Giải cho tui câu d thôi

`d)`

Ta có : `f(x) = 1 + \sin x; g(x) = e^x + 2`

Xét `g(x) > f(x)` (vì `e^x + 2 > 1 + \sin x`) nên:

`S = \int_0^{\frac{\pi}{2}} [ g(x) - f(x) ] dx`

`= \int_0^{\frac{\pi}{2}} ( e^x + 2 - 1 - \sin x ) dx`

`= \int_0^{\frac{\pi}{2}} (e^x - \sin x + 1) dx`

`->S = [ e^x + \cos x + x ]_0^{\frac{\pi}{2}}`

`= ( e^{\frac{\pi}{2}} + \cos \frac{\pi}{2} + \frac{\pi}{2}) - (1 + \cos 0 + 0)`

`= e^{\frac{\pi}{2}} + 0 + \frac{\pi}{2} - (1 + 1)`

`= e^{\frac{\pi}{2}} + \frac{\pi}{2} - 2`

 Theo đề bài ta có : 

`S = e^a + \frac{\pi}{b} - c`

`->a + b - c = \frac{\pi}{2} + 2 - 2 = \frac{\pi}{2}`

`->Sai.`