\(\int_{-π/2}^{π/2} f(x) \,dx\) =?
Where f(x) = sin |x| + cos |x|, x ∈ \((-\frac {π}{2}, \frac {π}{2})\)
We have f(x) = sin|x| + cos|x|
Then, f(x) =f(–x) Since, (f(x) is an even function.
I = \(\int_{-π/2}^{π/2}sin|x| + cos|x| \,dx\)
I = 2 \(\int_{0}^{π/2}sin|x| + cos|x| \,dx\)
I = 2[-cosx + sinx]π/20
I = 2[-cos \(\frac {π}{2}\) + sin \(\frac {π}{2}\) + cos 0 - sin 0]
I = 2[0 + 1 + 1–0]
I = 2x2
I = 4
Therefore, the correct option is (C) 4