Cho hai hàm số $y = f\left( x \right),\,\,y = g\left( x \right)$ là các hàm liên tục trên đoạn $\left[ {0;2} \right],$ có $\int\limits_0^1 {f\left( x \right){\rm{d}}x}  = 4,\,\,\int\limits_0^2 {g\left( x \right){\rm{d}}x}  =  - \,2$ và $\int\limits_1^2 {g\left( t \right){\rm{d}}t}  = 1.$ Tính $I = \int\limits_0^1 {\left[ {2f\left( x \right) - g\left( x \right)} \right]{\rm{d}}x} .$

\(\frac{{19}}{3}\).

\(\frac{{32}}{3}\).

\(\frac{{16}}{3}\).

\(\frac{{21}}{2}\).

$\int\limits_a^b {kf\left( x \right)dx} {\rm{\;}} = k\int\limits_a^b {f\left( x \right)dx} $

$\int\limits_a^b {xf\left( x \right)dx} {\rm{\;}} = x\int\limits_a^b {f\left( x \right)dx} $

$\int\limits_a^b {f\left( x \right)dx} {\rm{\;}} = {\rm{\;}} - \int\limits_b^a {f\left( x \right)dx} $

$\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} {\rm{\;}} = \int\limits_a^b {f\left( x \right)dx} {\rm{\;}} + \int\limits_a^b {g\left( x \right)dx} $

\(I =  - 12\)                          

\(I = 8\)                    

\(I = 12\)

\(I =  - 8\)  

\(f\left( x \right) =  - \dfrac{{{\pi ^x}}}{{\ln \pi }}\).

\(f(x) = \dfrac{{{\pi ^x}}}{{\ln \pi }}\).

\(f\left( x \right) = {\pi ^x}.\ln \pi \).

\(f\left( x \right) =  - {\pi ^x}.\ln \pi \).

$\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x = 2x\ln \left( {9x - 1} \right) + C.} $

$\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x = 6x\ln \left( {3x - 1} \right) + C.} $

$\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x = 6x\ln \left( {9x - 1} \right) + C.} $

$\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x = 3x\ln \left( {9x - 1} \right) + C.} $