[imath]\displaystyle \int^{1}_{-1}{f(x)}\mathrm{d}x + \displaystyle \int^{2}_{1}{f(x)}\mathrm{d}x = \displaystyle \int^{2}_{-1}{f(x)}\mathrm{d}x[/imath]
[imath]\Rightarrow \displaystyle \int^{1}_{-1}{f(x)}\mathrm{d}x =2-5=-3[/imath]
Đặt [imath]3-2x=t \Rightarrow \mathrm{d}t=-2 \mathrm{d}x \\
\Rightarrow \mathrm{d}x= \left ( \dfrac{-1}{2} \right ) \mathrm{d}t[/imath]
Đổi cận:
[imath]x=1 \Rightarrow t=1 \\
x=2 \Rightarrow t= -1[/imath]
Do đó ta có [imath] \left ( \dfrac{-1}{2} \right ) \displaystyle \int^{-1}_{1}{f(t)}\mathrm{d}t = \dfrac{1}{2} \displaystyle \int^{1}_{-1}{f(t)}\mathrm{d}t \\
= \dfrac{1}{2}(-3) = - \dfrac{3}{2}[/imath]