Cho hàm số f(x) có đạo hàm liên tục trên đoạn [0;1...

B

- Hướng dẫn giải

Đặt \(t = \sqrt x \Rightarrow {t^2} = x \Rightarrow {\rm{d}}x = 2t{\rm{d}}t\). Đổi cận \(x = 0 \Rightarrow t = 0;\,\,x = 1 \Rightarrow t = 1\)

Suy ra \(\int\limits_0^1 {f\left( {\sqrt x } \right){\rm{d}}x} = 2\int\limits_0^1 {t.f\left( t \right){\rm{d}}t} \) \( \Leftrightarrow \int\limits_0^1 {t.f\left( t \right){\rm{d}}t} = \frac{1}{5}\).

Do đó \( \Leftrightarrow \int\limits_0^1 {x.f\left( x \right){\rm{d}}x} = \frac{1}{5}\)

Mặt khác \(\int\limits_0^1 {x.f\left( x \right)} {\rm{d}}x = \left. {\frac{{{x^2}}}{2}f\left( x \right)} \right|_0^1 - \int\limits_0^1 {\frac{{{x^2}}}{2}f'\left( x \right){\rm{d}}x} = \frac{1}{2} - \int\limits_0^1 {\frac{{{x^2}}}{2}f'\left( x \right){\rm{d}}x} \).

Suy ra \(\int\limits_0^1 {\frac{{{x^2}}}{2}f'\left( x \right){\rm{d}}x} = \frac{1}{2} - \frac{1}{5} = \frac{3}{{10}} \Rightarrow \int\limits_0^1 {{x^2}f'\left( x \right){\rm{d}}x} = \frac{3}{5}\)

Ta tính được \(\int\limits_0^1 {{{\left( {3{x^2}} \right)}^2}} {\rm{d}}x = \frac{9}{5}\).

Do đó \(\int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}{\rm{d}}x - 2\int\limits_0^1 {3{x^2}f'\left( x \right){\rm{d}}x} } + \int\limits_0^1 {{{\left( {3{x^2}} \right)}^2}} {\rm{d}}x = 0 \Leftrightarrow \int\limits_0^1 {{{\left( {f'\left( x \right) - 3{x^2}} \right)}^2}{\rm{d}}x} = 0\)

\( \Leftrightarrow f'\left( x \right) - 3{x^2} = 0 \Leftrightarrow f'\left( x \right) = 3{x^2} \Leftrightarrow f\left( x \right) = {x^3} + C\).

Vì \(f\left( 1 \right) = 1\) nên \(f\left( x \right) = {x^3}\). Vậy \(I = \int\limits_0^1 {f\left( x \right)} {\rm{d}}x = \int\limits_0^1 {{x^3}} {\rm{d}}x = \frac{1}{4}\).