If f(x)=sinxsin2xsin3x034sinx0sinx1, then ∫0π/2f(x)dx=

f(x)=Sin xSin 2xSin 3x034 Sin x0Sinx1 =Sin x 3-4 Sin2x =3 Sin x-4 Sin3x=Sin 3x ∫0π2 f(x) dx=∫0π2Sin 3x dx =-13cos3x0π2 =13cos 0-cos 3π2 =131-0=13

If f(x)=sinxsin2xsin3x034sinx0sinx1, then ∫0π/2f(x)dx=

f(x)=Sin xSin 2xSin 3x034 Sin x0Sinx1 =Sin x 3-4 Sin2x =3 Sin x-4 Sin3x=Sin 3x ∫0π2 f(x) dx=∫0π2Sin 3x dx =-13cos3x0π2 =13cos 0-cos 3π2 =131-0=13

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$I f f (x) = |\begin{matrix} \sin x & \sin 2 x & \sin 3 x \\ 0 & 3 & 4 \sin x \\ 0 & \sin x & 1 \end{matrix}|, t h e n \int_{0}^{π / 2} f (x) d x =$

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$\frac{2}{3}$

$\frac{1}{3}$

answer is C.

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Detailed Solution

$f (x) = |\begin{matrix} S i n x & S in 2 x & S i n 3 x \\ 0 & 3 & 4 S i n x \\ 0 & S i n x & 1 \end{matrix}| = S i n x (3 - 4 S i n^{2} x) = 3 S i n x - 4 S i n^{3} x = S i n 3 x \int_{0}^{\frac{π}{2}} f (x) dx = \int_{0}^{\frac{π}{2}} S i n 3 x dx = {(\frac{- 1}{3} \cos 3 x)}_{0}^{\frac{π}{2}} = \frac{1}{3} [\cos 0 - \cos \frac{3 π}{2}] = \frac{1}{3} [1 - 0] = \frac{1}{3}$