Biết hàm số \(y = f\left( x \right)\) liên tục và có đạo hàm trên \(\left[ {0;\,\,2} \right],\)\(f\left( 0 \right) = \sqrt 5 ,\)\(f\left( 2 \right) = \sqrt {11} .\) Tích phân \(I = \int\limits_0^2 {f\left( x \right).f'\left( x \right)dx} \) bằng:

\(\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 0\)      

\(\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 1\)

\(\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  =  - 1\)  

\(\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 2\)

$\int\limits_{ - 1}^2 {f\left( {2x} \right)} d{\rm{x  =  2}}$

$\int\limits_{ - 3}^3 {f\left( {x + 1} \right)} d{\rm{x  =  2}}$

$\int\limits_{ - 1}^2 {f\left( {2x} \right)} d{\rm{x  =  1}}$

$\int\limits_0^6 {\dfrac{1}{2}f\left( {x - 2} \right)} d{\rm{x  =  1}}$

\(\int\limits_{ - a}^a {f\left( x \right)dx}  = 0\)

\(\int\limits_{ - a}^a {f\left( x \right)dx}  = 1\)

\(\int\limits_{ - a}^a {f\left( x \right)dx}  =  - 1\)

\(\int\limits_{ - a}^a {f\left( x \right)dx}  = a\)

\(I = \dfrac{{ - 1}}{2}\)

\(I =  - \dfrac{1}{4}\)

 \(I=\dfrac{1}{4}\)

$I =  - 2$

\(I =  - \dfrac{1}{4}{\pi ^4}\)

\(I =  - {\pi ^4}\)         

$I = 0  $

\(I =  - \dfrac{1}{4}\)

\(I = 2\int\limits_8^9 {\sqrt u du} \)

\(I = \dfrac{1}{2}\int\limits_8^9 {\sqrt u du} \)       

\(I = \int\limits_9^8 {\sqrt u du} \)

\(I = \int\limits_8^9 {\sqrt u du} \)

\(I = \left. {\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2\)                  

\(I = \left. {\dfrac{4}{3}\left( {{u^3} + u} \right)} \right|_1^2\)

\(I = \left. {2\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2\)

\(I = \left. {\dfrac{1}{3}\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2\)