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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn $4{{\left( f\left( x \right) \right)}^{3}}+7f\left( x \right)={{x}^{3}}+6{{x}^{2}}-16,\forall x\in \mathbb{R}$. Tích phân $\int\limits_{-2}^{-1}{x\left( x+4 \right)f\left( x \right)dx}$ thuộc khoảng nào dưới đây?
A. $\left( 0;\dfrac{1}{2} \right)$.
B. $\left( -\dfrac{1}{2};0 \right)$.
C. $\left( \dfrac{1}{2};2 \right)$.
D. $\left( 2;+\infty \right)$.
Đặt $t=f\left( x \right)\Rightarrow 4{{t}^{3}}+7t={{x}^{3}}+6{{x}^{2}}-16\Rightarrow \left( 3{{x}^{2}}+12x \right)dx=\left( 12{{t}^{2}}+7 \right)dt$
$\Rightarrow x\left( x+4 \right)dx=\dfrac{1}{3}\left( 12{{t}^{2}}+7 \right)dt$.
Đổi cận: $\left\{ \begin{aligned}
& x=-2\Rightarrow 4{{t}^{3}}+7t=0\Rightarrow t=0 \\
& x=-1\Rightarrow 4{{t}^{3}}+7t=-11\Rightarrow t=-1 \\
\end{aligned} \right.$.
Vậy $\int\limits_{-2}^{-1}{x\left( x+4 \right)f\left( x \right)dx}=\int\limits_{0}^{-1}{\dfrac{1}{3}\left( 12{{t}^{2}}+7 \right)dt}=\dfrac{13}{6}$.
A. $\left( 0;\dfrac{1}{2} \right)$.
B. $\left( -\dfrac{1}{2};0 \right)$.
C. $\left( \dfrac{1}{2};2 \right)$.
D. $\left( 2;+\infty \right)$.
Đặt $t=f\left( x \right)\Rightarrow 4{{t}^{3}}+7t={{x}^{3}}+6{{x}^{2}}-16\Rightarrow \left( 3{{x}^{2}}+12x \right)dx=\left( 12{{t}^{2}}+7 \right)dt$
$\Rightarrow x\left( x+4 \right)dx=\dfrac{1}{3}\left( 12{{t}^{2}}+7 \right)dt$.
Đổi cận: $\left\{ \begin{aligned}
& x=-2\Rightarrow 4{{t}^{3}}+7t=0\Rightarrow t=0 \\
& x=-1\Rightarrow 4{{t}^{3}}+7t=-11\Rightarrow t=-1 \\
\end{aligned} \right.$.
Vậy $\int\limits_{-2}^{-1}{x\left( x+4 \right)f\left( x \right)dx}=\int\limits_{0}^{-1}{\dfrac{1}{3}\left( 12{{t}^{2}}+7 \right)dt}=\dfrac{13}{6}$.
Đáp án D.