ক) AI-তৈরি নমুনা উত্তর
ধরি, I = ∫ x e − x d x I = \int x e^{-x} dx I = ∫ x e − x d x I = x ∫ e − x d x − ∫ ( d d x ( x ) ∫ e − x d x ) d x I = x \int e^{-x} dx - \int \left( \frac{d}{dx}(x) \int e^{-x} dx \right) dx I = x ∫ e − x d x − ∫ ( d x d ( x ) ∫ e − x d x ) d x = x ( − e − x ) − ∫ ( 1 ) ( − e − x ) d x = x(-e^{-x}) - \int (1)(-e^{-x}) dx = x ( − e − x ) − ∫ ( 1 ) ( − e − x ) d x = − x e − x + ∫ e − x d x = -x e^{-x} + \int e^{-x} dx = − x e − x + ∫ e − x d x = − x e − x − e − x + c = -x e^{-x} - e^{-x} + c = − x e − x − e − x + c = − ( x + 1 ) e − x + c = -(x + 1)e^{-x} + c = − ( x + 1 ) e − x + c c c c
খ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, P = 4 − x 2 P = 4 - x^2 P = 4 − x 2 4 − P = x 2 4 - P = x^2 4 − P = x 2 ∫ 4 − P P d x = ∫ x 2 4 − x 2 d x \int \frac{4-P}{P} dx = \int \frac{x^2}{4 - x^2} dx ∫ P 4 − P d x = ∫ 4 − x 2 x 2 d x = ∫ − ( 4 − x 2 ) + 4 4 − x 2 d x = \int \frac{-(4 - x^2) + 4}{4 - x^2} dx = ∫ 4 − x 2 − ( 4 − x 2 ) + 4 d x = ∫ ( − 1 + 4 4 − x 2 ) d x = \int \left( -1 + \frac{4}{4 - x^2} \right) dx = ∫ ( − 1 + 4 − x 2 4 ) d x = − ∫ d x + 4 ∫ 1 2 2 − x 2 d x = -\int dx + 4 \int \frac{1}{2^2 - x^2} dx = − ∫ d x + 4 ∫ 2 2 − x 2 1 d x = − x + 4 ⋅ 1 2 ( 2 ) ln ∣ 2 + x 2 − x ∣ + c = -x + 4 \cdot \frac{1}{2(2)} \ln\left|\frac{2 + x}{2 - x}\right| + c = − x + 4 ⋅ 2 ( 2 ) 1 ln 2 − x 2 + x + c = − x + ln ∣ 2 + x 2 − x ∣ + c = -x + \ln\left|\frac{2 + x}{2 - x}\right| + c = − x + ln 2 − x 2 + x + c c c c
গ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, P = 4 − x 2 P = 4 - x^2 P = 4 − x 2 4 − P = x 2 4 - P = x^2 4 − P = x 2 I = ∫ − 1 1 ( 4 − P ) P d x = ∫ − 1 1 x 2 4 − x 2 d x I = \int_{-1}^{1} (4-P)\sqrt{P}\, dx = \int_{-1}^{1} x^2 \sqrt{4 - x^2}\, dx I = ∫ − 1 1 ( 4 − P ) P d x = ∫ − 1 1 x 2 4 − x 2 d x f ( x ) = x 2 4 − x 2 f(x) = x^2\sqrt{4-x^2} f ( x ) = x 2 4 − x 2 f ( − x ) = f ( x ) f(-x) = f(x) f ( − x ) = f ( x ) I = 2 ∫ 0 1 x 2 4 − x 2 d x I = 2 \int_{0}^{1} x^2 \sqrt{4 - x^2}\, dx I = 2 ∫ 0 1 x 2 4 − x 2 d x
ধরি, x = 2 sin θ ⟹ d x = 2 cos θ d θ x = 2\sin\theta \implies dx = 2\cos\theta\, d\theta x = 2 sin θ ⟹ d x = 2 cos θ d θ x = 0 x = 0 x = 0 θ = 0 \theta = 0 θ = 0 x = 1 x = 1 x = 1 2 sin θ = 1 ⟹ sin θ = 1 2 ⟹ θ = π 6 2\sin\theta = 1 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6} 2 sin θ = 1 ⟹ sin θ = 2 1 ⟹ θ = 6 π
অতএব,I = 2 ∫ 0 π / 6 ( 4 sin 2 θ ) 4 − 4 sin 2 θ ⋅ ( 2 cos θ d θ ) I = 2 \int_{0}^{\pi/6} (4\sin^2\theta) \sqrt{4 - 4\sin^2\theta} \cdot (2\cos\theta\, d\theta) I = 2 ∫ 0 π /6 ( 4 sin 2 θ ) 4 − 4 sin 2 θ ⋅ ( 2 cos θ d θ ) = 2 ∫ 0 π / 6 4 sin 2 θ ⋅ ( 2 cos θ ) ⋅ 2 cos θ d θ = 2 \int_{0}^{\pi/6} 4\sin^2\theta \cdot (2\cos\theta) \cdot 2\cos\theta\, d\theta = 2 ∫ 0 π /6 4 sin 2 θ ⋅ ( 2 cos θ ) ⋅ 2 cos θ d θ = 16 ∫ 0 π / 6 ( 2 sin θ cos θ ) 2 d θ = 16 ∫ 0 π / 6 sin 2 ( 2 θ ) d θ = 16 \int_{0}^{\pi/6} (2\sin\theta\cos\theta)^2 d\theta = 16 \int_{0}^{\pi/6} \sin^2(2\theta) d\theta = 16 ∫ 0 π /6 ( 2 sin θ cos θ ) 2 d θ = 16 ∫ 0 π /6 sin 2 ( 2 θ ) d θ = 8 ∫ 0 π / 6 2 sin 2 ( 2 θ ) d θ = 8 ∫ 0 π / 6 ( 1 − cos 4 θ ) d θ = 8 \int_{0}^{\pi/6} 2\sin^2(2\theta) d\theta = 8 \int_{0}^{\pi/6} (1 - \cos 4\theta) d\theta = 8 ∫ 0 π /6 2 sin 2 ( 2 θ ) d θ = 8 ∫ 0 π /6 ( 1 − cos 4 θ ) d θ = 8 [ θ − sin 4 θ 4 ] 0 π / 6 = 8 \left[ \theta - \frac{\sin 4\theta}{4} \right]_{0}^{\pi/6} = 8 [ θ − 4 s i n 4 θ ] 0 π /6 = 8 ( π 6 − sin ( 2 π / 3 ) 4 ) − 0 = 8 \left( \frac{\pi}{6} - \frac{\sin(2\pi/3)}{4} \right) - 0 = 8 ( 6 π − 4 s i n ( 2 π /3 ) ) − 0 = 8 ( π 6 − 3 2 4 ) = 8 ( π 6 − 3 8 ) = 4 π 3 − 3 = 8 \left( \frac{\pi}{6} - \frac{\frac{\sqrt{3}}{2}}{4} \right) = 8 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right) = \frac{4\pi}{3} - \sqrt{3} = 8 ( 6 π − 4 2 3 ) = 8 ( 6 π − 8 3 ) = 3 4 π − 3
অতএব, নির্ণেয় মান 4 π 3 − 3 \frac{4\pi}{3} - \sqrt{3} 3 4 π − 3