Cho hàm số f(x) có f(2) = 0 và ${\text{f}}^{\text{'}}\left(\text{x}\right)=\frac{\text{x}+7}{\sqrt{2\text{x}-3}},\forall \text{x}\in \left(\frac{3}{2};+\infty \right)$. Biết rằng ${\int }_4^7\text{f}\left(\frac{\text{x}}{2}\right)\text{dx}=\frac{\text{a}}{\text{b}}$ ($\text{a},\text{b}\in \mathbb{Z},\text{b}>0,\frac{\text{a}}{\text{b}}$ là phân số tối giản). Khi đó a + b bằng

Ta có: ${\int }_4^{7}f\left(\frac{x}{2}\right)dx=2{\int }_2^{7/2}f\left(\frac{x}{2}\right)d\left(\frac{x}{2}\right)=2{\int }_2^{\frac{7}{2}}f\left(x\right)dx=2{\int }_2^{\frac{7}{2}}f\left(x\right)d\left(x-\frac{7}{2}\right)$

Đặt $\left\{\begin{array}{l}u=f\left(x\right)\\ dv=d\left(x-\frac{7}{2}\right)\end{array}\Rightarrow \left\{\begin{array}{l}du=f^{\text{'}}\left(x\right)dx\\ v=\left(x-\frac{7}{2}\right)\end{array}\right.\right.$

Khi đó: $2{\int }_2^{\frac{7}{2}}f\left(x\right)dx=2{\int }_2^{\frac{7}{2}}f\left(x\right)d\left(x-\frac{7}{2}\right)=2\left[{\left.\left(x-\frac{7}{2}\right)f\left(x\right)\right|}_0^{\frac{7}{2}}-{\int }_2^{\frac{7}{2}}\left(x-\frac{7}{2}\right)f^{\text{'}}\left(x\right)dx\right]$