Question

If $△\left(x\right)=\left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ 1+\sin x& \cos x& 1+\sin x-\cos x\\ \sin x& \sin x& 1\end{array}\right|$, then ${\int }_0^{\pi /2}△\left(x\right)dx$ is equal to

• A. $1/4$
• B. $1/2$
• C. $0$
• D. $-1/2$

Answer 1

The correct option is C $0$

$\Delta x=\left|\begin{array}{c}1\cos x1-\cos x\\ 1+\sin x\cos x1+\sin x-\cos x\\ \sin x\sin x1\end{array}\right|$

$R_3\to R_3-R_2+R_1,R_2\to R_2-R_3-R_1\Rightarrow \Delta x=\left|\begin{array}{c}1\cos x1-\cos x\\ 0-\sin x\sin x-1\\ 0\sin x1-\sin x\end{array}\right|$

Expanding with $R_1$

$\Rightarrow \left|\Delta x\right|=-\sin x\left(1-\sin x\right)+\sin x\left(1-\sin x\right)\Rightarrow \left|\Delta x\right|=-\sin x+{\sin }^{2}x+\sin x-{\sin }^{2}x$

$\left|\Delta x\right|=0$

Therefore,

${\int }_0^{\frac{\pi }{2}}0dx=0Ans.$

Hence, this is the answer.