Tích phân sau \(\int\limits_0^4 {\left( {3x - {e^{\dfrac{x}{2}}}} \right)dx = a + b{e^2}} \) khi đó a – 10b bằng:

* Hướng dẫn giải

Ta có: \(\int\limits_0^4 \left( {3x - {e^{\dfrac{x}{2}}}} \right)dx \)

\(= \left( {\dfrac{3}{2}{x^2}} \right) \left| \begin{array}{l}^4\\_0\end{array} \right. - 2\int\limits_0^4 {{e^{\dfrac{x}{2}}}\,d\left( {\dfrac{x}{2}} \right)}  \)

\(= 24 - 2\left( {{e^{\dfrac{x}{2}}}} \right)\left| \begin{array}{l}^4\\_0^{}\end{array} \right. \)

\(= 24 - 2\left( {{e^2} - 1} \right) = 26 - 2{e^2}\)

Khi đó ta có: \(\left\{ \begin{array}{l}a = 26\\b =  - 2\end{array} \right. \Rightarrow a - 10b = 26 + 20 = 46.\)

Chọn đáp án B.