Đáp án:

Đặt \( t = x - 2 \implies x = t + 2 \).  

\( f(4 - x) = f(x) \Leftrightarrow f(2 - t) = f(t + 2) \)  

Thay đổi t bởi -t hàm số f có giá trị không đổi:

\( \begin{cases} x = 1 & \Rightarrow t = -1 \\ x = 3 & \Rightarrow t = 1 \end{cases} \)

\(5 = \int_1^3 x f(x) \, dx  = \int_{-1}^1 (t + 2) f(t + 2) \, dt \)  

\( = \int_{-1}^1 t f(t + 2) \, dt + 2 \int_{-1}^1 f(t + 2) \, dt \)  

\(= 0 + \int_{-1}^1 t f(t + 2) \, dt \)  

(Vì \( h(t) = t f(t + 2) \) là hàm lẻ) 

\( = 2 \int_1^3 f(x) \, dx  \implies \int_1^3 f(x) \, dx = \frac{5}{2} \Rightarrow \boxed{A} \).

page6

\( \int_1^4 \frac{f(\sqrt{x})}{\sqrt{x}} \, dx = 2 \int_1^4 f(\sqrt{x}) \, d(\sqrt{x}) = 2 \int_1^2 f(x) \, dx  = 4. \)

\( \int_0^1 x^2 f(x^3) \, dx = \frac{1}{3} \int_0^1 f(x^3) \, d(x^3) = \frac{1}{3} \int_0^1 f(x) \, dx = 1. \)

page7

Đáp án:

\(3 = \int_1^2 f(x-1) dx = \int_1^2 f(x -1) d(x-1) = \int_0^1 f(t) dt\)

Xét \( I = \int_0^1 x^3 f'(x^2) dx \)

\( \begin{cases} u = x^2 \\ dv = x f'(x^2) dx \end{cases} \Rightarrow \begin{cases} du = 2x dx \\  v = \frac{1}{2} f(x^2)\end{cases} \)

\(\left( v = \int x f'(x^2) dx = \frac{1}{2} \int f'(x^2) d(x^2) = \frac{1}{2} f(x^2) \right)\)

\(I = \frac{1}{2} x^2 f(x^2) \big|_0^1 - \frac{1}{2} \int_0^1 2x f(x^2) dx\)

\(= 2 - \frac{1}{2} \int_0^1 f(x^2) dx^2 = 2 - \frac{1}{2} \int_0^1 f(t) dt = 2 - \frac{3}{2} = \frac{1}{2} \Rightarrow \boxed{B} \)

Gợi ý: \( \int f'(u) du = f(u)\)

page8

Đáp án:

Xét \( \int_0^{\frac{\pi}{4}} f(\tan x) \, dx \).

Đặt \( t = \tan x \implies dt = (1 + \tan^2 x) dx = (1 + t^2) dx \),  
\( \Rightarrow dx = \frac{1}{1 + t^2} \, dt. \)

 \( \begin{cases} x = 0 \Rightarrow t = 0 \\ x = \frac{\pi}{4} \Rightarrow t = 1 \end{cases}\).

\( 4 = \int_0^{\frac{\pi}{4}} f(\tan x) \, dx = \int_0^1 \frac{f(t)}{1 + t^2} \, dt. \)

\( 2 = \int_0^1 \frac{x^2 f(x)}{x^2 + 1} \, dx = \int_0^1 \frac{(x^2 + 1 - 1) f(x)}{x^2 + 1} \, dx. \)

\( = \int_0^1 f(x) \, dx - \int_0^1 \frac{f(x)}{x^2 + 1} \, dx. \)

\( \Rightarrow \int_0^1 f(x) \, dx = 2 + \int_0^1 \frac{f(x)}{x^2 + 1} \, dx = 2 + 4 = 6 \Rightarrow \boxed{B} \)

page9

Đáp án:

\( \frac{f'(x)}{f(x)} = \frac{1}{\sqrt{3x+1}} \implies \int \frac{f'(x)}{f(x)} \, dx = \int \frac{1}{\sqrt{3x+1}} \, dx. \)

\(= \frac{1}{3} \int 3(3x+1)^{-\frac{1}{2}} \, dx = \frac{2}{3} \sqrt{3x+1} + C.\)

\( \Rightarrow \ln(f(x))  = \frac{2}{3} \sqrt{3x+1} + C. \)

\( x = 1 \implies  \frac{4}{3}  + C = 0 \implies C = -\frac{4}{3} \)

\(\Rightarrow \ln(f(x)) = \frac{2}{3} \sqrt{3x+1} - \frac{4}{3}. \)

\( \Rightarrow f(x) = e^{\frac{2}{3} \sqrt{3x+1} - \frac{4}{3}}. \)

\(\Rightarrow f(5) \approx 3.79366 \Rightarrow \boxed{C} \)

page10