2. ## Re: Can someone please check where my mistake is

You only need to find:

$\int ^{\frac {\pi}{6}} _{-\frac {\pi}{2}} (cos x - sin x) dx = \left [ sin x + cos x \right ]^* _* = (\frac {1}{2} + \frac {\sqrt 3}{2} ) - (-1 + 0 ) = \frac {\sqrt 3}{2} + \frac {3}{2}$

3. ## Re: Can someone please check where my mistake is

You did it correctly but in the end, when you were doing the integral for (cosx - sinx), you forgot to attach -(sin0 + cos0) in your third step. That gives you -1, which in the end would give you the correct answer

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