Hoặc
A. ∫−122x2−2x−4dx
B. ∫−12(−2x+2)dx
C. ∫−12(2x−2)dx
D. ∫−12−2x2+2x+4dx
Ta thấy ∀x∈[−1;2] thì −x2+3≥x2−2x−1 nên S=∫−12−x2+3−x2−2x−1dx=∫−12−2x2+2x+4dx.