Đáp án:
$B.\ I = 31$
Giải thích các bước giải:
Xét $I_1 = 2\displaystyle\int\limits_0^{\tfrac{\pi}{2}}f(\sin x)\cos xdx$
Đặt $u = \sin x$
$\Rightarrow du = \cos xdx$
Đổi cận:
\(\begin{array}{c|ccc}
x&0&&&\dfrac{\pi}{2}\\
\hline
u&0&&&1
\end{array}\)
Ta được:
$\quad I_1 = 2\displaystyle\int\limits_0^1f(u)du$
$\Leftrightarrow I_1 = 2\displaystyle\int\limits_0^1f(x)dx$
Xét $I_2 = 3\displaystyle\int\limits_0^1f(3-2x)dx$
Đặt $t = 3 - 2x$
$\Rightarrow dt = - 2dx$
Đổi cận:
\(\begin{array}{c|ccc}
x&0&&&1\\
\hline
t&3&&&1
\end{array}\)
Ta được:
$\quad I_2 = -\dfrac32\displaystyle\int\limits_3^1f(t)dt$
$\Leftrightarrow I_2 = \dfrac32\displaystyle\int\limits_1^3f(t)dt$
$\Leftrightarrow I_2 = \dfrac32\displaystyle\int\limits_1^3f(x)dx$
Khi đó:
$\quad I = I_1 + I_2$
$\Leftrightarrow I = 2\displaystyle\int\limits_0^1f(x)dx + \dfrac32\displaystyle\int\limits_1^3f(x)dx$
$\Leftrightarrow I = 2\displaystyle\int\limits_0^1(5-x)dx + \dfrac32\displaystyle\int\limits_1^3(x^2 + 3)dx$
$\Leftrightarrow I = 2\left(5x - \dfrac{x^2}{2}\right)\Bigg|_0^1 + \dfrac32\left(\dfrac{x^3}{3} + 3x\right)\Bigg|_1^3$
$\Leftrightarrow I = 2\cdot \dfrac92 + \dfrac32\cdot \dfrac{44}{3}$
$\Leftrightarrow I = 31$
