જો f(x) = begin{vmatrix} 1 + sin x + sin 2x + | vedclass

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(B) આપેલ છે કે, $f(x) = \begin{vmatrix} 1 + \sin x + \sin 2x + \sin 3x & \frac{3 + \sin 2x}{2} & \frac{-2 + \sin 3x}{3} \ 3 + 4 \sin x & \frac{3}{2}

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$\frac{-1}{9}$

Vedclass · Answer

(B) આપેલ છે કે, $f(x) = \begin{vmatrix} 1 + \sin x + \sin 2x + \sin 3x & \frac{3 + \sin 2x}{2} & \frac{-2 + \sin 3x}{3} \ 3 + 4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \ 1 + \sin x & \frac{1}{2} \sin x & \frac{1}{3} \end{vmatrix}$.સ્તંભ પ્રક્રિયા $C_1 ightarrow C_1 - 2C_2 - 3C_3$ લાગુ પાડતા, આપણને મળે છે:$f(x) = \begin{vmatrix} \sin x & \frac{3 + \sin 2x}{2} & \frac{-2 + \sin 3x}{3} \ 0 & \frac{3}{2} & \frac{4}{3} \sin x \ 0 & \frac{1}{2} \sin x & \frac{1}{3} \end{vmatrix}$.પ્રથમ સ્તંભની સાપેક્ષમાં વિસ્તરણ કરતા:$f(x) = \sin x \left( \frac{3}{2} \cdot \frac{1}{3} - \frac{4}{3} \sin x \cdot \frac{1}{2} \sin x ight) = \sin x \left( \frac{1}{2} - \frac{2}{3} \sin^2 x ight) = \frac{1}{6} (3 \sin x - 4 \sin^3 x) = \frac{\sin 3x}{6}$.હવે, $f^{\prime}(x) = \frac{d}{dx} \left( \frac{\sin 3x}{6} ight) = \frac{3 \cos 3x}{6} = \frac{\cos 3x}{2}$.આપણે $I = \int_0^{\pi / 2} (f(x) + f^{\prime}(x)) dx = \int_0^{\pi / 2} \left( \frac{\sin 3x}{6} + \frac{\cos 3x}{2} ight) dx$ ની ગણતરી કરવાની છે.$I = \left[ \frac{-\cos 3x}{18} + \frac{\sin 3x}{6} ight]_0^{\pi / 2}$.$I = \left( \frac{-\cos(3\pi/2)}{18} + \frac{\sin(3\pi/2)}{6} ight) - \left( \frac{-\cos(0)}{18} + \frac{\sin(0)}{6} ight)$.$I = \left( 0 - \frac{1}{6} ight) - \left( -\frac{1}{18} + 0 ight) = -\frac{1}{6} + \frac{1}{18} = \frac{-3 + 1}{18} = -\frac{2}{18} = -\frac{1}{9}$.

જો $f(x) = \begin{vmatrix} 1 + \sin x + \sin 2x + \sin 3x & \frac{3 + \sin 2x}{2} & \frac{-2 + \sin 3x}{3} \\ 3 + 4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1 + \sin x & \frac{1}{2} \sin x & \frac{1}{3} \end{vmatrix}$ હોય, તો $\int_0^{\pi / 2} (f(x) + f^{\prime}(x)) dx =$

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