Đáp án:

\( 9 f''(x) + [f'(x) - x]^2 = 9\)

\(\Rightarrow 9  f''(x) = -[f'(x) - x]^2. \)

\( \int \frac{f''(x) - 1}{[f'(x) - x]^2}  = -\frac{1}{9}  \Rightarrow  \int \frac{f''(x) - 1}{[f'(x) - x]^2} \, dx =  -\frac{1}{9} x + C. \)

\(\Rightarrow - \frac{1}{f'(x) - x} = -\frac{1}{9} x + C  \)

- \( f'(0) = 9 \implies C = -\frac{1}{9} \)

\( \frac{1}{f'(x) - x} = \frac{1}{9} (x + 1) \implies f'(x) = \frac{9}{x + 1} + x. \)

\( f(1) - f(0) = \int_0^1 f'(x) \, dx = \int_0^1 \left( x + \frac{9}{x+1} \right) \, dx. \)

\( = \frac{x^2}{2} + 9 \ln(x+1) \big|_0^1 = \frac{1}{2} + 9 \ln 2  \Rightarrow \boxed{C} \)

page11

Đáp án:

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

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

\( \Rightarrow 2 \int_0^1 f(x) \, dx - 3 \int_0^1 f(x) \, dx = \int_0^1 \sqrt{1-x^2} \, dx = \frac{\pi}{4}. \)

\( 5I = \frac{\pi}{4} \implies I = \frac{\pi}{20} \Rightarrow \boxed{C}. \)

Đọc nhanh kết quả:

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

2. \( \int_{-1}^2 f(x) \, dx = 8 \) và \( \int_{-1}^{-3} f(-2x) \, dx = 3 \). Tính \( I = \int_1^6 f(x) \, dx \).

\( 3 = \int_{-1}^{-3} f(-2x) \, dx = -\frac{1}{2} \int_{-1}^{-3} f(-2x) \, d(-2x) = -\frac{1}{2} \int_2^6 f(x) \, dx. \)

\(\Rightarrow \int_2^6 f(x) \, dx = -6 \Rightarrow I = 2-6 =-4\)

Do đó:

\( I = \int_1^6 f(x) \, dx = 2 - 6 = -4. \)

page12

Đáp án:

\( \int_1^4 f(x) \, dx = \int_1^4 \frac{f(2\sqrt{x} - 1)}{\sqrt{x}} \, dx + \int_1^4 \frac{\ln x}{x} \, dx. \)

\({\int_1^4 f(2\sqrt{x} - 1) \, d(2\sqrt{x} - 1)  + \frac{\ln^2 x}{2} }\big|_1^4\)

\(= \int_1^3 f(x) \, dx + \frac{\ln^2 4}{2}\)

\(\Rightarrow \int_3^4 f(x) \, dx = \int_1^4 f(x) \, dx - \int_1^3 f(x) \, dx = 2\ln^2 2  \Rightarrow \boxed{A}\)

page13

Đáp án:

Xét \( \int_0^1 x^3 f(x) dx. \)
\( \begin{cases} 
u = f(x)  \\ 
dv = x^3 dx 
\end{cases} \Rightarrow \begin{cases} du = f'(x)  \\ v = \frac{x^4}{4} \end{cases}\)

\( \frac{1}{2} = \int_0^1 x^3 f(x) dx =  \frac{x^4}{4} f(x) \big|_0^1 - \int_0^1 \frac{x^4}{4} f'(x) dx. \)

\( \Rightarrow \int_0^1 \frac{x^4}{4} f'(x) dx = -1. \)

Tìm \( k \) sao cho:

\( \int_0^1 \left[ f'(x) - kx^4 \right]^2 dx = 0 \Leftrightarrow \int_0^1 \left( f'(x) \right)^2 dx - 2k \int_0^1 x^4 f'(x) dx + k^2 \int_0^1 x^8 dx = 0. \)

\( \Rightarrow 9 + 2k  +   \frac{k^2}{9} = 0 \Leftrightarrow k^2 + 18k + 81 = 0 \Leftrightarrow  k=-9\)

\( \Rightarrow f'(x) = -9x^4 \implies f(x) = -\frac{9x^5}{5} + C. \)

\( f(1) = 1 \Rightarrow C = 1 + \frac{9}{5} = \frac{14}{5}. \)

\( \Rightarrow\int_0^1 f(x) dx = \int_0^1 \left( -\frac{9x^5}{5} + \frac{14}{5} \right) dx = \frac{5}{2}. \)

page14

Đáp án:

\( \frac{2}{5} = \int_0^1 f(\sqrt{x}) dx = \int_0^1 f(\sqrt{x}) \cdot 2\sqrt{x} \, d(\sqrt{x}) = 2 \int_0^1 t f(t) dt. \)

\( \begin{cases}
u = f(t)  \\
dv = t dt
\end{cases}  \implies \begin{cases} du = f'(t) dt \\ v = \frac{t^2}{2} \end{cases}\)

\( \frac{1}{5} = \int_0^1 t f(t) dt = \frac{t^2}{2} f(t) \big|_0^1 - \int_0^1 \frac{t^2}{2} f'(t) dt \)

\( = \frac{1}{2} - \int_0^1 \frac{t^2}{2} f'(t) dt \implies \int_0^1 t^2 f'(t) dt = 2 \left( \frac{1}{2} - \frac{1}{5} \right) = \frac{3}{5}. \)

\(\implies \int_0^1 x^2 f'(x) dx = \frac{3}{5}\)

Tìm \( k \) sao cho:

\( \int_0^1 \left[ f'(x) - kx^2 \right]^2 dx = 0 \Leftrightarrow \int_0^1 \left( f'(x) \right)^2 dx - 2k \int_0^1 x^2 f'(x) dx + \int_0^1 k^2 x^4 dx = 0. \)

\( \frac{9}{5} - \frac{6k}{5} + \frac{k^2}{5} = 0 \Leftrightarrow (k-3)^3=0 \Leftrightarrow k = 3. \)

Suy ra: \( f'(x) = 3x^2 \implies f(x) = x^3 + C. \)

\( f(1) = 1 \Rightarrow C = 0. \)

\( f(x) = x^3 \Rightarrow \int_0^1 f(x) dx = \frac{1}{4} \implies \boxed{D} \)

page15