Câu 48: (MH Toan 2020) Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ 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 $\displaystyle\int^0_{-1}f(x)\mathrm{d}x$ bằng

Câu 48: (MH Toan 2020) Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ 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 $\displaystyle\int^0_{-1}f(x)\mathrm{d}x$ bằng
A. \( – \frac{{17}}{{20}}\).
B. \( – \frac{{13}}{4}\).
C. \(\frac{{17}}{4}\).
D. \( – 1\).

Đáp án: B
Ta có \(xf\left( {{x^3}} \right) + f\left( {1 – {x^2}} \right) = – {x^{10}} + {x^6} – 2x \Leftrightarrow {x^2}f\left( {{x^3}} \right) + xf\left( {1 – {x^2}} \right) = – {x^{11}} + {x^7} – 2{x^2}\)
\( \Rightarrow \int\limits_{ – 1}^0 {{x^2}} f\left( {{x^3}} \right){\rm{d}}x + \int\limits_{ – 1}^0 x f\left( {1 – {x^2}} \right){\rm{d}}x = \int\limits_{ – 1}^0 {\left( { – {x^{11}} + {x^7} – 2{x^2}} \right)} {\rm{d}}x\)
\( \Leftrightarrow \int\limits_{ – 1}^0 {{x^2}} f\left( {{x^3}} \right){\rm{d}}x + \int\limits_{ – 1}^0 x f\left( {1 – {x^2}} \right){\rm{d}}x = – \frac{{17}}{{24}}\)
Đặt \({x^3} = t\)
\( \Rightarrow 3{x^2}{\rm{d}}x = {\rm{d}}t \Leftrightarrow {x^2}{\rm{d}}x = \frac{1}{3}{\rm{d}}t\)
\( \Rightarrow \) Khi \(x = 0\) thì \(t = 0\); \(x = – 1\) thì \(t = – 1\)
\( \Rightarrow \int\limits_{ – 1}^0 {{x^2}} f\left( {{x^3}} \right){\rm{d}}x = \frac{1}{3}\int\limits_{ – 1}^0 f (t){\rm{d}}t\)
Đặt \(1 – {x^2} = t\)
\( \Rightarrow – 2x{\rm{d}}x = {\rm{d}}t\)
Khi \(x = 0\) thì \(t = 1\), \(x = – 1\) thì \(t = 0\)
\( \Rightarrow \int\limits_{ – 1}^0 x f\left( {1 – {x^2}} \right){\rm{d}}x = – \frac{1}{2}\int\limits_0^1 f (t){\rm{d}}t\)
\( \Rightarrow \frac{1}{3}\int\limits_{ – 1}^0 f (t){\rm{d}}t – \frac{1}{2}\int\limits_0^1 f (t){\rm{d}}t = – \frac{{17}}{{24}}\)
Ta có: \({x^2}f\left( {{x^3}} \right) + xf\left( {1 – {x^2}} \right) = – {x^{11}} + {x^7} – 2{x^2}\)
\( \Rightarrow \int\limits_0^1 {{x^2}} f\left( {{x^3}} \right){\rm{d}}x + \int\limits_0^1 x f\left( {1 – {x^2}} \right){\rm{d}}x = – \frac{5}{8}\)
Đặt \({x^3} = t\)
\( \Rightarrow 3{x^2}{\rm{d}}x = {\rm{d}}t \Leftrightarrow {x^2}{\rm{d}}x = \frac{1}{3}{\rm{d}}t\)
\( \Rightarrow \) Khi \(x = 0\) thì \(t = 0\); \(x = 1\) thì \(t = 1\)
\( \Rightarrow \int\limits_0^1 {{x^2}} f\left( {{x^3}} \right){\rm{d}}x = \frac{1}{3}\int\limits_0^1 f (t){\rm{d}}t\)
Đặt \(1 – {x^2} = t\)
\( \Rightarrow – 2x{\rm{d}}x = {\rm{d}}t\)
Khi \(x = 0\) thì \(t = 1\), \(x = 1\) thì \(t = 0\)
\( \Rightarrow \int\limits_0^1 x f\left( {1 – {x^2}} \right){\rm{d}}x = – \frac{1}{2}\int\limits_1^0 f (t){\rm{d}}t = \frac{1}{2}\int\limits_0^1 f (t){\rm{d}}t\)
\( \Rightarrow \frac{1}{3}\int\limits_0^1 f (t){\rm{d}}t + \frac{1}{2}\int\limits_0^1 f (t){\rm{d}}t = – \frac{5}{8}\)\( \Leftrightarrow \frac{5}{6}\int\limits_0^1 f (t){\rm{d}}t = – \frac{5}{8} \Leftrightarrow \int\limits_0^1 f (t){\rm{d}}t = – \frac{3}{4}\)
\( \Rightarrow \frac{1}{3}\int\limits_{ – 1}^0 f (t){\rm{d}}t – \frac{1}{2}\int\limits_0^1 f (t){\rm{d}}t = – \frac{{17}}{{24}} \Leftrightarrow \frac{1}{3}\int\limits_{ – 1}^0 f (t){\rm{d}}t + \frac{1}{2}.\frac{3}{4} = – \frac{{17}}{{24}} \Leftrightarrow \frac{1}{3}\int\limits_{ – 1}^0 f (t){\rm{d}}t = – \frac{{17}}{{12}} \Leftrightarrow \int\limits_{ – 1}^0 f (t){\rm{d}}t = – \frac{{13}}{4}\)