Tích phân

$I = \int\limits_0^2 {{x^3}dx}  = \left. {\dfrac{{{x^4}}}{4}} \right|_0^2 = 4$ và $J = \int\limits_0^2 {xdx}  = \left. {\dfrac{{{x^2}}}{2}} \right|_0^2 = 2$.

Suy ra $I.J = 8$.

Ta có:

$\begin{array}{l}\int\limits_a^d {f\left( x \right)dx}  = \int\limits_a^c {f\left( x \right)dx}  + \int\limits_c^b {f\left( x \right)dx}  + \int\limits_b^d {f\left( x \right)dx} \\ \Rightarrow \int\limits_c^b {f\left( x \right)dx}  = \int\limits_a^d {f\left( x \right)dx}  - \int\limits_a^c {f\left( x \right)dx}  - \int\limits_b^d {f\left( x \right)dx} \\ \Rightarrow \int\limits_c^b {f\left( x \right)dx}  = 10 - 7 - 18 =  - 15 \Rightarrow \int\limits_b^c {f\left( x \right)dx}  = 15\end{array}$

Ta có: $\int\limits_1^4 {\left[ {f\left( x \right) + g\left( x \right)} \right]dx}  = \int\limits_1^4 {f\left( x \right)dx}  + \int\limits_1^4 {f\left( x \right)dx}  = 10$ nên A đúng.

$\int\limits_1^4 {f\left( x \right)dx}  = \int\limits_1^3 {f\left( x \right)dx}  + \int\limits_3^4 {f\left( x \right)dx}  \Rightarrow \int\limits_3^4 {f\left( x \right)dx}  = \int\limits_1^4 {f\left( x \right)dx}  - \int\limits_1^3 {f\left( x \right)dx}  = 3 - \left( { - 2} \right) = 5$ nên C đúng, B sai.

$\int\limits_1^4 {\left[ {4f\left( x \right) - 2g\left( x \right)} \right]dx}  = 4\int\limits_1^4 {f\left( x \right)dx}  - 2\int\limits_1^4 {g\left( x \right)dx}  =  - 2$ nên D đúng.

$\dfrac{{4{{\sin }^3}x}}{{1 + \cos x}} = \dfrac{{4{{\sin }^3}x(1 - \cos x)}}{{{{\sin }^2}x}} = 4\sin x - 4\sin x\cos x = 4\sin x - 2\sin 2x$

$\Rightarrow I = \int_0^{\dfrac{\pi }{2}} {(4\sin x - 2\sin 2x)} dx = 2.$

Ta có: $\int\limits_0^2 {f\left( x \right)dx}  = 4 \Leftrightarrow \int\limits_0^2 {\left( {A\sin \pi x + B{x^2}} \right)dx}  = 4$

$\Leftrightarrow \left. {\left( { - \dfrac{A}{\pi }\cos \pi x + \dfrac{B}{3}{x^3}} \right)} \right|_0^2 = 4 \Leftrightarrow \dfrac{B}{3}{.2^3} = 4 \Leftrightarrow B = \dfrac{3}{2}$

Phương pháp tự luận

$\begin{array}{c}I = \int\limits_0^{2\pi } {\sqrt {{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}^2}} } dx = \int\limits_0^{2\pi } {\left| {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right|} dx = \sqrt 2 \int\limits_0^{2\pi } {\left| {\sin \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)} \right|} dx\\ = \sqrt 2 \left[ {\int\limits_0^{\dfrac{{3\pi }}{2}} {\sin \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)} dx - \int\limits_{\dfrac{{3\pi }}{2}}^{2\pi } {\sin \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)dx} } \right] = 4\sqrt 2 \end{array}$

Ta có: $\int\limits_0^1 {\left[ {{f^2}\left( x \right) - f\left( x \right)} \right]dx = 5}$

