Cho hàm số f(x) có f(0) = -1 và

B

- Hướng dẫn giải

Ta có: \(f'\left( x \right) = x\left( {6 + 12x + {e^{ - x}}} \right),\forall x \in R\) nên f(x) là một nguyên hàm của f'(x)

\(\int {f'\left( x \right){\rm{d}}x = \int {x\left( {6 + 12x + {e^{ - x}}} \right){\rm{d}}x} } = \int {\left( {6x + 12{x^2}} \right){\rm{d}}x + \int {x{e^{ - x}}{\rm{d}}x} } \)

Mà \(\int {\left( {6x + 12{x^2}} \right){\rm{d}}x = 3{x^2} + 4{x^3}} + C\)

Xét \(\int {x{e^{ - x}}{\rm{d}}x} \): Đặt \(\left\{ \begin{array}{l}u = x\\{\rm{d}}v = {e^{ - x}}{\rm{d}}x\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\rm{d}}u = {\rm{d}}x\\v = - {e^{ - x}}\end{array} \right.\)

\(\int {x{e^{ - x}}{\rm{d}}x = - x{e^{ - x}} + \int {{e^{ - x}}} } {\rm{d}}x = - x{e^{ - x}} - {e^{ - x}} + C = - \left( {x + 1} \right){e^{ - x}} + C\)

Suy ra \(f\left( x \right) = 3{x^2} + 4{x^3} - \left( {x + 1} \right){e^{ - x}} + C,\forall x \in R\).

Mà \(f\left( 0 \right) = - 1 \Rightarrow C = 0\) nên \(f\left( x \right) = 3{x^2} + 4{x^3} - \left( {x + 1} \right){e^{ - x}},\forall x \in R\).

Ta có

\(\int\limits_0^1 {f\left( x \right)} {\rm{d}}x = \int\limits_0^1 {\left( {3{x^2} + 4{x^3} - \left( {x + 1} \right){e^{ - x}}} \right)} {\rm{d}}x = \left. {\left( {{x^3} + {x^4}} \right)} \right|_0^1 - \int\limits_0^1 {\left( {x + 1} \right){e^{ - x}}{\rm{d}}x} = 2 - \int\limits_0^1 {\left( {x + 1} \right){e^{ - x}}{\rm{d}}x} \)

Xét \(\int\limits_0^1 {\left( {x + 1} \right){e^{ - x}}{\rm{d}}x} \): Đặt \(\left\{ \begin{array}{l}u = x + 1\\{\rm{d}}v = {e^{ - x}}{\rm{d}}x\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\rm{d}}u = {\rm{d}}x\\v = - {e^{ - x}}\end{array} \right.\)

\(\int\limits_0^1 {\left( {x + 1} \right){e^{ - x}}{\rm{d}}x} = \left. { - \left( {x + 1} \right){e^{ - x}}} \right|_0^1 + \int\limits_0^1 {{e^{ - x}}{\rm{d}}x} = - 2{e^{ - 1}} + 1 - \left. {{e^{ - x}}} \right|_0^1 = - 2{e^{ - 1}} + 1 - {e^{ - 1}} + 1 = 2 - 3{e^{ - 1}}\)

Vậy \(\int\limits_0^1 {f\left( x \right){\rm{d}}x} = 3{e^{ - 1}}\).