Cho 2 ∫ 1 f ( x ) d x = − 1 , 2 ∫ 1 ( f ( x ) + g ( x ) ) d x = 2 . Khi đó 2 ∫ 1 g ( x ) d x bằng

2/22

Cho \(\int\limits_1^2 {f\left( x \right)dx = - 1} \), \(\int\limits_1^2 {\left( {f\left( x \right) + g\left( x \right)} \right)dx = 2} \). Khi đó \(\int\limits_1^2 {g\left( x \right)dx} \) bằng

\( - 1\).

\(1\).

\( - 3\).

\(3\).

Giải thích

Đáp án đúng là: D

Ta có \(\int\limits_1^2 {\left( {f\left( x \right) + g\left( x \right)} \right)dx = 2} \)\( \Leftrightarrow \int\limits_1^2 {f\left( x \right)dx + \int\limits_1^2 {g\left( x \right)dx} } = 2\)

\( \Leftrightarrow - 1 + \int\limits_1^2 {g\left( x \right)dx} = 2 \Leftrightarrow \int\limits_1^2 {g\left( x \right)dx} = 3\).