Giúp mình với ạ.....

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
x \in \left[ {1;2} \right] \Rightarrow 3x - 1 > 0\\
I = \int\limits_1^2 {\left| {3x - 1} \right|dx}  = \int\limits_1^2 {\left( {3x - 1} \right)dx}  = \mathop {\left. {\left( {\frac{3}{2}{x^2} - x} \right)} \right|}\nolimits_1^2 \\
 = \left( {\frac{3}{2}{{.2}^2} - 2} \right) - \left( {\frac{3}{2}{{.1}^2} - 1} \right) = 4 - \frac{1}{2} = \frac{7}{2}\\
2,\\
x \in \left[ {2;5} \right] \Rightarrow 6x - 2 > 0\\
I = \int\limits_2^5 {\left| {6x - 2} \right|dx}  = \int\limits_2^5 {\left( {6x - 2} \right)dx}  = \mathop {\left. {\left( {3{x^2} - 2x} \right)} \right|}\nolimits_2^5 \\
 = \left( {{{3.5}^2} - 2.5} \right) - \left( {{{3.2}^2} - 2.2} \right) = 65 - 8 = 57\\
3,\\
t = \sin x \Rightarrow \left\{ \begin{array}{l}
dt = \left( {\sin x} \right)'dx = \cos xdx\\
x = 0 \Rightarrow t = 0\\
x = \frac{\pi }{2} \Rightarrow t = 1
\end{array} \right.\\
I = \int\limits_0^{\frac{\pi }{2}} {2{{\sin }^3}x.\left( {\cos xdx} \right)}  = \int\limits_0^1 {2{t^3}dt}  = \mathop {\left. {\left( {2.\frac{{{t^4}}}{4}} \right)} \right|}\nolimits_0^1  = \frac{{{1^4}}}{2} - \frac{{{0^4}}}{2} = \frac{1}{2}\\
4,\\
I = \int\limits_{ - 2}^3 {{{\left( {2x - 1} \right)}^2}{e^x}dx} \\
\left\{ \begin{array}{l}
u = {\left( {2x - 1} \right)^2}\\
v' = {e^x}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 2.\left( {2x - 1} \right)'.\left( {2x - 1} \right) = 4.\left( {2x - 1} \right)\\
v = {e^x}
\end{array} \right.\\
 \Rightarrow I = \mathop {\left. {{{\left( {2x - 1} \right)}^2}.{e^x}} \right|}\nolimits_{ - 2}^3  - \int\limits_{ - 2}^3 {4.\left( {2x - 1} \right).{e^x}dx} \\
 = {\left( {2.3 - 1} \right)^2}.{e^3} - {\left( {2.\left( { - 2} \right) - 1} \right)^2}.{e^{ - 2}} - 4\int\limits_{ - 2}^3 {\left( {2x - 1} \right){e^x}dx} \\
 = 25\left( {{e^3} - {e^{ - 2}}} \right) - 4.\int\limits_{ - 2}^3 {\left( {2x - 1} \right){e^x}dx} \\
{I_1} = \int\limits_{ - 2}^3 {\left( {2x - 1} \right){e^x}dx} \\
\left\{ \begin{array}{l}
u = 2x - 1\\
v' = {e^x}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 2\\
v = {e^x}
\end{array} \right.\\
 \Rightarrow {I_1} = \mathop {\left. {\left( {2x - 1} \right).{e^x}} \right|}\nolimits_{ - 2}^3  - \int\limits_{ - 2}^3 {2{e^x}dx} \\
 = \left( {2.3 - 1} \right).{e^3} - \left( {2.\left( { - 2} \right) - 1} \right).{e^{ - 2}} - \mathop {\left. {\left( {2{e^x}} \right)} \right|}\nolimits_{ - 2}^3 \\
 = 5{e^3} + 5{e^{ - 2}} - 2{e^3} + 2{e^{ - 2}}\\
 = 3{e^3} + 7{e^{ - 2}}\\
 \Rightarrow I = 25\left( {{e^3} - {e^{ - 2}}} \right) - 4.\left( {3{e^3} + 7{e^{ - 2}}} \right) = 13{e^3} - 53{e^{ - 2}}
\end{array}\)