Xét cái: [tex]\displaystyle \int ^1_0 f(\sqrt{x})dx=\frac{2}{5}[/tex]
Đặt $t= \sqrt{x}$ có [tex]dt=\frac{1}{2\sqrt{x}}dx\Leftrightarrow 2t.dt=dx[/tex]
Có: [tex]\displaystyle \int ^1_0 2t f(t)dt=\frac{2}{5}\\ \displaystyle \Leftrightarrow f(t).t^2|^1_0 - \int ^1_0 t^2 f'(t)dt=\frac{2}{5}\\ \displaystyle \Leftrightarrow 1-\int ^1_0 t^2 f'(t)dt=\frac{2}{5}\\ \displaystyle \Leftrightarrow \int ^1_0 t^2 f'(t)dt=\frac{3}{5}\\\displaystyle \Leftrightarrow \int ^1_0 3t^2 f'(t)dt=\frac{9}{5}\\\displaystyle \Leftrightarrow \displaystyle \int ^1_0 3x^2 f'(x)dx=\frac{9}{5}[/tex]
Theo đề mình có cái: $\displaystyle \int_0^1 \Big[f'(x)\Big]\, \mathrm dx=\dfrac{9}5$
Nên chọn hàm $f'(x)=3x^2$ có $f(x)=x^3+C$, có $f(1)=1$ nên $C=0$
Vậy [tex]\displaystyle \int ^1_0 f(x)dx= \displaystyle \int ^1_0 x^3 dx=\frac{1}{4}[/tex]