Cho hàm số \(f\left( x \right)\) có \(f\left( 2 \right) = 0\) và \(f'\left( x \right) = \dfrac{{x + 7}}{{\sqrt {2x - 3} }}\), \(\forall x \in \left( {\dfrac{3}{2}; + \infty } \right)\) . Biết rằng \(\int\limits_4^7 {f\left( {\dfrac{x}{2}} \right)dx}  = \dfrac{a}{b}\) (\(a,\,\,b \in \mathbb{Z}\), \(b > 0\), \(\dfrac{a}{b}\) là phân số tối giản). Khi đó \(a + b\) bằng:

Xét tích phân \(\int\limits_4^7 {f\left( {\dfrac{x}{2}} \right)dx}  = \dfrac{a}{b}\).

Đặt \(t = \dfrac{x}{2} \Rightarrow dt = \dfrac{1}{2}dx \Leftrightarrow dx = 2dt\). Đổi cận: \(\left\{ \begin{array}{l}x = 4 \Rightarrow t = 2\\x = 7 \Rightarrow t = \dfrac{7}{2}\end{array} \right.\).

Khi đó ta có: \(I = 2\int\limits_2^{\dfrac{7}{2}} {f\left( t \right)dt} \).

Đặt \(\left\{ \begin{array}{l}u = f\left( t \right)\\dv = dt\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = f'\left( t \right)dt\\v = t - \dfrac{7}{2}\end{array} \right.\), khi đó ta có:

\(\begin{array}{l}I = 2\left( {\left. {\left( {t - \dfrac{7}{2}} \right)f\left( t \right)} \right|_2^{\dfrac{7}{2}} - \int\limits_2^{\dfrac{7}{2}} {\left( {t - \dfrac{7}{2}} \right)f'\left( t \right)dt} } \right)\\I = 2\left( {\dfrac{7}{2}f\left( 0 \right) - \int\limits_2^{\dfrac{7}{2}} {\left( {x - \dfrac{7}{2}} \right)f'\left( x \right)dx} } \right)\\I = 2\left( {\dfrac{7}{2}f\left( 2 \right) - \int\limits_2^{\dfrac{7}{2}} {\left( {x - \dfrac{7}{2}} \right).\dfrac{{x + 7}}{{\sqrt {2x - 3} }}dx} } \right)\\I =  - 2\int\limits_2^{\dfrac{7}{2}} {\left( {x - \dfrac{7}{2}} \right).\dfrac{{x + 7}}{{\sqrt {2x - 3} }}dx} \\I = \dfrac{{236}}{{15}}\\ \Rightarrow a = 236,\,\,b = 15.\end{array}\)

Vậy \(a + b = 236 + 15 = 251\) .