1-\frac{x^2}{9}\\ \Rightarrow \frac{y^2}{4}& =\frac{9-x^2}{9}\\ \Rightarrow y^2& =\frac{4}{9}(9-x^2)\end{array}$

আবদ্ধক্ষেত্রেরক্ষেত্রফলবর্গএকক$\begin{array}{cc}& \Rightarrow y=\sqrt{\frac{4}{9}(9-x^2)}\\ & =\frac{2}{3}\sqrt{(9-x^2)}\\ & \therefore {\int }_0^{a}ydx=\frac{2}{3}{\int }_0^3\sqrt{9-x^2}dx/2)\\ & =\frac{2}{3}{[\frac{x\sqrt{3^2-x^2}}{2}+\frac{9}{2}{\sin }^{-1}\left(\frac{x}{3}\right)]}_0^3\\ & =\frac{2}{3}[0+\frac{9}{2}{\sin }^{-1}\left(\frac{3}{2}\right)-(0+0\left)\right]\\ & =\frac{2}{3}{[\frac{9}{2}\times \frac{\pi }{2}]}^{20}\times 1=\\ & =\frac{3\pi }{2}\\ & \therefore \text{ আবদ্ধ ক্ষেত্রের ক্ষেত্রফল }=4\times \frac{3\pi }{2}\\ & =6\pi \text{ বর্গ একক }\end{array}$