গণিত প্রশ্ন

AI-তৈরি নমুনা উত্তর

ধরি, $I = \int_{-1}^{1} x^2 \sqrt{4-x^2}\,dx$

যেহেতু $f(-x) = (-x)^2 \sqrt{4-(-x)^2} = x^2 \sqrt{4-x^2} = f(x)$, অর্থাৎ ফাংশনটি জোড় (even),
সুতরাং,
$I = 2 \int_{0}^{1} x^2 \sqrt{4-x^2}\,dx$

ধরি, $x = 2\sin\theta$
$\implies dx = 2\cos\theta\,d\theta$

সীমা পরিবর্তন:
যখন $x = 0$, তখন $\theta = 0$
যখন $x = 1$, তখন $2\sin\theta = 1 \implies \theta = \frac{\pi}{6}$

অতএব,
$I = 2 \int_{0}^{\pi/6} (4\sin^2\theta) \sqrt{4 - 4\sin^2\theta} \cdot (2\cos\theta)\,d\theta$
$= 2 \int_{0}^{\pi/6} (4\sin^2\theta) (2\cos\theta) (2\cos\theta)\,d\theta$
$= 32 \int_{0}^{\pi/6} \sin^2\theta \cos^2\theta\,d\theta$
$= 8 \int_{0}^{\pi/6} (2\sin\theta\cos\theta)^2\,d\theta$
$= 8 \int_{0}^{\pi/6} \sin^2(2\theta)\,d\theta$
$= 4 \int_{0}^{\pi/6} (1 - \cos 4\theta)\,d\theta$
$= 4 \left[ \theta - \frac{\sin 4\theta}{4} \right]_{0}^{\pi/6}$
$= 4 \left[ \frac{\pi}{6} - \frac{\sin(2\pi/3)}{4} \right] - 0$
$= 4 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right)$
$= \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$

উত্তর: $\frac{2\pi}{3} - \frac{\sqrt{3}}{2}$