Question

If $f\left(x\right)\left|\begin{array}{ccc}x& cosx& e^{x^2}\\ sinx& x^2& secx\\ tanx& x^4& 2x^2\end{array}\right|$ then ${\int }_{-\pi /2}^{\pi /2}f\left(x\right)dx=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$f\left(x\right)=\left|\begin{array}{ccc}x& \cos x& e^{x^2}\\ \sin x& x^2& \sec x\\ \tan x& x^4& 2x^2\end{array}\right|$

$=x[2x^4-x^4\sec x]-\cos x[2x^2\sin x+\sec x\tan x]+e^{x^2}[x^4\sin x-x^2\tan x]$

$=x^5[2-\sec x]-2x^2\sin x\cos x+\tan x+e^{x^2}{x}^4\sin x-e^{x^2}\tan x$

${\int }_{-\pi /2}^{\pi /2}\left[x^5\right[2-\sec x]-2x^2\sin x\cos x+\tan x+e^{x^2}{x}^4\sin x-e^{x^2}{x}^2\tan x]$

$[\frac{2x^6}{6}=x^5\sec x-2x^2\sin x\cos x-\tan x+e^{x^2}{x}^4\sin x-e^{x^2}{x}^2\tan x]$

$=0$