Cho hàm số \(f(x)\) liên tục trên \(\mathbb{R}\) và thỏa mãn \(xf\left(x^3\right)+f\left(1-x^2\right)=-x^{10}+x^6-2x\), \(\forall x\in\mathbb{R}\). Khi đó \(\displaystyle\int\limits_{-1}^0f(x)\mathrm{\,d}x\) bằng
 | \(-\dfrac{17}{20}\) |
 | \(-\dfrac{13}{4}\) |
 | \(\dfrac{17}{4}\) |
 | \(-1\) |
Chọn phương án B.
\(\begin{aligned}
xf\left(x^3\right)+f\left(1-x^2\right)&=-x^{10}+x^6-2x\\
\Leftrightarrow\,x^2f\left(x^3\right)+xf\left(1-x^2\right)&=-x^{11}+x^7-2x^2\;(1)\\
\Leftrightarrow\displaystyle\int\limits_{-1}^{0}x^2f\left(x^3\right)\mathrm{\,d}x+\displaystyle\int\limits_{-1}^{0}xf\left(1-x^2\right)\mathrm{\,d}x&=\displaystyle\int\limits_{-1}^{0}\left(-x^{11}+x^7-2x^2\right)\mathrm{\,d}x\\
\Leftrightarrow\displaystyle\int\limits_{-1}^{0}x^2f\left(x^3\right)\mathrm{\,d}x+\displaystyle\int\limits_{-1}^{0}xf\left(1-x^2\right)\mathrm{\,d}x&=-\dfrac{17}{24}.
\end{aligned}\)
- Đặt \(u=x^3\Rightarrow\mathrm{\,d}u=3x^2\mathrm{\,d}x\).
Ta có \(x=-1\Rightarrow u=-1\), \(x=0\Rightarrow u=0\).
\(\begin{aligned}\Rightarrow\displaystyle\int\limits_{-1}^{0}x^2f\left(x^3\right)\mathrm{\,d}x&=\dfrac{1}{3}\displaystyle\int\limits_{-1}^{0}f(u)\mathrm{\,d}u\\ &=\dfrac{1}{3}\displaystyle\int\limits_{-1}^{0}f(x)\mathrm{\,d}x.\end{aligned}\) - Đặt \(v=1-x^2\Rightarrow\mathrm{\,d}v=-2x\mathrm{\,d}x\).
Ta có \(x=-1\Rightarrow v=0\), \(x=0\Rightarrow v=1\).
\(\begin{aligned}\Rightarrow\displaystyle\int\limits_{-1}^{0}xf\left(1-x^2\right)\mathrm{\,d}x&=-\dfrac{1}{2}\displaystyle\int\limits_{0}^{1}f(v)\mathrm{\,d}v\\ &=-\dfrac{1}{2}\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x.\end{aligned}\)
Suy ra \(\dfrac{1}{3}\displaystyle\int\limits_{-1}^{0}f(x)\mathrm{\,d}x-\dfrac{1}{2}\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x=-\dfrac{17}{24}\) (2)
Cũng từ (1) ta có: $$\begin{aligned}
\displaystyle\int\limits_{0}^{1}x^2f\left(x^3\right)\mathrm{\,d}x+\displaystyle\int\limits_{0}^{1}xf\left(1-x^2\right)\mathrm{\,d}x&=\displaystyle\int\limits_{0}^{1}\left(-x^{11}+x^7-2x^2\right)\mathrm{\,d}x\\
\Leftrightarrow\displaystyle\int\limits_{0}^{1}x^2f\left(x^3\right)\mathrm{\,d}x+\displaystyle\int\limits_{0}^{1}xf\left(1-x^2\right)\mathrm{\,d}x&=-\dfrac{5}{8}.
\end{aligned}$$
- Đặt \(u=x^3\Rightarrow\mathrm{\,d}u=3x^2\mathrm{\,d}x\).
Ta có \(x=0\Rightarrow u=0\), \(x=1\Rightarrow u=1\).
\(\begin{aligned}\Rightarrow\displaystyle\int\limits_{0}^{1}x^2f\left(x^3\right)\mathrm{\,d}x&=\dfrac{1}{3}\displaystyle\int\limits_{0}^{1}f(u)\mathrm{\,d}u\\ &=\dfrac{1}{3}\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x.\end{aligned}\) - Đặt \(v=1-x^2\Rightarrow\mathrm{\,d}v=-2x\mathrm{\,d}x\).
Ta có \(x=0\Rightarrow v=1\), \(x=1\Rightarrow v=0\).
\(\begin{aligned}\Rightarrow\displaystyle\int\limits_{0}^{1}xf\left(1-x^2\right)\mathrm{\,d}x&=-\dfrac{1}{2}\displaystyle\int\limits_{1}^{0}f(v)\mathrm{\,d}v\\ &=-\dfrac{1}{2}\displaystyle\int\limits_{1}^{0}f(x)\mathrm{\,d}x.\\ &=\dfrac{1}{2}\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x.\end{aligned}\)
Suy ra $$\begin{aligned}
\dfrac{1}{3}\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x+\dfrac{1}{2}\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x&=-\dfrac{5}{8}\\
\Leftrightarrow\dfrac{5}{6}\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x&=-\dfrac{5}{8}\\
\Leftrightarrow\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x&=-\dfrac{3}{4}.
\end{aligned}$$
Thay vào (2) ta có $$\begin{aligned}
\dfrac{1}{3}\displaystyle\int\limits_{-1}^{0}f(x)\mathrm{\,d}x-\dfrac{1}{2}\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x&=-\dfrac{17}{24}\\
\Leftrightarrow\,\dfrac{1}{3}\displaystyle\int\limits_{-1}^{0}f(x)\mathrm{\,d}x-\dfrac{1}{2}\left(-\dfrac{3}{4}\right)&=-\dfrac{17}{24}\\
\Leftrightarrow\,\displaystyle\int\limits_{-1}^{0}f(x)\mathrm{\,d}x&=-\dfrac{13}{4}.
\end{aligned}$$