Question

$\int_{-π/2}^{π/2} f(x) \,dx$ =?
Where f(x) = sin |x| + cos |x|, x ∈ $(-\frac {π}{2}, \frac {π}{2})$

• A. 0
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (C)

4

Answer 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