Cho hàm số f(x) có đạo hàm liên tục trên đoạn [0;1] thỏa mãn f(1) = 0

* Hướng dẫn giải

Đáp án A.

Đặt $\left\{\begin{array}{l}u=f\left(x\right)\\ dv=3x^{2}dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=f\text{'}\left(x\right)dx\\ v=x^3\end{array}\right.,$ 

khi đó $\underset{0}{\overset{1}{\int }}3x^{2}f\left(x\right)dx=x^{3}f\left(x\right)\left|\begin{array}{l}{}^1\\ {}_0\end{array}\right.-\underset{0}{\overset{1}{\int }}{x}^{3}f\text{'}\left(x\right)dx.$

$$\begin{gathered} \Rightarrow 1=f\left(1\right)-\underset{0}{\overset{1}{\int }}{x}^{3}f\text{'}\left(x\right)dx\Rightarrow \underset{0}{\overset{1}{\int }}{x}^{3}f\text{'}\left(x\right)dx=-1 \\ \iff \underset{0}{\overset{1}{\int }}14x^{3}f\text{'}\left(x\right)dx=-7. \end{gathered}$$

Mà $\underset{0}{\overset{1}{\int }}49x^{6}dx=7$

$$\begin{gathered} suyra\underset{0}{\overset{1}{\int }}{\left[f\text{'}\left(x\right)\right]}^{2}dx+\underset{0}{\overset{1}{\int }}7\underset{0}{\overset{1}{\int }}{x}^{3}f\text{'}\left(x\right)dx+\underset{0}{\overset{1}{\int }}49x^{6}dx=0 \\ \iff \underset{0}{\overset{1}{\int }}{\left[f\text{'}\left(x\right)+7x^3\right]}^{2}dx=0. \end{gathered}$$

Vậy 

$f\text{'}\left(x\right)+7x^3=0\Rightarrow 0\Rightarrow f\left(x\right)=-\frac{7}{4}{x}^4+C$