Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}{e^x} + m\,\,\,khi\,\,x \ge 0\\2x\sqrt {3 + {x^2}} \,\,khi\,\,x...

* Hướng dẫn giải

Hàm số \(f\left( x \right)\) liên tục trên \(\mathbb{R}\)  \( \Leftrightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right)\)

\( \Leftrightarrow \mathop {\lim }\limits_{x \to {0^ + }} \left( {{e^x} + m} \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {2x\sqrt {3 + {x^2}} } \right) \Leftrightarrow 1 + m = 0 \Leftrightarrow m =  - 1\)

Khi đó:

\(\begin{array}{l}\int\limits_{ - 1}^1 {f\left( x \right)dx}  = \int\limits_{ - 1}^0 {f\left( x \right)dx}  + \int\limits_0^1 {f\left( x \right)dx} \\ = \int\limits_{ - 1}^0 {2x\sqrt {3 + {x^2}} dx}  + \int\limits_0^1 {\left( {{e^x} - 1} \right)dx}  = \int\limits_{ - 1}^0 {\sqrt {3 + {x^2}} d\left( {3 + {x^2}} \right)}  + \left. {\left( {{e^x} - x} \right)} \right|_0^1\\ = \left. {\dfrac{2}{3}\left( {3 + {x^2}} \right)\sqrt {3 + {x^2}} } \right|_{ - 1}^0 + \left. {\left( {{e^x} - x} \right)} \right|_0^1 = \dfrac{2}{3}.3.\sqrt 3  - \dfrac{2}{3}.4.2 + \left( {e - 1 - 1} \right) = e + 2\sqrt 3  - \dfrac{{22}}{3}\\ \Rightarrow a = 1,\,\,b = 2,\,\,c =  - \dfrac{{22}}{3} \Rightarrow T = a + b + 3c = 1 + 2 - 22 =  - 19\end{array}\)

Chọn: C