Cho hàm số y=f(x) liên tục trên đoạn

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Cho hàm số y=f(x) liên tục trên đoạn \[\left[ {0;1} \right],\;\]có \[\mathop \smallint \limits_0^1 \left[ {3 - 2f\left( x \right)} \right]{\rm{d}}x = 5.\]. Tính \[\mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x\].

\[\mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x = - \,1.\]

\[\mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x = 1.\]

\[\mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x = 2.\]

\[\mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x = - \,2.\]

Giải thích

Ta có

\[\mathop \smallint \limits_0^1 \left[ {3 - 2f\left( x \right)} \right]{\rm{d}}x = \mathop \smallint \limits_0^1 3dx - 2\mathop \smallint \limits_0^1 f(x)dx = 3x|_0^1 - 2\mathop \smallint \limits_0^1 f(x)dx = 3 - 2\mathop \smallint \limits_0^1 f(x)dx\]

Mặt khác

\[\mathop \smallint \limits_0^1 \left[ {3 - 2f\left( x \right)} \right]{\rm{d}}x = 5 \Rightarrow 3 - 2\,\mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x = 5 \Leftrightarrow \mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x = - \,1.\]

Đáp án cần chọn là: A