Cho hàm số \(f\left(x\right)\) có đạo hàm liên tục trên đoạn [0;1] thỏa mãn \(f\left(1\right)=0;\int_0^1\left[f'\left(x\right)\right]^2\text{d}x=7\) và \(\int_0^1x^2f\left(x\right)\text{d}x=\dfrac{1}{3}\). Tích phân \(\int_0^1f\left(x\right)\text{d}x\) bằng
\(\dfrac{7}{5}\).\(1\).\(\dfrac{7}{4}\).\(4\).Hướng dẫn giải:\(\int_0^1x^2f\left(x\right)\text{d}x=\dfrac{1}{3}\)
Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=x^2\text{d}x\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)\text{d}x\\v=\dfrac{x^3}{3}\end{matrix}\right.\)
Áp dụng công thức tích phân từng phần ta có:
\(\int_0^1x^2f\left(x\right)\text{d}x=\dfrac{1}{3}x^3.f\left(x\right)|^1_0-\dfrac{1}{3}\int_0^1x^3f'\left(x\right)\text{d}x=-\dfrac{1}{3}\int_0^1x^3f'\left(x\right)\text{d}x\) (Vì f(1) = 0)
\(\int_0^1x^3f'\left(x\right)\text{d}x=-3\int_0^1x^3f'\left(x\right)\text{d}x=-1\)
Ta cũng có: \(\left\{{}\begin{matrix}\int_0^1\left[f'\left(x\right)\right]^2\text{d}x=7\\\int_0^114x^3f'\left(x\right)\text{d}x=-14\\\int_0^149x^6\text{d}x=7\end{matrix}\right.\)
\(\Rightarrow\int_0^1\left[f'\left(x\right)\right]^2\text{d}x+\int_0^114x^3f'\left(x\right)\text{d}x+\int_0^149x^6\text{d}x=0\)
\(\Leftrightarrow\int_0^1\left[f'\left(x\right)+7x^3\right]^2\text{d}x=0\)
Mà \(\int_0^1\left[f'\left(x\right)+7x^3\right]^2\text{d}x\ge0\) nên đẳng thức xảy ra khi \(f'\left(x\right)+7x^3=0\Leftrightarrow f'\left(x\right)=-7x^3\)
\(\Rightarrow f\left(x\right)=-\dfrac{7x^4}{4}+C\)
Ta có \(f\left(1\right)=0\Leftrightarrow C=\dfrac{7}{4}\Rightarrow f\left(x\right)=\dfrac{7}{4}\left(1-x^4\right)\)
\(\Rightarrow\int_0^1f\left(x\right)\text{d}x=\dfrac{7}{4}\int_0^1\left(1-x^4\right)\text{d}x=\dfrac{7}{4}\left(x-\dfrac{x^5}{5}\right)\bigg|^1_0=\dfrac{7}{4}\left(1-\dfrac{1}{5}\right)=\dfrac{7}{5}\).