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 $I = \int\limits_0^1 {\left[ {2f\left( x \right) - g\left( x \right)} \right]{\rm{d}}x} .$

$\int\limits_1^2 {{e^x}dx}$.

$\int\limits_0^1 {2dx}$.

$\int\limits_0^{\dfrac{\pi }{2}} {\sin xdx}$.

$\int\limits_0^1 {xdx}$.

 $\int{{{e}^{x}}\,\text{d}x}={{e}^{x}}+C.$         

 $\int{0\,\text{d}x}=C.$             

$\int{\dfrac{1}{x}\,\text{d}x}=\ln x+C.$

 $\int{\text{dx}}=x+C.$

 $\frac{61}{9}$

 $4.$

$61.$

 $\frac{61}{3}$.

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

$\int\limits_a^b {kdx}  = k\left( {b - a} \right)$

$\int\limits_a^b {f\left( x \right)dx}  + \int\limits_b^c {f\left( x \right)dx}  = \int\limits_a^c {f\left( x \right)dx}$ với $b \in \left[ {a;c} \right]$ 

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

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

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

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

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

$y = \sin x + 1$         

$y = \cos x$   

$y = \cot x$

$y =  - \cos x$