Chọn D
Hàm số có đạo hàm tại \(\Leftrightarrow f\left(2\right)={\mathop{\lim }\limits_{x\to 2^{+} }} f\left(x\right)={\mathop{\lim }\limits_{x\to 2^{-} }} f\left(x\right)\Leftrightarrow 4+2a+b=-2\Leftrightarrow 2a+b=-6. \left(1\right)\)
Có \({\mathop{\lim }\limits_{x\to 2^{-} }} \frac{f\left(x\right)-f\left(2\right)}{x-2} ={\mathop{\lim }\limits_{x\to 2^{-} }} \frac{x^{3} -x^{2} -8x+10-4-2a-b}{x-2} \, \, \, =\, \, \, {\mathop{\lim }\limits_{x\to 2^{-} }} \frac{x^{3} -x^{2} -8x+12}{x-2} \)
\(={\mathop{\lim }\limits_{x\to 2^{-} }} \frac{\left(x-2\right)^{2} \left(x+3\right)}{x-2} ={\mathop{\lim }\limits_{x\to 2^{-} }} \left[\left(x-2\right)\left(x+3\right)\right]=0;\)
\({\mathop{\lim }\limits_{x\to 2^{+} }} \frac{f\left(x\right)-f\left(2\right)}{x-2} ={\mathop{\lim }\limits_{x\to 2^{+} }} \frac{x^{2} +ax+b-4-2a-b}{x-2} ={\mathop{\lim }\limits_{x\to 2^{+} }} \frac{\left(x-2\right)\left(x+2+a\right)}{x-2} ={\mathop{\lim }\limits_{x\to 2^{+} }} \left(x+a+2\right)=a+4.\)
Hàm số có đạo hàm tại x=2 nên hàm số liên tục tại x=2
suy ra \({\mathop{\lim }\limits_{x\to 2^{+} }} \frac{f\left(x\right)-f\left(2\right)}{x-2} =\, {\mathop{\lim }\limits_{x\to 2^{-} }} \frac{f\left(x\right)-f\left(2\right)}{x-2} \Leftrightarrow a+4=0\Leftrightarrow a=-4. \left(2\right)\)
Từ \(\left(1\right) và \left(2\right)\), suy ra a=-4 và b=2.
Khi đó \(f\left(x\right)=\left\{\begin{array}{l} {x^{2} -4x+2\, \, {\rm khi}\, \, \, x\ge 2} \\ {x^{3} -x^{2} -8x+10\, \, \, \, {\rm khi}\, \, \, x<2} \end{array}\right. .\)
\(I=\int _{0}^{4}f\left(x\right) dx=\int _{0}^{2}f\left(x\right) dx+\int _{2}^{4}f\left(x\right) dx\)
\(\begin{array}{l} {=\int _{0}^{2}\left(x^{3} -x^{2} -8x+10\right) dx+\int _{2}^{4}\left(x^{2} -4x+2\right) dx} \\ {=\left(\frac{x^{4} }{4} -\frac{x^{3} }{3} -4x^{2} +10x\right)\left|\begin{array}{l} {2} \\ {0} \end{array}\right. +\left(\frac{x^{3} }{3} -2x^{2} +2x\right)\left|\begin{array}{l} {4} \\ {2} \end{array}\right. =\frac{16}{3} -\frac{4}{3} =4} \end{array}\)
Vậy I=4.