Đáp án:
$A.\ -6$
Giải thích các bước giải:
$\quad I= \displaystyle\int\limits_{\ln\tfrac12}^{\ln2}\left(e^x -\dfrac12\right)e^xf\left(2e^x\right)dx$
Đặt $t = 2e^x\longrightarrow e^x =\dfrac t2$
$\to dt = 2e^xdx$
Đổi cận:
$x\quad\Big|\quad \ln\dfrac12\qquad \ln2$
$\overline{\ t\quad\Big|\qquad 1\qquad\quad 4\quad}$
Ta được:
$\quad I =\dfrac12\displaystyle\int\limits_1^4\left(\dfrac t2 - \dfrac12\right)f(t)dt$
$\to I= \dfrac14\displaystyle\int\limits_1^4(t-1)f(t)dt$
Đặt $\begin{cases}u = f(t)\\dv = (t-1)dt\end{cases}\longrightarrow \begin{cases}du = f'(t)dt\\v =\dfrac12(t-1)^2\end{cases}$
Ta được:
$\quad I = \dfrac18(t-1)^2f(t)\Bigg|_1^4 - \dfrac18\displaystyle\int\limits_1^4(t-1)^2f'(t)dt$
$\to 8I = (4-1)^2f(4) - (1-1)^2f(1) - \displaystyle\int\limits_1^4(t-1)^2f'(t)dt$
$\to 8.3 = 3^2.2 - \displaystyle\int\limits_1^4(t-1)^2f'(t)dt$
$\to \displaystyle\int\limits_1^4(t-1)^2f'(t)dt = - 6$
$\to \displaystyle\int\limits_1^4(x-1)^2f'(x)dx = - 6$