$\begin{array}{l}\int\limits_0^1 {{{\left[ {f\left( x \right) + 1} \right]}^2}dx = 36}  \Leftrightarrow \int\limits_0^1 {\left[ {{f^2}\left( x \right) + 2f\left( x \right) + 1} \right]} dx = 36\\ \Rightarrow \int\limits_0^1 {\left[ {{f^2}\left( x \right) + 2f\left( x \right) + 1} \right]} dx - \int\limits_0^1 {\left[ {{f^2}\left( x \right) - f\left( x \right)} \right]dx}  = 36 - 5\\ \Leftrightarrow \int\limits_0^1 {\left[ {3f\left( x \right) + 1} \right]dx}  = 31 \Leftrightarrow 3\int\limits_0^1 {f\left( x \right)dx}  + \int\limits_0^1 {dx}  = 31\\ \Leftrightarrow 3\int\limits_0^1 {f\left( x \right)dx}  + \left. x \right|_0^1 = 31 \Leftrightarrow 3\int\limits_0^1 {f\left( x \right)dx}  + 1 = 31\\ \Leftrightarrow 3\int\limits_0^1 {f\left( x \right)dx}  = 30 \Leftrightarrow \int\limits_0^1 {f\left( x \right)dx}  = 10.\end{array}$

$\begin{array}{c}\int\limits_{ - 1}^5 {\left| {{x^2} - 2x - 3} \right|dx}  = \int\limits_{ - 1}^5 {\left| {(x - 3)(x + 1)} \right|dx}  =  - \int\limits_{ - 1}^3 {\left( {{x^2} - 2x - 3} \right)dx}  + \int\limits_3^5 {\left( {{x^2} - 2x - 3} \right)dx} \\ =  - \left. {\left( {\dfrac{{{x^3}}}{3} - {x^2} - 3x} \right)} \right|_{ - 1}^3 + \left. {\left( {\dfrac{{{x^3}}}{3} - {x^2} - 3x} \right)} \right|_3^5 = \dfrac{{64}}{3}.\end{array}$

Ta có:

$\begin{array}{l}\int\limits_1^b {\left( {2x - 6} \right)dx}  = 0 \Leftrightarrow \int\limits_1^b {2xdx}  - \int\limits_1^b {6dx}  = 0 \Leftrightarrow \left. {{x^2}} \right|_1^b - \left. {6x} \right|_1^b = 0\\ \Leftrightarrow {b^2} - 1 - 6b + 6 = 0 \Leftrightarrow {b^2} - 6b + 5 = 0 \Leftrightarrow \left[ \begin{array}{l}b = 1\\b = 5\end{array} \right.\end{array}$

$\int\limits_2^3 {\dfrac{{{x^2} - x + 4}}{{x + 1}}dx}  = \int\limits_2^3 {\left( {x - 2 + \dfrac{6}{{x + 1}}} \right)dx}  = \left. {\left( {\dfrac{{{x^2}}}{2} - 2x + 6\ln \left| {x + 1} \right|} \right)} \right|_2^3 = \dfrac{1}{2} + 6\ln \dfrac{4}{3}$.

Ta có:

$\int\limits_0^a {\left( {\cos x + \sin x} \right)dx} {\rm{\;}} = 0$ $\Leftrightarrow \left. {\sin x} \right|_0^a - \left. {\cos x} \right|_0^a = 0$ $\Leftrightarrow \sin a - \cos a + 1 = 0$

$\begin{array}{l} \Leftrightarrow \sin a - \cos a =  - 1\\ \Leftrightarrow \dfrac{1}{{\sqrt 2 }}.\sin a - \dfrac{1}{{\sqrt 2 }}.\cos a =  - \dfrac{1}{{\sqrt 2 }}\\ \Leftrightarrow \sin a.\cos \dfrac{\pi }{4} - \cos a.\sin \dfrac{\pi }{4} =  - \dfrac{1}{{\sqrt 2 }}\end{array}$

$\Leftrightarrow \sin \left( {a - \dfrac{\pi }{4}} \right) = {\rm{\;}} - \dfrac{1}{{\sqrt 2 }}$

$\Leftrightarrow \sin \left( {a - \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{{ - \pi }}{4}} \right)$

$\Leftrightarrow \left[ {\begin{array}{*{20}{l}}{a - \dfrac{\pi }{4} = {\rm{\;}} - \dfrac{\pi }{4} + k2\pi }\\{a - \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi }\end{array}} \right.$

$\Leftrightarrow a = \dfrac{{3\pi }}{2}\left( {0 < a < 2\pi } \right)$

Ta có:

$\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$

$= \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right]$$\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]$

