Tích phân

Ta có

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

$ \Rightarrow I = \int\limits_0^1 {\left[ {{{(x + 1)}^{ - 2}} - {{(x + 1)}^{ - 3}}} \right]} dx$$ = \left. {\left[ { - {{\left( {x + 1} \right)}^{ - 1}} + \dfrac{1}{2}{{\left( {x + 1} \right)}^{ - 2}}} \right]} \right|_0^1 = \dfrac{1}{8}$

Ta có:

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

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

Suy ra \(I.J = 8\).

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

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 $

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

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

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

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

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

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:

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

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

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

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

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:

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

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

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

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 $

$\begin{array}{l}f(x) = A\sin\pi x + B \Rightarrow f'(x) = A\pi \cos \pi x\\f'(1) = 2 \Rightarrow A\pi \cos \pi = 2 \Rightarrow A = - \dfrac{2}{\pi }\end{array}$
\(\int\limits_0^2 {f(x)dx = 4} \Rightarrow \int\limits_0^2 {(A\sin\pi x + B)dx = 4} \) \(\Rightarrow - \dfrac{A}{\pi }\cos 2\pi + 2B + \dfrac{A}{\pi }\cos 0 = 4 \Rightarrow B = 2\)

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

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

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

Do hàm số $f(x) = \sqrt {1 - \cos 2x}  = \sqrt 2 \left| {\sin x} \right|$là hàm liên tục và tuần hoàn với chu kì $T = \pi $ nên ta có

\(\begin{array}{l}\int\limits_0^T {f\left( x \right)dx = \int\limits_T^{2T} {f\left( x \right)dx} }  = \int\limits_{2T}^{3T} {f\left( x \right)dx} \\ = ... = \int\limits_{\left( {n - 1} \right)T}^{nT} {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^{nT} {f\left( x \right)dx}  = \int\limits_0^T {f\left( x \right)dx}  \\+ \int\limits_T^{2T} {f\left( x \right)dx}+ \int\limits_{2T}^{3T} {f\left( x \right)dx}   + ... + \\ \int\limits_{\left( {n - 1} \right)T}^{nT} {f\left( x \right)dx}  = n\int\limits_{0}^{T} {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^{2017\pi } {\sqrt {1 - \cos 2x} dx}  \\= 2017\int\limits_0^\pi  {\sqrt {1 - \cos 2x} dx} \\ = 2017\sqrt 2 \int\limits_0^\pi  {\sin x dx = 4034\sqrt 2 } \end{array}\)

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 \(a = \dfrac{1}{2};b = \dfrac{2}{3} \Rightarrow 2a + 3b = 3.\)

Ta có \(g\left( x \right) = 1 + 2\int\limits_0^x {f\left( t \right)dt} \)  suy ra \(\left\{ \begin{array}{l}g\left( x \right) - 1 = 2\int\limits_0^x {f\left( t \right)dt} \\g\left( 0 \right) = 1 + \int\limits_0^0 {f\left( t \right)dt} \end{array} \right. \)\(\Leftrightarrow \left\{ \begin{array}{l}g'\left( x \right) = 2f\left( x \right) \Rightarrow f\left( x \right) = \dfrac{{g'\left( x \right)}}{2}\\g\left( 0 \right) = 1\end{array} \right.\)

Mà

\(g\left( x \right) \ge {\left[ {f\left( x \right)} \right]^3} \Leftrightarrow g\left( x \right) \ge {\left[ {\dfrac{{g'\left( x \right)}}{2}} \right]^3} \)\(\Leftrightarrow \sqrt[3]{{g\left( x \right)}} \ge \dfrac{{g'\left( x \right)}}{2} \Leftrightarrow \dfrac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}} \le 2\)

Với \(t \in \left[ {0;1} \right]\), Lấy tích phân hai vế ta được

\(\begin{array}{l}\int\limits_0^t {\dfrac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}}} dx \le \int\limits_0^t {2dx} \\ \Leftrightarrow \int\limits_0^t {{{\left[ {g\left( x \right)} \right]}^{\dfrac{{ - 1}}{3}}}} d\left( {g\left( x \right)} \right) \le 2t\\ \Leftrightarrow 2t \ge \dfrac{3}{2}\left. {{{\left[ {g\left( x \right)} \right]}^{\dfrac{2}{3}}}} \right|_0^t \\\Leftrightarrow \dfrac{4}{3}t \ge \sqrt[3]{{{g^2}\left( t \right)}} - \sqrt[3]{{{g^2}\left( 0 \right)}}\end{array}\)

Mà \(g\left( 0 \right) = 1\) nên \(\sqrt[3]{{{g^2}\left( t \right)}} \le \dfrac{4}{3}t + 1 \Rightarrow \sqrt[3]{{{g^2}\left( x \right)}} \le \dfrac{4}{3}x + 1\)

Từ đó ta có \(\int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \int\limits_0^1 {\left( {\dfrac{4}{3}x + 1} \right)dx} \)\( \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \left. {\left( {\dfrac{2}{3}{x^2} + x} \right)} \right|_0^1\\ \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \dfrac{5}{3}\)

Hay giá trị lớn nhất cần tìm là \(\dfrac{5}{3}.\)

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

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