Question

[T. Hữu Lộc]

Cho hàm số FX có đạo hàm liên tục trên đoạn [0;1] bằng mãn f (1) = 1 và (f'(x))²+4(6x²-1).f(x)=40x⁶-44x⁴+32x²-4,Tích phân cân từ 0 đến 1 của xf(x)dx bằng

A.-13/15

B.5/12

C.13/15

B.-5/12

Answer 1

[Nguyễn Việt Lâm]

Lấy tích phân 2 vế:

\(\int\limits^1_0\left[f'\left(x\right)\right]^2dx+\int\limits^1_04\left(6x^2-1\right)f\left(x\right)dx=\int\limits^1_0\left(40x^6-44x^4+32x^2-4\right)dx=\frac{376}{105}\)

Xét \(I=\int\limits^1_0\left(6x^2-1\right)f\left(x\right)dx\)

Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=\left(6x^2-1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=2x^3-x\end{matrix}\right.\)

\(\Rightarrow I=\left(2x^3-x\right)f\left(x\right)|^1_0-\int\limits^1_0\left(2x^3-x\right)f'\left(x\right)dx=1-\int\limits^1_0\left(2x^3-x\right)f'\left(x\right)dx\)

\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+4-4\int\limits^1_0\left(2x^3-x\right)f'\left(x\right)dx=\frac{376}{105}\)

\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)-2\left(2x^3-x\right)\right]^2dx-\int\limits^1_04\left(2x^3-x\right)^2dx=-\frac{44}{105}\)

\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)-2\left(2x^3-x\right)\right]^2dx-\frac{44}{105}=-\frac{44}{105}\)

\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)-\left(4x^3-2x\right)\right]^2dx=0\)

\(\Rightarrow f'\left(x\right)=4x^3-2x\Rightarrow f\left(x\right)=x^4-x^2+C\)

\(f\left(1\right)=1\Rightarrow1-1+C=1\Rightarrow C=1\)

\(\Rightarrow f\left(x\right)=x^4-x^2+1\)

\(\Rightarrow\int\limits^1_0x\left(x^4-x^2+1\right)dx=\frac{5}{12}\)