Cho hàm số \(f\left( x \right)\) có đạo hàm xác định trên \(\mathbb{R}.\) Biết \(f\left( 1 \right) = 2\) và \(\int\limits_0^1 {{x^2}f'\left( x \right){\rm{d}}x}  = \int\limits_1^4 {\frac{{1 + 3\sqrt x }}{{2\sqrt x }}f\left( {2 - \sqrt x } \right){\rm{d}}x}  = 4.\) Giá trị của \(\int\limits_0^1 {f\left( x \right){\rm{d}}x} \) bằng

Đặt \(t = 2 - \sqrt x  \Leftrightarrow {\rm{d}}t =  - \frac{1}{{2\sqrt x }}\;{\rm{d}}x \Leftrightarrow  - {\rm{d}}t = \frac{1}{{2\sqrt x }}\;{\rm{d}}x\)

Đổi cận: \(x = 1 \Rightarrow t = 1\,;\,\,x = 4 \Rightarrow t = 0.\)

Do đó \(\int\limits_1^4 {\frac{{1 + 3\sqrt x }}{{2\sqrt x }}f\left( {2 - \sqrt x } \right){\rm{d}}x}  = 4 \Leftrightarrow \int\limits_1^4 {\left( {1 + 3\sqrt x } \right)f\left( {2 - \sqrt x } \right)\frac{{{\rm{d}}x}}{{2\sqrt x }} = 4} \)

\( \Leftrightarrow \int\limits_1^0 {\left( {7 - 3t} \right)f\left( t \right)\left( { - {\rm{d}}t} \right)}  = 4 \Leftrightarrow \int\limits_0^1 {\left( {7 - 3x} \right)f\left( x \right){\rm{d}}x}  = 4.\)

Ta có \(\int\limits_0^1 {{x^2}f'\left( x \right){\rm{d}}x}  = 4 \Leftrightarrow \int\limits_0^1 {{x^2}{\rm{d}}\left( {f\left( x \right)} \right)}  = \left. {{x^2}f\left( x \right)} \right|_0^1 - \int\limits_0^1 {2x \cdot f\left( x \right){\rm{d}}x = 4} \)

\[ \Leftrightarrow f\left( 1 \right) - 2 \cdot \int\limits_0^1 {x \cdot f\left( x \right){\rm{d}}x = 4}  \Leftrightarrow \int\limits_0^1 {x \cdot f\left( x \right){\rm{d}}x =  - 1} \].

Suy ra \(\int\limits_0^1 {\left( {7 - 3x} \right)\,} f\left( x \right){\rm{d}}x = 4 \Leftrightarrow 7 \cdot \int\limits_0^1 {f\left( x \right)} \,{\rm{d}}x - 3 \cdot \int\limits_0^1 {x \cdot f\left( x \right)} \,{\rm{d}}x = 4\)

\( \Leftrightarrow 7 \cdot \int\limits_0^1 {f\left( x \right)} \,{\rm{d}}x - 3 \cdot \left( { - 1} \right) = 4 \Leftrightarrow \int\limits_0^1 {f\left( x \right)} \,{\rm{d}}x = \frac{1}{7}\). Chọn D.