`7.` `B.` `2`
`*`
`+)`$\displaystyle\int_0^2 {f(x)}dx=1$
`+)`$\displaystyle\int_1^2 {f(t)}dx =-\int_2^1 {f(t)}dx$
$=-\displaystyle\int_2^1 {f(x)}dx =3$
`*`
$I=\displaystyle\int_0^1 {f(x)}dx=\displaystyle\int_0^2 {f(x)}dx+\displaystyle\int_2^1{f(x)}dx$
`=5-3=2`