$= \left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right)\left( {1 - \dfrac{3}{4}{{\sin }^2}2x} \right)$$= 1 - \dfrac{5}{4}{\sin ^2}2x + \dfrac{3}{8}{\left( {{{\sin }^2}2x} \right)^2}$$= 1 - \dfrac{5}{4}.\dfrac{{1 - \cos 4x}}{2} + \dfrac{3}{8}.{\left( {\dfrac{{1 - \cos 4x}}{2}} \right)^2}$$= \dfrac{3}{8} + \dfrac{5}{8}\cos 4x + \dfrac{3}{{32}}\left( {1 - 2\cos 4x + {{\cos }^2}4x} \right)$$= \dfrac{{15}}{{32}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{32}}{\cos ^2}4x$$= \dfrac{{15}}{{32}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{32}}.\dfrac{{1 + \cos 8x}}{2}$$= \dfrac{{33}}{{64}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{64}}\cos 8x$

Do đó $I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{33}}{{64}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{64}}\cos 8x} \right)dx}$$= \left. {\dfrac{{33}}{{64}}x} \right|_0^{\dfrac{\pi }{2}} + \left. {\dfrac{7}{{64}}\sin 4x} \right|_0^{\dfrac{\pi }{2}} + \left. {\dfrac{3}{{512}}\sin 8x} \right|_0^{\dfrac{\pi }{2}}$$= \dfrac{{33}}{{128}}\pi$

Ta có: $\int\limits_1^2 {\dfrac{{dx}}{{x + 3}}}  = \left. {\ln \left| {x + 3} \right|} \right|_1^2 = \ln 5 - \ln 4 = \ln \dfrac{5}{4}$

Do đó $a = 5,b = 4$.

Khi đó: $3a - b = 3.5 - 4 = 11 < 12$ nên A đúng.

$a + 2b = 5 + 2.4 = 13$ nên B đúng.

$a - b = 5 - 4 = 1 < 2$ nên C sai.

${a^2} + {b^2} = {5^2} + {4^2} = 41$ nên D đúng.

Ta có: $\int\limits_{ - 1}^0 {\left( {x + 1 + \dfrac{2}{{x - 1}}} \right)dx}  = \left. {\left( {\dfrac{{{x^2}}}{2} + x + 2\ln \left| {x - 1} \right|} \right)} \right|_{ - 1}^0$

$= \dfrac{1}{2} - 2\ln 2 \Rightarrow \left\{ \begin{array}{l}a = \dfrac{1}{2}\\b =  - 2\end{array} \right. \Rightarrow a + b =  - \dfrac{3}{2}$

Ta có: $\int\limits_2^4 {2xdx}  = \left. {{x^2}} \right|_2^4 = 12$

$\int\limits_1^2 {\left[ {{a^2} + (4 - 4a)x + 4{x^3}} \right]dx = }$$= \left. {\left[ {{a^2}x + (2 - 2a){x^2} + {x^4}} \right]} \right|_1^2 = {a^2} - 6a + 21$

$\Rightarrow {a^2} - 6a + 21 = 12 \Leftrightarrow a = 3.$

Ta có

$I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\cos 2x}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx}$

Đặt $\sin x - \cos x + 3 = t \Leftrightarrow \left\{ \begin{array}{l}\left( {\cos x + \sin x} \right)dx = dt\\\cos x - \sin x = 3 - t\end{array} \right.$

Đổi cận $x = 0 \Rightarrow t = 2;\,x = \dfrac{\pi }{4} \Rightarrow t = 3$

Suy ra $I = \int\limits_2^3 {\dfrac{{\left( {3 - t} \right)dt}}{{{t^2}}} = \int\limits_2^3 {\left( {\dfrac{3}{{{t^2}}} - \dfrac{1}{t}} \right)dt}  = \left. {\left( { - \dfrac{3}{t} - \ln \left| t \right|} \right)} \right|_2^3}  = \dfrac{1}{2} + \ln 2 - \ln 3 = \dfrac{1}{2} + \ln \dfrac{2}{3}$

Hay $a = \dfrac{1}{2};b = \dfrac{2}{3} \Rightarrow 2a + 3b = 3.$

Theo đề bài ta có

 $\begin{array}{l}\left\{ \begin{array}{l}f\left( 1 \right) = 10\\f\left( 2 \right) = 20\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 + a + b + c = 10\\{2^3} + {2^2}.a + 2b + c = 20\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a + b + c = 9\\4a + 2b + c = 12\end{array} \right. \Rightarrow 3a + b = 3\end{array}$

