Cho hàm số $\Large f(x)$ có $\Large f(1) = 0$ và $\Large f{}'(x) = 201

Cho hàm số $\Large f(x)$ có $\Large f(1) = 0$ và $\Large f{}'(x) = 2019.2020.x(x - 1)^{2018}$, $\Large \forall x \in \mathbb{R}$. Khi đó $\Large \int_{0}^{1}f(x) dx$ bằng

• A.

  A. $\Large \dfrac{2}{2021}$

• B.

  B. $\Large \dfrac{1}{1011}$

• C.

  C. $\Large -\dfrac{2}{2021}$

• D.

  D. $\Large -\dfrac{1}{1011}$

Chọn C

Cần nhớ: $\Large \int f{}'(x)dx = f(x) + C$

và $\Large \int (ax + b)^{\alpha }dx = \dfrac{1}{a}.\dfrac{(ax + b)^{a + 1}}{\alpha + 1} + C (\alpha \neq -1)$

Ta có:

$\Large f(x) = \int f{}'(x)dx$ 

$\Large = \int 2019.2020.x.(x-1)^{2018}dx$

$\Large = 2019.2020.\int x(x-1)^{2018} dx$

Đặt $\Large t = x - 1 => dt = dx$ và $\Large x = t + 1$.

Suy ra:

$\Large f(x) = 2019.2020.\int (t+1).t^{2018}dt=2019.2020.\int t^{2019}+t^{2018}dt$ 

$\Large = 2019 . 2020 . \left ( \dfrac{t^{2020}}{2020} + \dfrac{t^{2019}}{2019}\right ) + C$ 

$\Large = 2019t^{2020} + 2020t^{2019} + C$.

Từ đó: $\Large f(x) = 2019(x-1)^{2020} + 2020(x-1)^{2019} + C$

Mà: $\Large f(1) = 0$

$\Large \Leftrightarrow 2019(1-1)^{2020} + 2020(1-1)^{2019} + C = 0$

$\Large \Leftrightarrow C = 0$

Suy ra:

$\Large f(x) = 2019(x -1)^{2020} + 2020(x-1)^{2019}$.

Vậy:

$\Large \int_{0}^{1}f(x)dx = \int_{0}^{1}\left [ 2019(x -1)^{2020} + 2020(x-1)^{2019} \right ]dx$ 

$\Large = \left [ 2019.\dfrac{(x-1)^{2021}}{2021} + 2020.\dfrac{(x-1)^{2020}}{2020}\right ]\bigg|_{0}^{1}$ 

$\Large = -\left ( -\dfrac{2019}{2021} + 1\right )$ 

$\Large = -\dfrac{2}{2021}$