Biết pi/ 2 ∫ 0 ( 2 x − 1 − sin x ) dx = pi (pi/ a − 1 /b ) − 1 ; a , b ∈ Z . Tính a + b .

\(\int\limits_0^{\frac{\pi }{2}} {\left( {2x - 1 - \sin x} \right)dx} = \left. {\left( {{x^2} - x + \cos x} \right)} \right|_0^{\frac{\pi }{2}} = \frac{{{\pi ^2}}}{4} - \frac{\pi }{2} - 1 = \pi \left( {\frac{\pi }{4} - \frac{1}{2}} \right) - 1\).

Suy ra \(a = 4;b = 2\). Suy ra \(a + b = 6\).

Trả lời: 6.