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AI-তৈরি নমুনা উত্তর

(a) ধরি,
$I = \int_{0}^{\pi/2} \frac{dx}{\sin x + \cos x}$
আমরা জানি, $\sin x + \cos x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x\right) = \sqrt{2}\cos\left(x - \frac{\pi}{4}\right)$
অতএব,
$I = \frac{1}{\sqrt{2}} \int_{0}^{\pi/2} \sec\left(x - \frac{\pi}{4}\right) dx$
$= \frac{1}{\sqrt{2}} \left[ \ln\left|\sec\left(x - \frac{\pi}{4}\right) + \tan\left(x - \frac{\pi}{4}\right)\right| \right]_{0}^{\pi/2}$
$= \frac{1}{\sqrt{2}} \left[ \ln\left|\sec\frac{\pi}{4} + \tan\frac{\pi}{4}\right| - \ln\left|\sec\left(-\frac{\pi}{4}\right) + \tan\left(-\frac{\pi}{4}\right)\right| \right]$
$= \frac{1}{\sqrt{2}} [ \ln(\sqrt{2} + 1) - \ln(\sqrt{2} - 1) ]$
$= \frac{1}{\sqrt{2}} \ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) = \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1)^2 = \sqrt{2}\ln(\sqrt{2}+1)$

(b) প্রদত্ত সমীকরণদ্বয়: $y^2 = x$ এবং $y = x$
ছেদবিন্দুর জন্য, $x^2 = x \implies x(x - 1) = 0 \implies x = 0, 1$
সীমিত অঞ্চলের ক্ষেত্রফল:
$A = \int_{0}^{1} (\sqrt{x} - x) dx = \left[ \frac{2}{3}x^{3/2} - \frac{x^2}{2} \right]_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6} \text{ বর্গ একক}$

উত্তর: (a) $\sqrt{2}\ln(\sqrt{2}+1)$ এবং (b) $\frac{1}{6}$ বর্গ একক।