Đáp án:
$A.\dfrac{13}{24}$
Giải thích các bước giải:
$\displaystyle\int\limits^2_{-2} xf'(2x) \, dx\\ =\dfrac{1}{4}\displaystyle\int\limits^2_{-2} 2xf'(2x) \, d(2x)\\ =\dfrac{1}{4}\displaystyle\int\limits^1_{-1} xf'(x) \, dx\\ =\dfrac{1}{4}\left(\displaystyle\int\limits^0_{-1} xf'(x) \, dx+\displaystyle\int\limits^1_{0} xf'(x) \, dx\right)\\ =\dfrac{1}{4}\left(\displaystyle\int\limits^0_{-1} x(x^2-x-1)' \, dx+\displaystyle\int\limits^1_{0} x(2x-1)' \, dx\right)\\ =\dfrac{1}{4}\left(\displaystyle\int\limits^0_{-1} (2x^2-x) \, dx+2\displaystyle\int\limits^1_{0} x \, dx\right)\\ =\dfrac{1}{4}\left(\dfrac{2}{3}x^3-\dfrac{x^2}{2}\right)\Bigg\vert^0_{-1}+\dfrac{1}{2}.\dfrac{x^2}{2}\Bigg\vert^1_{0} \\ =\dfrac{13}{24}$
