Đáp án

Bài tập: Cho hàm số \(y = f(x)\) liên tục trên \(\mathbb{R}\) thỏa mãn: \(2x[1 + f(x)] = [f'(x)]^3, \, \forall x \in \mathbb{R}, \, f(0) = -1, \, f(x) > -1, \, \forall x \neq 0.\) Tính \(I = \int_{0}^{1} f(x) dx.\)
A. \(\frac{1}{4} \quad\)  B. \(-\frac{5}{6}\quad\)  C. \(\frac{1}{3}\quad\)  D. \(-\frac{2}{3}\)

Đáp án:

\(2x(1 + f(x)) = [f'(x)]^3 \Leftrightarrow f'(x) = \sqrt[3]{2x(1 + f(x))}\)

\(\Rightarrow \frac{f'(x)}{\sqrt[3]{1 + f(x)}} = \sqrt[3]{2x} \)

\(\Rightarrow \quad \int f'(x)(1 + f(x))^{-\frac{1}{3}} dx = \sqrt[3]{2} \int x^{\frac{1}{3}} dx\)

\(\Rightarrow \frac{3}{2}(1 + f(x))^{\frac{2}{3}} = \sqrt[3]{2}\frac{3}{4} x^{\frac{4}{3}} + C\)

\(\Rightarrow (1 + f(x))^{\frac{2}{3}} = \frac{\sqrt[3]{2}}{2} x^{\frac{4}{3}} + C\)

\(f(0) = -1 \Rightarrow C = 0\)

\(\Rightarrow (1 + f(x))^2 = \frac{1}{4} x^4 \quad \Rightarrow \quad 1+ f(x) = \frac{1}{2}x^2 \)

\(\Rightarrow f(x) = \frac{1}{2}x^2 -1 \)

\(\int_{0}^{1} f(x) dx = \int_{0}^{1} \left(\frac{1}{2}x^2 - 1\right) dx = -\frac{5}{6}\)