Cho hàm số f(x) có đạo hàm liên tục trên [0;6] và thỏa mãn $f\left(0\right)=\mathrm{2,}\underset{0}{\overset{2}{\int }}\left(2x-4\right)f^{\text{'}}\left(x\right)\text{d}x=4.$ Tính $\underset{0}{\overset{6}{\int }}f\left(\frac{x}{3}\right)\text{d}x.$

Đáp án đúng là: D

Đặt $\frac{x}{3}=t\Rightarrow \text{d}x=3\text{d}t.$ Đổi cận: $\left\{\begin{array}{l}x=0\Rightarrow t=0\\ x=6\Rightarrow t=2\end{array}\right.\Rightarrow \underset{0}{\overset{6}{\int }}f\left(\frac{x}{3}\right)\text{d}x=3\underset{0}{\overset{2}{\int }}f\left(t\right)\text{d}t$.

Đặt: $\left\{\begin{array}{l}2x-4=u\\ f^{\text{'}}\left(x\right)\text{d}x=\text{d}v\end{array}\right.\Rightarrow \left\{\begin{array}{l}2\text{d}x=\text{d}u\\ f\left(x\right)=v\end{array}\right.$.

$\Rightarrow \underset{0}{\overset{2}{\int }}\left(2x-4\right)f^{\text{'}}\left(x\right)\text{d}x={\left.\left(2x-4\right)f\left(x\right)\right|}_0^2-2\underset{0}{\overset{2}{\int }}f\left(x\right)\text{d}x$$=4f\left(0\right)-2\underset{0}{\overset{2}{\int }}f\left(x\right)\text{d}x=4$.

$\Rightarrow 4\cdot 2-2\underset{0}{\overset{2}{\int }}f\left(x\right)\text{d}x=4\Rightarrow \underset{0}{\overset{2}{\int }}f\left(x\right)\text{d}x=2$.

Nên: $\underset{0}{\overset{6}{\int }}f\left(\frac{x}{3}\right)\text{\hspace{0.17em}d}x=3\underset{0}{\overset{2}{\int }}f\left(x\right)\text{d}x=6$.