Cho hàm số $f(x)$ có đạo hàm $f'(x)$ liên tục trên $\mathbb{R}$ và thỏa mãn $\displaystyle\displaystyle\int\limits_{0}^{1}(3x+1)f'(x)\mathrm{\,d}x=2022$ và $4f(1)-f(0)=2028$. Giá trị của $I=\displaystyle\displaystyle\int\limits_{0}^{\tfrac{1}{4}}f(4x)\mathrm{\,d}x$ là
 | $2$ |
 | $\dfrac{2022}{3}$ |
 | $\dfrac{1}{2}$ |
 | $\dfrac{1}{4}$ |
Chọn phương án C.
u=3x+1\\ v'=f'(x)
\end{cases}\Rightarrow\begin{cases}
u'=3\\ v=f(x).
\end{cases}$
$\begin{array}{lcl}
\Rightarrow&\displaystyle\int\limits_{0}^{1}(3x+1)f'(x)\mathrm{\,d}x&=(3x+1)f(x)\bigg|_0^1-\int\limits_{0}^{1}3f(x)\mathrm{\,d}x\\
\Rightarrow&2022&=\big[4f(1)-f(0)\big]-3\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x\\
\Rightarrow&2022&=2028-3\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x\\
\Rightarrow&\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x&=2.
\end{array}$
Đặt $t=4x\Rightarrow\mathrm{d}t=4\mathrm{d}x$ hay $\mathrm{d}x=\dfrac{1}{4}\mathrm{d}t$.
- $x=\dfrac{1}{4}\Rightarrow t=1$
- $x=0\Rightarrow t=0$
Ta có $\begin{aligned}[t]
I&=\displaystyle\int\limits_{0}^{\tfrac{1}{4}}f(4x)\mathrm{\,d}x=\int\limits_0^1f(t)\cdot\dfrac{1}{4}\mathrm{\,d}t\\
&=\dfrac{1}{4}\displaystyle\int\limits_0^1f(t)\mathrm{\,d}t=\dfrac{1}{4}\int\limits_0^1f(x)\mathrm{\,d}x\\
&=\dfrac{1}{4}\cdot2=\dfrac{1}{2}.
\end{aligned}$