Given,f(x)=0x2−sin ⁡xcos ⁡x−2sin ⁡x−x201−2x2−cos⁡ x2x−10' then ∫f(x)dx is equal to

We have, ⇒ f(x)=0x2−sin ⁡xcos ⁡x−2sin ⁡x−x201−2x2−cos ⁡x2x−10f(x)=0sin⁡ x−x22−cos ⁡ xx2−sin ⁡x02x−1cos ⁡x−21−2x0[interchanging rows and columns]⇒f(x)=(−1)30x2−sin ⁡xcos⁡ x−2sin⁡ x−x201−2x2−cos ⁡x2x−10[taking (-1) common from each column]⇒ f(x)=−f(x)⇒f(x)=0⇒ ∫f(x)dx=c

Given,f(x)=0x2−sin ⁡xcos ⁡x−2sin ⁡x−x201−2x2−cos⁡ x2x−10' then ∫f(x)dx is equal to

We have, ⇒ f(x)=0x2−sin ⁡xcos ⁡x−2sin ⁡x−x201−2x2−cos ⁡x2x−10f(x)=0sin⁡ x−x22−cos ⁡ xx2−sin ⁡x02x−1cos ⁡x−21−2x0[interchanging rows and columns]⇒f(x)=(−1)30x2−sin ⁡xcos⁡ x−2sin⁡ x−x201−2x2−cos ⁡x2x−10[taking (-1) common from each column]⇒ f(x)=−f(x)⇒f(x)=0⇒ ∫f(x)dx=c

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Given, $f (x) = |\begin{array}{ccc} 0 & x^{2} - \sin x & \cos x - 2 \\ \sin x - x^{2} & 0 & 1 - 2 x \\ 2 - \cos x & 2 x - 1 & 0 \end{array}|$ ' then $\int f (x) dx$ is equal to

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$\frac{x^{3}}{3} - x^{2} \sin x + \sin 2 x + C$

$\frac{x^{3}}{3} - x^{2} \sin x - \cos 2 x + C$

$\frac{x^{3}}{3} - x^{2} \cos x - \cos 2 x + C$

None of the above

answer is D.

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Detailed Solution

We have,

$\Rightarrow$ $\begin{aligned} f (x) = |\begin{array}{ccl} 0 & x^{2} - \sin x & \cos x - 2 \\ \sin x - x^{2} & 0 & 1 - 2 x \\ 2 - \cos x & 2 x - 1 & 0 \end{array}| \\ f (x) = |\begin{array}{ccc} 0 & \sin x - x^{2} & 2 - \cos x \\ x^{2} - \sin x & 0 & 2 x - 1 \\ \cos x - 2 & 1 - 2 x & 0 \end{array}| \end{aligned}$

[interchanging rows and columns]

$\Rightarrow$ $f (x) = (- 1)^{3} |\begin{array}{ccc} 0 & x^{2} - \sin x & \cos x - 2 \\ \sin x - x^{2} & 0 & 1 - 2 x \\ 2 - \cos x & 2 x - 1 & 0 \end{array}|$