Question

Đã trả lời bởi chuyên gia

Answer 1

Đáp án A

Đặt $\{\begin{cases} u = f (x) \\ d v = (2 x - 1) d x \end{cases} \Rightarrow \{\begin{cases} d u = f (x) d x \\ v = x^{2} - x \end{cases}$

$\Rightarrow - \frac{1}{30} = \int_{0}^{1} (2 x - 1) f (x) d x = (x^{2} - x) f (x) |\begin{array}{l} ^{1} \\ _{0} \end{array} - \int_{0}^{1} (x^{2} - x) f^{'} (x) d x = - \int_{0}^{1} (x^{2} - x) f^{'} (x) d x$

$\Rightarrow \int_{0}^{1} (x^{2} - x) f^{'} (x) d x = \frac{1}{30}$ .

Ta có: $\int_{0}^{1} {(x^{2} - x)}^{2} d x = \int_{0}^{1} (x^{4} - 2 x^{3} + x^{2}) d x = (\frac{x^{5}}{5} - \frac{x^{4}}{2} + \frac{x^{3}}{3}) |\begin{array}{l} ^{1} \\ _{0} \end{array} = \frac{1}{30}$ .

Do đó, $\int_{0}^{1} {[f^{'} (x) - (x^{2} - x)]}^{2} d x = \int_{0}^{1} {[f^{'} (x)]}^{2} d x - 2 \int_{0}^{1} (x^{2} - x) f (x) d x + \int_{0}^{1} {(x^{2} - x)}^{2} d x = 0$

$\Rightarrow f^{'} (x) = x^{2} - x \Rightarrow f (x) = \frac{x^{3}}{3} - \frac{x^{2}}{2} + C$ , mà f(0)=1 nên $C = 1 \Rightarrow f (x) = \frac{x^{3}}{3} - \frac{x^{2}}{2} + 1$

Vậy $\int_{0}^{1} f (x) d x = \int_{0}^{1} (\frac{x^{3}}{3} - \frac{x^{2}}{2} + 1) d x = (\frac{x^{4}}{12} - \frac{x^{3}}{6} + x) |\begin{array}{l} ^{1} \\ _{0} \end{array} = \frac{11}{12}$

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Answer 2