Cho hàm số f(x) có f(0) = -1 và \(f’\left( x \right) = x\left( {6 + 12x + {e^{ – x}}} \right),\forall x \in R\). Khi đó \(\int\limits_0^1 {f\left( x \right)} {\rm{d}}x\) bằng

Cho hàm số f(x) có f(0) = -1 và \(f’\left( x \right) = x\left( {6 + 12x + {e^{ – x}}} \right),\forall x \in R\). Khi đó \(\int\limits_0^1 {f\left( x \right)} {\rm{d}}x\) bằng

A. 3e

B. 3e^-1

C. 4-3e^-1

D. -3e^-1

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