Đáp án:

\( f(x) \ln(f(x)) + x f'(x) = x f'(x) f(x) \)

\( \Leftrightarrow \ln(f(x)) + x \cdot \frac{f'(x)}{f(x)} = x f'(x) \)

\( \Leftrightarrow \left( x \ln(f(x)) \right)' = x f'(x) \)

\( \Rightarrow x \ln(f(x)) \big|_{0}^{1} = \int_{0}^{1} x f'(x) dx = x f(x) \big|_{0}^{1} - \int_{0}^{1} f(x) dx \)

\( \Rightarrow \int_{0}^{1} f(x) dx = 1 \Rightarrow \boxed{D} \)

page61

​​​​​​​Đáp án:

\( 3 f(x) +f(x)+ x f'(x) = x^{2017} \)

\( \Leftrightarrow 3 f(x) + \left( x f(x) \right)' = x^{2017} \)

\( \Leftrightarrow 3 \int_{0}^{1} f(x) dx + x f(x) \big|_{0}^{1} = \int_{0}^{1} x^{2017} dx \)

\( \Rightarrow 3 \int_{0}^{1} f(x) dx + f(1) = \frac{x^{2018}}{2018} \Big|_{0}^{1} \)

\( f(1) = \frac{1}{2021} \Rightarrow I = \text{A} \)

\( 4 f(x) + x f'(x) = x^{2017} \) (Nhân cả hai vế với \( x^3 \))

\( \Rightarrow 4 x^3 f(x) + x^4 f'(x) = x^{2020} \)

\( \Rightarrow (x^4 f(x))' = x^{2020} \)

\( \Rightarrow x^4 f(x) = \frac{x^{2021}}{2021} + C \)

\( f(1) = \frac{1}{2021}  \Rightarrow C = 0 \)

\( \Rightarrow f(x) = \frac{x^{2017}}{2021} \Rightarrow  \int_{0}^{1} f(x) dx = \frac{1}{2018 \cdot 2021}\Rightarrow \boxed{A}\)

page62

​​​​​​​Đáp án:

Đặt \( x = 2018 - t \Rightarrow dx = -dt \),

\( \begin{cases} x = 0 \Rightarrow t = 2018 \\ x = 2018 \Rightarrow t = 0 \end{cases} \),

\( I = \int_{2018}^{0} \frac{1}{1 + f(2018 - t)} (-dt) = \int_{0}^{2018} \frac{dt}{1 + f(2018 - t)} \),

\( = \int_{0}^{2018} \frac{1}{1 + \frac{1}{f(t)}} dt = \int_{0}^{2018} \frac{f(t)}{f(t) + 1} dt = \int_{0}^{2018} dt - \int_{0}^{2018} \frac{1}{1 + f(t)} dt \),

\( \Rightarrow 2I = 2018 \Rightarrow I = 1009 \Rightarrow \boxed{C} \).

page63

​​​​​​​Đáp án:

Đặt \( x = -t \Rightarrow dx = -dt \)

\( \begin{cases} x = -1 \Rightarrow t = 1 \\ x = 1 \Rightarrow t = -1 \end{cases} \)

\( I = \int_{-1}^{1} \frac{f(x)}{1 + 2018^x} dx = - \int_{1}^{-1} \frac{f(-t)}{1 + 2018^{-t}} (dt) = \int_{-1}^{1} \frac{f(t) \cdot 2018^t}{1 + 2018^t} dt \),

\( = \int_{-1}^{1} \frac{f(t)(2018^t + 1 - 1)}{1 + 2018^t} dt = \int_{-1}^{1} f(t) dt - I \),

\( \Rightarrow 2I = 6 \Rightarrow I = 3  \Rightarrow \boxed{B} \).

page64

​​​​​​​Đáp án:

\( 2^{f(x)} + f(x) = x + 1 \Rightarrow f'(x) 2^{f(x)} + f'(x)f(x) = x f'(x) + f'(x) \)

\( \Rightarrow \int_{0}^{2} f'(x) \cdot 2^{f(x)} dx + \int_{0}^{2} f'(x) \cdot f(x) dx = \int_{0}^{2} x f'(x) dx + \int_{0}^{2} f'(x) dx \)

\(  \frac{2^{f(x)}}{\ln 2} \Big|_{0}^{2} + \frac{(f(x))^2}{2} \Big|_{0}^{2} -  f(x) \Big|_{0}^{2}  = \int_{0}^{2} f(x) dx \)

Đặt: \( \begin{cases} u = x \\ dv = f'(x) dx \end{cases} \Rightarrow \begin{cases} du=dx \\ v= f(x) \end{cases} \)

\( \int_{0}^{2} x f'(x) dx = x f(x) \Big|_{0}^{2} - \int_{0}^{2} f(x) dx \)

\( \Rightarrow \int_{0}^{2} f(x) dx = x f(x) \Big|_{0}^{2} - \frac{2^{f(x)}}{\ln 2} \Big|_{0}^{2} - \frac{(f(x))^2}{2} \Big|_{0}^{2} + f(x) \Big|_{0}^{2} \)

+ \( 2^{f(x)} + f(x) = x + 1, \forall x \in \mathbb{R} \)

\( x = 0: 2^{f(0)} + f(0) = 1 \Leftrightarrow f(0) = 0 \)

\( x = 2: 2^{f(2)} + f(2) = 3 \Leftrightarrow f(2) = 1 \)

\( \Rightarrow \int_{0}^{2} f(x) dx = 2 - \frac{1}{\ln 2} (2 - 1) - \frac{1}{2} + 1 \)

\( = \frac{5}{2} - \frac{1}{\ln 2} \Rightarrow \begin{cases} a = 5\\ b = -1 \end{cases} \Rightarrow a + b = 4 \Rightarrow \boxed{A} \)

page65