I =[X + 2F (X) − 3G (X)] DX−12Z 2G (X) DXF (X) DX − 3+ 2= X22= 32 + 2

Câu 33. Ta có: I =

[x + 2f (x) − 3g (x)] dx

−1

2

Z

2

g (x) dx

f (x) dx − 3

+ 2

= x

2

2

= 3

2 + 2.2 − 3 (−1) = 17

Chọn đáp án C