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 \(b \in \left[ {a;c} \right]\) 

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

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

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

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

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

$y = \sin x + 1$         

\(y = \cos x\)   

\(y = \cot x\)

\(y =  - \cos x\)