সৃজনশীলঃ ‘f(x)=cot−1⁡x এবং y2=4ax,x2=4ay.…’ মাইলস্টোন কলেজ ২০২৫ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

[বি. দ্র.: $g\left(x\right)$ এর পরিবর্তে $f\left(x\right)$ হবে)

দেওয়া আছে,

$\begin{array}{cc}& g\left(x\right)={\cot }^{-1}x\\ & \therefore {\int }_1^{\sqrt{3}}xg\left(x\right)dx={\int }_1^{\sqrt{3}}x{\cot }^{-1}xdx\\ & =[{\cot }^{-1}x{\int }_{1}xdx-{\left\{\frac{d}{dx}\right(\cot x\left){\int }_{1}xdxdx\right]}_1^{\sqrt{3}}\\ & ={[\frac{x^2}{2}{\cot }^{-1}x+{\int }_1^2\frac{x^2}{2+x^2)}dx]}_1^{\sqrt{3}}\\ & ={[\frac{x^2}{2}{\cot }^{-1}x+\frac{1}{2}{\int }_1^2\frac{1}{1+x^2}dx]}_1^{\sqrt{3}}\\ & ={[\frac{x^2}{2}{\cot }^{-1}x+\frac{1}{2}{\int }_1^{2}1-\frac{1}{1+x^2}dx]}_1^{\sqrt{3}}\end{array}$

$\begin{array}{cc}& ={[\frac{x^2}{2}{\cot }^{-1}x+\frac{x}{2}+\frac{1}{2}{\cot }^{-1}x+c]}_1^{\sqrt{3}}\\ & \therefore {\int }_1^{\sqrt{3}}x{\cot }^{-1}xdx\\ & ={[\frac{x^2}{2}{\cot }^{-1}x+\frac{x}{2}+\frac{1}{2}{\cot }^{-1}x]}_1^{\sqrt{3}}\\ & =\frac{1}{2}{[x^2{\cot }^{-1}x]}_1^{\sqrt{3}}+\frac{1}{2}[x{]}_1^{\sqrt{3}}+\frac{1}{2}{[{\cot }^{-1}x]}_1^{\sqrt{3}}\\ & =\frac{1}{2}[3{\cot }^{-1}\sqrt{3}-{\cot }^{-1}\frac{1}{11}]+\frac{1}{2}[\sqrt{3}-1]\\ & +\frac{1}{2}({\cot }^{-1}\sqrt{3}-{\cot }^{-1}1)\\ & =\frac{1}{2}(3\cdot \frac{\pi }{6}-\frac{\pi }{4})+\frac{1}{2}(\sqrt{3}-1)+\frac{1}{2}(\frac{\pi }{6}-\frac{\pi }{4})\\ & =\frac{\pi }{4}-\frac{\pi }{8}+\frac{1}{2}(\sqrt{3}-1)+\frac{\pi }{12}-\frac{\pi }{8}\\ & =\frac{1}{2}(\sqrt{3}-1)+\frac{\pi }{12}\\ & =\frac{1}{2}(\sqrt{3}-1+\frac{\pi }{6})\end{array}$

ইহাই নির্ণেয় মান।  

এখানে, $y^2=4\mathrm{ax}$

বা, $x=\frac{y^2}{4a}$

আবার, $x^2=4ay$

বা, ${\left(\frac{y^2}{4a}\right)}^2=4ay$

বা, $y^4=64a^{3}y$ বা, $y(y^3-64a^3)=0$

$\therefore$ হয়, $y=0$

অথবা, $y^3-64a^3=0$

বা, $y^3=(4a{)}^3$

$\therefore y=4a$

$y=4a$ হলে, $x^2=16a^2$

$\therefore x=4a$

$\therefore$ ছেদবিন্দু C (4 a, 4 a ̀)

নির্ণেয় ক্ষেত্রফল = ক্ষেত্র $OACN$ – ক্ষেত্র $OBCN$

$={\int }_0^{4a}(2a^{\frac{1}{2}}{x}^{\frac{1}{2}}-\frac{x^2}{4a})dx=2a^{\frac{1}{2}}{\int }_0^{4a}{x}^{\frac{1}{2}}dx-\frac{1}{4a}{\int }_0^{4a}{x}^{2}dx$

$=2a^{\frac{1}{2}}\cdot \frac{2}{3}{\left[\left(x^{3/2}\right)\right]}_0^{4a}-\frac{1}{4a}{\left[\frac{x^3}{3}\right]}_0^{4a}$

$=\frac{4a^{\frac{1}{2}}}{3}\cdot (4a{)}^{\frac{3}{2}}-\frac{1}{4a}\cdot \frac{1}{3}(4a{)}^3=\frac{32}{3}{a}^{\frac{1}{2}}\cdot a^{\frac{3}{2}}-\frac{16}{3}{a}^2$

$=\frac{32}{3}{a}^2-\frac{16}{3}{a}^2=\frac{32a^2-16a^2}{3}=\frac{16}{3}{a}^2$

নির্ণেয় ক্ষেত্রফল $\frac{16}{3}{a}^2$ বর্গ একক।