Question

${\int }_{-\pi /2}^{\pi /2}cosxdx$

• A. 1 unit
• B. 2 units
• C. 0 unit
• D. None of these

Answer 1

The correct option is B

2 units

$I={\int }_{-\pi /2}^{\pi /2}cosxdx=sinx{|}_{-\pi /2}^{\pi /2}=sin\frac{\pi }{2}-sin(-\frac{\pi }{2})=1+1=2\because {\int }_a^{b}f\left(x\right)dx=I\left(x\right){|}_a^b=I\left(b\right)-I\left(a\right)$