Đáp án:

Xét \(\int_{0}^{1} (x^2 + x) f''(x) dx \quad \)  Đặt \(\begin{cases} u = x^2 + x \\ dv = f''(x) dx \end{cases} \Rightarrow \begin{cases} du = (2x + 1) dx \\ v = f'(x) \end{cases}\)  

\(\frac{5}{2} = \int_{0}^{1} (x^2 + x) f''(x) dx = [x^2 + x] f'(x) \big|_{0}^{1} - \int_{0}^{1} (2x + 1) f'(x) dx\)  

\(\frac{5}{2} = 9 - \int_{0}^{1} (2x + 1) f'(x) dx \Rightarrow \int_{0}^{1} (2x + 1) f'(x) dx = 9 - \frac{5}{2} = \frac{13}{2}\).  

Tìm \(k\) sao cho: \(\int_{0}^{1} \left[ f'(x) - k(2x + 1) \right]^2 dx = 0\)  

\(\Leftrightarrow \int_{0}^{1} (f'(x))^2 dx - 2k \int_{0}^{1} (2x + 1) f'(x) dx + k^2 \int_{0}^{1} (2x + 1)^2 dx = 0\)  

\(\Leftrightarrow \frac{39}{4} - 13k + \frac{13}{3}k^2 = 0 \Rightarrow \frac{k^2}{3} - k + \frac{3}{4} = 0 \Leftrightarrow k = \frac{3}{2}\).

Suy ra: \(f'(x) = \frac{3}{2} (2x + 1)\) \(\Rightarrow f(x) = \frac{3x^2}{2} + \frac{3x}{2} + C.\)  

\(f(0) = 0 \Rightarrow C = 0\) \(\Rightarrow f(x) = \frac{3x^2}{2} + \frac{3x}{2}\)  

\(I = \int_{0}^{2} f(x) dx = 7 \Rightarrow \boxed{D}\)  

\(!\) Gặp: Cho \(\int_{a}^{b} (f(x))^2 dx = ...\) hoặc \(\int_{a}^{b} g(x) f(x) dx = ... \), thì tìm \(k\) sao cho: \(\int_{a}^{b} \left[ f(x) - k g(x) \right]^2 dx = 0\) \(\Rightarrow f(x) = k g(x).\)

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Đáp án:

Xét \(\int_{0}^{\frac{\pi}{4}} f'(x) \sin{2x} dx\).  

Đặt: \(\begin{cases} u = \sin{2x} \\ dv = f'(x) dx \end{cases} \Rightarrow \begin{cases} du = 2 \cos{2x} dx \\ v = f(x) \end{cases}\)  

\(-\frac{\pi}{4} = \int_{0}^{\frac{\pi}{4}} f'(x) \sin{2x} dx =  f(x) \sin{2x} \big|_{0}^{\frac{\pi}{4}} - 2\int_{0}^{\frac{\pi}{4}} f(x) \cdot \cos{2x} dx\)  

\(\Rightarrow \int_{0}^{\frac{\pi}{4}} f(x) \cos{2x} dx = \frac{\pi}{8}.\)  

Tìm \(k\) sao cho: \(\int_{0}^{\frac{\pi}{4}} \left[f(x) - k \cos{2x}\right]^2 dx = 0\)  

\(\Leftrightarrow \int_{0}^{\frac{\pi}{4}} (f(x))^2 dx - 2k \int_{0}^{\frac{\pi}{4}} f(x) \cos{2x} dx + k^2 \int_{0}^{\frac{\pi}{4}} \cos^2{2x} dx = 0\)  

\(\Leftrightarrow \frac{\pi}{8} -  \frac{\pi}{4}k + \frac{\pi}{8} k^2 = 0 \iff k^2 - 2k + 1 = 0 \iff k = 1.\)  

Suy ra: \(f(x) = \cos{2x}\).  

\(I = \int_{0}^{\frac{\pi}{8}} f(2x) dx = \int_{0}^{\frac{\pi}{8}} \cos{4x} dx = \frac{1}{4} \Rightarrow \boxed{B}\)

page32

Đáp án:

Đặt \(h(x) = \frac{f'(x)}{[f(x)]^2}\).  

