Cho hàm số \(f\left( x \right)\) liên tục trên \(\mathbb{R}.\) Gọi \(F\left( x \right),\,\,G\left( x \right)\) là hai nguyên hàm của \(f\left( x \right)\) trên \(\mathbb{R}\) thỏa mãn \(F\left( 8 \right) + G\left( 8 \right) = 8\) và \[F\left( 0 \right) + G\left( 0 \right) = - 2.\] Khi đó \(\int\limits_{ - 2}^0 {f\left( { - 4x} \right)} \,dx\) bằng

Ta có \(\int\limits_{ - 2}^0 {f\left( { - 4x} \right)} \,dx =  - \frac{1}{4}\int\limits_{ - 2}^0 {f\left( { - 4x} \right)} \,{\rm{d}}\left( { - 4x} \right)\)

\( =  - \frac{1}{4}\int\limits_8^0 {f\left( x \right)} \,{\rm{d}}x = \frac{1}{4}\int\limits_8^0 {f\left( x \right)} \,{\rm{d}}x = \frac{1}{4}\left[ {F\left( 8 \right) - F\left( 0 \right)} \right]\).

Vì \(F\left( x \right),\,\,G\left( x \right)\) là hai nguyên hàm của \(f\left( x \right)\) nên \(G\left( x \right) = F\left( x \right) + C\).

Khi đó \[\left\{ {\begin{array}{*{20}{l}}{G\left( 8 \right) = F\left( 8 \right) + C}\\{G\left( 0 \right) = F\left( 0 \right) + C}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{G\left( 8 \right) = F\left( 8 \right) + C}\\{G\left( 0 \right) = F\left( 0 \right) + C}\end{array}} \right.} \right.\]

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{8 - F\left( 8 \right) = F\left( 8 \right) + C}\\{ - 2 - F\left( 0 \right) = F\left( 0 \right) + C}\end{array}} \right.\)\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{F\left( 8 \right) = \frac{2}}\\{F\left( 0 \right) = \frac{{ - 2 - C}}{2}}\end{array} \Rightarrow F\left( 8 \right) - F\left( 0 \right) = 5.} \right.\)

Vậy \(\int\limits_{ - 2}^0 {f\left( { - 4x} \right)} \,dx = \frac{1}{4}\left[ {F\left( 8 \right) - F\left( 0 \right)} \right] = \frac{5}{4}.\) Chọn A.