Nếu tích phân [ 4 f(x) + 3x^2 ] dx = 7 thì f(x) dx bằng

Ta có:

\(\begin{array}{l}\int\limits_0^3 {\left[ {4f\left( x \right) + 3{x^2}} \right]{\rm{d}}x}  = 7 \Leftrightarrow \int\limits_0^3 {4f\left( x \right){\rm{d}}x + \int\limits_0^3 {3{x^2}{\rm{d}}x} }  = 7\\ \Leftrightarrow 4\int\limits_0^3 {f\left( x \right){\rm{d}}x + \int\limits_0^3 {3{x^2}{\rm{d}}x} }  = 7 \Leftrightarrow 4\int\limits_0^3 {f\left( x \right){\rm{d}}x + 27}  = 7 \Leftrightarrow 4\int\limits_0^3 {f\left( x \right){\rm{d}}x}  =  - 20 \Leftrightarrow \int\limits_0^3 {f\left( x \right){\rm{d}}x}  =  - 5\end{array}\).