যোগজীকরণের সাহয্যে [গণিত] দ্বারা আবদ্ধ ক্ষেত্রের ক্ষেত্রফল নির্ণয় কর?
উত্তর: দেওয়া আছে. [গণিত] এখানে, [গণিত] আবদ্ধ ক্ষেত্রের ক্ষেত্রফল, [গণিত] এক চতর্থাংশের ক্ষেত্রফল, [গণিত]
HSC Higher Math 1st Paper Chapter 10: যোগজীকরণ CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 10: যোগজীকরণ · সৃজনশীল
যোগজীকরণের সাহয্যে দ্বারা আবদ্ধ ক্ষেত্রের ক্ষেত্রফল নির্ণয় কর?
উদ্দীপক
tanx.
Comilla · 2022
উত্তর
দেওয়া আছে,
let,
\begin{aligned}& \int \frac{h^{\prime}(x)}{\{1+h(x)\}^2} d x \\= & \int \frac{\sec^2 x}{(1+\tan x}^2} d x \\= & \int \frac{d z}{(1+z}^2} \\= & \frac{(1+z)^{-2+1}}{-2+1}+c \\= & -\frac{1}{1+z^2}+c=-\frac{1}{1+\tan x}+c\end{aligned}
উত্তর
দেওয়া আছে,
প্রদত্ত রাশি,
\begin{aligned}& \int_0^{\frac{\pi}{4}} \frac{30 \mathrm{~h}^{\prime}(\mathrm{x}) \mathrm{dx}}{\left[\{\mathrm{h}(\mathrm{x})\}^2-9\right]\{\mathrm{h}(\mathrm{x})-2\}} \\& =\int_0^{\frac{\pi}{4}} \frac{3 \sec^2 x d x}{\left(\tan^2 x-9\right}(\tan x-2)}\end{aligned}
\begin{aligned} & =\int_0^1 \frac{30 d y}{\left(y^2-9\right}(y-2)} \\ & =30 \int_0^1 \frac{1 d y}{(y+3}(y-3)(y-2)} \\ & =30 \int_0^1\left[\frac{1}{(y+3}(-3-3)(-3-2)}+\frac{1}{(y-3}(3+3)(3-2)}+\frac{1}{(y-2}(2+3)(2-3)}\right] d y \\ & =30 \int_0^1\left[\frac{1}{30(y+3)}+\frac{1}{6(y-3)}-\frac{1}{5(y-2)}\right] d y \\ & \left.=[\ln (y+z)]_0^1+5[\ln (y-3)]\right]_0-6[\ln (y-2)]_0^1 \\ & =\ln 4-\ln 3+5[(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 \left(\frac{4}{3} \times\left(\frac{2}{3}\right)^5 \div \frac{1}{2^6}\right) \\ & =\ln \left(\frac{4}{3} \times\left(\frac{2}{5}\right)^5 \times 2^6\right) \\ & =\ln \frac{2^{13}}{3^6} \\ & =\ln 2^{13}-\ln 3^6 \\ & =13 \ln 2-6 \ln 3 \\ & \end{aligned}
\begin{aligned} & =\int_0^1 \frac{30 d y}{\left(y^2-9\right}(y-2)} \\ & =30 \int_0^1 \frac{1 d y}{(y+3}(y-3)(y-2)} \\ & =30 \int_0^1\left[\frac{1}{(y+3}(-3-3)(-3-2)}+\frac{1}{(y-3}(3+3)(3-2)}+\frac{1}{(y-2}(2+3)(2-3)}\right] d y \\ & =30 \int_0^1\left[\frac{1}{30(y+3)}+\frac{1}{6(y-3)}-\frac{1}{5(y-2)}\right] d y \\ & \left.=[\ln (y+z)]_0^1+5[\ln (y-3)]\right]_0-6[\ln (y-2)]_0^1 \\ & =\ln 4-\ln 3+5[(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 \left(\frac{4}{3} \times\left(\frac{2}{3}\right)^5 \div \frac{1}{2^6}\right) \\ & =\ln \left(\frac{4}{3} \times\left(\frac{2}{5}\right)^5 \times 2^6\right) \\ & =\ln \frac{2^{13}}{3^6} \\ & =\ln 2^{13}-\ln 3^6 \\ & =13 \ln 2-6 \ln 3 \\ & \end{aligned}উত্তর
দেওয়া আছে.
\[\begin{aligned} & f(x, y)=4 x^2+9 y^2 \\ & \therefore \quad 36=4 x^2+9 y^2 \\ & \Rightarrow \frac{4}{36} x^2+\frac{9}{36} y^2=\frac{36}{36} \\ & \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1 \end{aligned}\]
এখানে,
আবদ্ধ ক্ষেত্রের ক্ষেত্রফল, এক চতর্থাংশের ক্ষেত্রফল,
\text{ এখন, } \[\begin{aligned}\frac{y^2}{4} & =1-\frac{x^2}{9} \Rightarrow \frac{y^2}{4} & =\frac{9-x^2}{9} \Rightarrow y^2 & =\frac{4}{9}\left(9-x^2\right)\end{aligned}\]