$\begin{array}{l} \Rightarrow \int\limits_0^3 {f'\left( x \right)dx}  = \left. {f\left( x \right)} \right|_0^3 = f\left( 3 \right) - f\left( 0 \right)\\ = {3^3} + {3^2}.a + 3b + c - c = 27 + 9a + 3b\\ = 27 + 3\left( {3a + b} \right) = 27 + 3.3 = 36.\end{array}$

Ta có:  $2f\left( x \right) + xf\left( {\dfrac{1}{x}} \right) = x$, với $x = \dfrac{1}{t}$ ta có $2f\left( {\dfrac{1}{t}} \right) + \dfrac{1}{t}f\left( t \right) = \dfrac{1}{t}$ $\Rightarrow f\left( {\dfrac{1}{t}} \right) = \dfrac{1}{2}\left( {\dfrac{1}{t} - \dfrac{1}{t}f\left( t \right)} \right)$

$\Rightarrow f\left( {\dfrac{1}{x}} \right) = \dfrac{1}{2}\left( {\dfrac{1}{x} - \dfrac{1}{x}f\left( x \right)} \right)$ 

Khi đó ta có

$\begin{array}{l}2f\left( x \right) + \dfrac{1}{2}x\left( {\dfrac{1}{x} - \dfrac{1}{x}f\left( x \right)} \right) = x\\ \Leftrightarrow 2f\left( x \right) + \dfrac{1}{2} - \dfrac{1}{2}f\left( x \right) = x\\ \Leftrightarrow \dfrac{3}{2}f\left( x \right) = x - \dfrac{1}{2}\\ \Leftrightarrow \dfrac{3}{2}\int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \int\limits_{\frac{1}{2}}^2 {\left( {x - \frac{1}{2}} \right)dx} \\ \Leftrightarrow \dfrac{3}{2}\int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \dfrac{9}{8} \Leftrightarrow \int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \dfrac{3}{4}\end{array}$

Ta có $f\left( x \right) = \int {f'\left( x \right)dx}  = \int {{{\sin }^4}xdx}$

$\begin{array}{l} = \int {{{\left( {\dfrac{{1 - \cos 2x}}{2}} \right)}^2}dx} \\ = \dfrac{1}{4}\int {\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)dx} \\ = \dfrac{1}{4}\int {\left( {1 - 2\cos 2x + \dfrac{{1 + \cos 4x}}{2}} \right)dx} \\ = \dfrac{1}{4}\left( {x - \sin 2x + \dfrac{1}{2}x + \dfrac{1}{2}.\dfrac{{\sin 4x}}{4}} \right) + C\\ = \dfrac{{3x}}{8} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}} + C\end{array}$ 

Theo bài ra ta có $f\left( 0 \right) = 0 \Leftrightarrow C = 0$ $\Rightarrow f\left( x \right) = \dfrac{{3x}}{8} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}}$.

Vậy $\int\limits_0^{\frac{\pi }{2}} {f\left( x \right)dx}  = \int\limits_0^{\frac{\pi }{2}} {\left( {\dfrac{{3x}}{8} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}}} \right)dx}  = \dfrac{{3{\pi ^2} - 16}}{{64}}$ (sử dụng MTCT).

Ta có $v\left( t \right) = \int {a\left( t \right)dt}  = \int {\left( {6 - 3t} \right)dt}  = 6t - \dfrac{{3{t^2}}}{2} + C$.

Theo bài ra ta có: Ô tô đang đứng yên và bắt đầu chuyển động, do đó $v\left( 0 \right) = 0$ $\Rightarrow C = 0$.

Khi đó ta có $v\left( t \right) = 6t - \dfrac{3}{2}{t^2}$, đây là một parabol có bề lõm hướng xuống, đạt giá trị lớn nhất tại $t = \dfrac{{ - b}}{{2a}} = \dfrac{{ - 6}}{{2.\left( { - \dfrac{3}{2}} \right)}} = 2$.

Vậy quãng đường ô tô đi được từ khi chuyển động đến khi vận tốc của ô tô đạt giá trị lớn nhất là:

$S = \int\limits_0^2 {v\left( t \right)dt}  = \int\limits_0^2 {\left( {6t - \dfrac{3}{2}{t^2}} \right)dt}  = 8\,\,\left( m \right).$