\(\Rightarrow \int_{4}^{8} h(x) dx = \int_{4}^{8} \frac{f'(x)}{[f(x)]^2} dx = -\frac{1}{f(x)} \big|_{4}^{8} = -2 + 4 = 2.\)  

\(\int_{4}^{8} [h(x)]^2 dx = 1.\)  

Tìm \(k\) sao cho: \(\int_{4}^{8} \left[h(x) - k\right]^2 dx = 0.\)  

\(\Leftrightarrow \int_{4}^{8} [h(x)]^2 dx - 2k \int_{4}^{8} h(x) dx + k^2 \int_{4}^{8} dx = 0\)  

\(\Leftrightarrow 1 - 4k + 4k^2 = 0 \iff (2k - 1)^2 = 0 \iff k = \frac{1}{2}.\)  

\(\Rightarrow h(x) =k \iff h(x) = \frac{1}{2} \iff \frac{f'(x)}{[f(x)]^2} = \frac{1}{2}\).  

\(\Rightarrow \frac{-1}{[f(x)]^2} =  \frac{1}{2} x + C.\)  

\(f(4) = \frac{1}{4} \Rightarrow  -4 = 2 + C \Rightarrow C = -6.\)  

\(\Rightarrow f(x) = \frac{-1}{\frac{x}{2}- 6} \Rightarrow f(6) = \frac{1}{3} \Rightarrow \boxed{D}.\)

page33

Đáp án:

\(\int_{0}^{1} \left[\frac{f'(x)}{e^{\frac{x}{2}}}\right]^2 dx = \frac{1}{e - 1}.\)

Đặt \(h(x) = \frac{f'(x)}{e^{\frac{x}{2}}} \Rightarrow f'(x) = e^{\frac{x}{2}} h(x)\).

Tìm \(k\) để: \(\int_{0}^{1} \left[h(x) - k e^{\frac{x}{2}}\right]^2 dx = 0.\)

\(\Leftrightarrow \int_{0}^{1} [h(x)]^2 dx - 2k \int_{0}^{1} f'(x)  dx + k^2 \int_{0}^{1} e^x dx = 0.\)

\(\Leftrightarrow \frac{1}{e - 1} - 2k f(x) \big|_{0}^{1} + k^2 e^x \big|_{0}^{1} = 0.\)

\(\Leftrightarrow (e - 1)k^2 - 2k + \frac{1}{e - 1} = 0 \quad \Leftrightarrow \quad k = \frac{1}{e - 1}\)
\(h(x) = \frac{e^{\frac{x}{2}}}{e - 1} \Rightarrow \frac{f'(x)}{e^{\frac{x}{2}}} = \frac{e^{\frac{x}{2}}}{e - 1}\)

\(f'(x) = \frac{e^{x}}{e - 1} \quad \Rightarrow \quad f(x) =  \frac{e^{x}}{e - 1} + C.\)

\(f(0) = 0 \quad \Rightarrow \quad C = -\frac{1}{e - 1}.\)

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

\(\int_{0}^{1} f(x) dx = \int_{0}^{1} \frac{e^{x} - 1}{e - 1} dx = \frac{e - 2}{e - 1} \Rightarrow \boxed{A}.\)

page34

Đáp án:

\(6 = \int_{0}^{2} \left(\frac{f'(x)}{\sqrt{f(x)}}\right)^2 dx.\)

Tìm \(k\) sao cho: \(\int_{0}^{2} \left[\frac{f'(x)}{\sqrt{f(x)}} - k\right]^2 dx = 0.\)

\(\Leftrightarrow \int_{0}^{2} \frac{[f'(x)]^2}{f(x)} dx - 2k \int_{0}^{2} \frac{f'(x)}{\sqrt{f(x)}} dx + k^2 \int_{0}^{2} dx = 0.\)

\(\Rightarrow 18 - 4k \sqrt{f(x)}\big|_{0}^{2} + k^2 x\big|_{0}^{2} = 0.\)

\(\Rightarrow 2k^2 - 12k + 18 = 0 \quad \Leftrightarrow k^2 -6k +9 =0  \quad \Leftrightarrow k = 3.\)

Suy ra: \(\frac{f'(x)}{\sqrt{f(x)}} = 3 \quad \Rightarrow \quad 2 \sqrt{f(x)} = 3x + C.\)

\(f(0) = 0 \Rightarrow C = 0.\)

\(\Rightarrow \sqrt{f(x)} = \frac{3x}{2} \quad \Rightarrow \quad \sqrt{f(1)}= \frac{3}{2} \Rightarrow f(1) = \frac{9}{4} \Rightarrow \boxed{A}\)

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